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Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling 
Exercise 3.1. The gradient operator in Cartesian coordinates (x, y, z) is: 
€ 
∇ = ex ∂ ∂x( ) + ey ∂ ∂y( ) + ez ∂ ∂z( ) where 
€ 
ex , 
€ 
ey , and 
€ 
ez are the unit vectors. In cylindrical polar 
coordinates (R, ϕ, z) having the same origin, (see Figure 3.3b), coordinates and unit vectors are 
related by: 
€ 
R = x 2 + y 2 , 
€ 
ϕ = tan−1 y x( ) , and z = z; and 
€ 
eR = ex cosϕ + ey sinϕ , 
€ 
eϕ = −ex sinϕ + ey cosϕ , and 
€ 
ez = ez . Determine the following in the cylindrical polar coordinate 
system. 
a) 
€ 
∂eR ∂ϕ and 
€ 
∂eϕ ∂ϕ 
b) the gradient operator ∇ 
c) the divergence of the velocity field ∇⋅u 
d) the Laplacian operator
€ 
∇ ⋅∇ ≡ ∇2 
e) the advective acceleration term (u⋅∇)u 
 
Solution 3.1. The Cartesian unit vectors do not depend on the coordinates so the unit vectors 
from the cylindrical coordinate system can be differentiated when they are written in terms of ex, 
ey, and ez. 
a) First work with eR, use the given unit vector definition, and proceed with straightforward 
differentiation. The variables R and z do not appear in the formula for eR, so 
€ 
∂eR ∂R = ∂eR ∂z = 0 . However eR does depend on the angle ϕ. Thus, 
€ 
∂eR
∂ϕ
=
∂
∂ϕ
ex cosϕ + ey sinϕ( ) = −ex sinϕ + ey cosϕ = eϕ . 
Proceed to determine the derivatives of eϕ. Again note that the variables R and z do not appear in 
the formula for eϕ, so 
€ 
∂eϕ ∂R = ∂eϕ ∂z = 0. However, like eR, eϕ does depend on the angle ϕ. 
Thus, 
€ 
∂eϕ
∂ϕ
=
∂
∂ϕ
−ex sinϕ + ey cosϕ( ) = −ex cosϕ − ey sinϕ = −eR . 
The third unit vector, ez, is the same as the Cartesian unit vector and does not depend on the 
coordinates. 
b) Start by constructing the expressions for ex, ey, and ez in terms of eR, eϕ, and ez. This can be 
done my inverting the linear system 
€ 
cosϕ sinϕ 0
−sinϕ cosϕ 0
0 0 1
$ 
% 
& 
' 
& 
( 
) 
& 
* 
& 
ex
ey
ez
$ 
% 
& 
' 
& 
( 
) 
& 
* 
& 
=
eR
eϕ
ez
$ 
% 
& 
' 
& 
( 
) 
& 
* 
& 
 to find 
 
€ 
ex = eR cosϕ − eϕ sinϕ , 
€ 
ey = eR sinϕ + eϕ cosϕ , and 
€ 
ez = ez (1,2,3) 
The next step is to use the coordinate definitions: 
€ 
R = x 2 + y 2 , 
€ 
ϕ = tan−1 y x( ) , and z = z (4,5,6) 
to transform the Cartesian partial derivatives. 
 
€ 
∂
∂x
=
∂R
∂x
∂
∂R
+
∂ϕ
∂x
∂
∂ϕ
+
∂z
∂x
∂
∂z
=
x
x 2 + y 2
∂
∂R
−
1
1+ y x( )2
−
y
x 2
% 
& 
' 
( 
) 
* 
∂
∂ϕ
+ (0) ∂
∂z
 
 
€ 
= cosϕ ∂
∂R
−
sinϕ
R
∂
∂ϕ
 
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling 
 
€ 
∂
∂y
=
∂R
∂y
∂
∂R
+
∂ϕ
∂y
∂
∂ϕ
+
∂z
∂y
∂
∂z
=
y
x 2 + y 2
∂
∂R
+
1
1+ y x( )2
1
x
$ 
% 
& 
' 
( 
) 
∂
∂ϕ
+ (0) ∂
∂z
 
 
 
€ 
= sinϕ ∂
∂R
+
cosϕ
R
∂
∂ϕ
 
 
€ 
∂
∂z
=
∂R
∂z
∂
∂R
+
∂ϕ
∂z
∂
∂ϕ
+
∂z
∂z
∂
∂z
= (0) ∂
∂R
+ (0) ∂
∂ϕ
+ (1) ∂
∂z
=
∂
∂z
 
Now reassemble the gradient operator 
€ 
∇ = ex
∂
∂x
+ ey
∂
∂y
+ ez
∂
∂z
using the cylindrical coordinate 
unit vectors and differentiation definitions: 
€ 
∇ = eR cosϕ − eϕ sinϕ( ) cosϕ
∂
∂R
−
sinϕ
R
∂
∂ϕ
& 
' 
( 
) 
* 
+ + eR sinϕ + eϕ cosϕ( ) sinϕ
∂
∂R
+
cosϕ
R
∂
∂ϕ
& 
' 
( 
) 
* 
+ + ez
∂
∂z
. 
Collect all of the terms with like unit vectors and differential operators together: 
€ 
∇ = eR cos
2ϕ + sin2ϕ( ) ∂
∂R
+ eR −cosϕ sinϕ + sinϕ cosϕ( )
1
R
∂
∂ϕ
 
 
€ 
eϕ −sinϕ cosϕ + cosϕ sinϕ( )
∂
∂R
+ eϕ sin
2ϕ + cos2ϕ( ) 1R
∂
∂ϕ
+ ez
∂
∂z
 
The terms in (,)-parentheses are either +1 or 0. When evaluated they produce: 
€ 
∇ = eR
∂
∂R
+ eϕ
1
R
∂
∂ϕ
+ ez
∂
∂z
. 
c) In cylindrical coordinates, the divergence of the velocity is: 
€ 
∇ ⋅u = eR
∂
∂R
+ eϕ
1
R
∂
∂ϕ
+ ez
∂
∂z
& 
' 
( 
) 
* 
+ ⋅ eRuR + eϕuϕ + ezuz( ). 
Further simplification requires that both the unit vectors and the u's be differentiated. 
Completing this task term by term produces: 
 
€ 
eR
∂
∂R
⋅ eRuR( ) = eR ⋅ eR
∂uR
∂r
+ uReR ⋅
∂eR
∂R
=
∂uR
∂R
, 
 
€ 
eR
∂
∂R
⋅ eϕuϕ( ) = eR ⋅ eϕ
∂uϕ
∂R
+ uϕeR ⋅
∂eϕ
∂R
= 0 , 
 
€ 
eR
∂
∂R
⋅ ezuz( ) = eR ⋅ ez
∂uz
∂R
+ uzeR ⋅
∂ez
∂r
= 0 , 
 
€ 
eϕ
R
∂
∂ϕ
⋅ eRuR( ) =
eϕ ⋅ eR
R
∂uR
∂ϕ
+
uR
R
eϕ ⋅
∂eR
∂ϕ
= 0 + uR
R
eϕ ⋅ eϕ =
uR
R
, 
 
€ 
eϕ
R
∂
∂ϕ
⋅ eϕuϕ( ) =
eϕ ⋅ eϕ
R
∂uϕ
∂ϕ
+
uϕ
R
eϕ ⋅
∂eϕ
∂ϕ
=
1
R
∂uϕ
∂ϕ
−
uϕ
R
eϕ ⋅ eR =
1
R
∂uϕ
∂ϕ
, 
 
€ 
eϕ
R
∂
∂ϕ
⋅ ezuz( ) =
eϕ ⋅ ez
R
∂uϕ
∂ϕ
+
uz
R
eϕ ⋅
∂ez
∂ϕ
= 0 
 
€ 
ez
∂
∂z
⋅ eRuR( ) = ez ⋅ eR
∂uR
∂z
+ uRez ⋅
∂eR
∂z
= 0, 
 
€ 
ez
∂
∂z
⋅ eϕuϕ( ) = ez ⋅ eϕ
∂uϕ
∂z
+ uϕez ⋅
∂eϕ
∂z
= 0 , and 
 
 
€ 
ez
∂
∂z
⋅ ezuz( ) = ez ⋅ ez
∂uz
∂z
+ uzez ⋅
∂ez
∂z
=
∂uz
∂z
.
 
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling 
Reassembling the equation produces: 
€ 
∇ ⋅u = ∂uR
∂R
+
uR
R
+
1
R
∂uϕ
∂ϕ
+
∂uz
∂z
. 
The 1st & 2nd terms on the right side are commonly combined to yield: 
€ 
∇ ⋅u = 1
R
∂
∂R
RuR( ) +
1
R
∂uϕ
∂ϕ
+
∂uz
∂z
. (10) 
d) The Laplacian operator is 
€ 
∇2 ≡ ∇ ⋅∇, and its form in cylindrical coordinates can be found by 
evaluating the dot product. Fortunately, the results of part c) can be used via the following 
replacements for the second gradient operator of the dot product: 
€ 
uR ↔
∂
∂R
 , 
€ 
uϕ ↔
1
R
∂
∂ϕ
 , and 
€ 
uz ↔
∂
∂z
. (7,8,9) 
Inserting the replacements (7,8,9) into (10) produces: 
€ 
∇2 =
1
R
∂
∂R
R ∂
∂R
$ 
% 
& 
' 
( 
) +
1
R2
∂ 2
∂ϕ 2
+
∂ 2
∂z2
. 
e) Start with the answer to part b) and compute the first dot product to find: 
€ 
u ⋅ ∇ = uR
∂
∂R
+ uϕ
1
R
∂
∂ϕ
+ uz
∂
∂z
 
This is the scalar operator applied to u = uReR +uϕeϕ +uzez to find the advective acceleration: 
€ 
u ⋅ ∇( )u = uR
∂
∂R
+ uϕ
1
R
∂
∂ϕ
+ uz
∂
∂z
& 
' 
( 
) 
* 
+ uReR + uϕeϕ + uzez( ) . 
Here the components of u and the unit vectors eR and eϕ depend on the angular coordinate. 
€ 
u ⋅ ∇( )u = eR uR
∂uR
∂R
+ uϕ
1
R
∂uR
∂ϕ
+ uz
∂uR
∂z
& 
' 
( 
) 
* 
+ + uRuϕ
1
R
∂eR
∂ϕ
+ 
 eϕ uR
∂uϕ
∂R
+uϕ
1
R
∂uϕ
∂ϕ
+uz
∂uϕ
∂z
!
"
#
$
%
&+uϕ
2 1
R
∂eϕ
∂ϕ
+ ez uR
∂uz
∂R
+uϕ
1
R
∂uz
∂ϕ
+uz
∂uz
∂z
!
"
#
$
%
& 
Use the results of part a) to evaluate the unit vector derivatives. 
€ 
u ⋅ ∇( )u = eR uR
∂uR
∂R
+ uϕ
1
R
∂uR
∂ϕ
+ uz
∂uR
∂z
& 
' 
( 
) 
* 
+ + uRuϕ
eϕ
R
+ 
 
€ 
eϕ uR
∂uϕ
∂R
+ uϕ
1
R
∂uϕ
∂ϕ
+ uz
∂uϕ
∂z
$ 
% 
& 
' 
( 
) − uϕ
2 eR
R
+ ez uR
∂uz
∂R
+ uϕ
1
R
∂uz
∂ϕ
+ uz
∂uz
∂z
$ 
% 
& 
' 
( 
) 
Collect components 
€ 
u ⋅ ∇( )u = eR uR
∂uR
∂R
+ uϕ
1
R
∂uR
∂ϕ
+ uz
∂uR
∂z
−
uϕ
2
R
' 
( 
) 
* 
+ 
, + 
 
€ 
eϕ uR
∂uϕ
∂R
+ uϕ
1
R
∂uϕ
∂ϕ
+ uz
∂uϕ
∂z
+
uRuϕ
R
$ 
% 
& 
' 
( 
) + ez uR
∂uz
∂R
+ uϕ
1
R
∂uz
∂ϕ
+ uz
∂uz
∂z
$ 
% 
& 
' 
( 
) 
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling 
Exercise 3.2. Consider Cartesian coordinates (as given in Exercise no. 1) and spherical polar 
coordinates (r, θ, ϕ) having the same origin (see Figure 3.3c). Here coordinates and unit vectors 
are related by: 
€ 
r = x 2 + y 2 + z2 , 
€ 
θ = tan−1 x 2 + y 2 z( ) , and 
€ 
ϕ = tan−1 y x( ) ; and 
€ 
er = ex cosϕ sinθ + ey sinϕ sinθ + ez cosθ , 
€ 
eθ = ex cosϕ cosθ + ey sinϕ cosθ − ez sinθ , and 
€ 
eϕ = −ex sinϕ + ey cosϕ . In the spherical polar coordinate system, determine the following items. 
a) 
€ 
∂er ∂θ , 
€ 
∂er ∂ϕ , 
€ 
∂eθ ∂θ , 
€ 
∂eθ ∂ϕ , and 
€ 
∂eϕ ∂ϕ 
b) the gradient operator ∇ 
c) the divergence of the velocity field ∇⋅u 
d) the Laplacian 
€ 
∇ ⋅∇ ≡ ∇2 
e) the advective acceleration term (u⋅∇)u 
 
Solution 3.2. The Cartesian unit vectors do not depend on the coordinatesso the unit vectors 
from the spherical coordinate system can be differentiated when they are written in terms of ex, 
ey, and ez. 
a) First work with er, use the given unit vector definition, and proceed with straightforward 
differentiation. The variable r doesn’t even appear in the formula for er, so 
€ 
∂er ∂r = 0. However 
er does depend on both angles. Thus, 
€ 
∂er
∂θ
=
∂
∂θ
ex cosϕ sinθ + ey sinϕ sinθ + ez cosθ( ) = ex cosϕ cosθ + ey sinϕ cosθ − ez sinθ = eθ 
and, 
€ 
∂er
∂ϕ
=
∂
∂ϕ
ex cosϕ sinθ + ey sinϕ sinθ + ez cosθ( ) = −ex sinϕ sinθ + ey cosϕ sinθ = eϕ sinθ . 
Proceed to determine the derivatives of eθ. Again note that the variable r doesn’t appear in its 
formula, so 
€ 
∂eθ ∂r = 0 . However, like er, eθ does depend on both angles. Thus, 
€ 
∂eθ
∂θ
=
∂
∂θ
ex cosϕ cosθ + ey sinϕ cosθ − ez sinθ( ) = −ex cosϕ sinθ − ey sinϕ sinθ − ez cosθ = −er 
and, 
€ 
∂eθ
∂ϕ
=
∂
∂ϕ
ex cosϕ cosθ + ey sinϕ cosθ − ez sinθ( ) = −ex sinϕ cosθ + ey cosϕ cosθ = eϕ cosθ . 
Now consider eϕ and note that the variables r and θ don’t appear in its formula, so 
€ 
∂eϕ ∂r = ∂eϕ ∂θ = 0 . However, eϕ does depend on ϕ. Thus, 
€ 
∂eϕ
∂ϕ
=
∂
∂ϕ
−ex sinϕ + ey cosϕ( ) = −ex cosϕ − ey sinϕ . ($) 
The question now is how to relate the right side of this equation back to er, and eθ [note: because 
€ 
eϕ ⋅ ∂eϕ ∂ϕ( ) = 0 , 
€ 
∂eϕ ∂ϕ can only be a linear combination of er and eθ]. Assuming 
€ 
∂eϕ ∂ϕ = aer + beθ , then ($) and the unit vector definitions require: 
€ 
acosϕ sinθ + bcosϕ cosθ = −cosϕ , 
€ 
asinϕ sinθ + bsinϕ cosθ = −sinϕ , and 
€ 
acosθ − bsinθ = 0 . 
After dividing out common factors, the first two equations are the same: 
€ 
asinθ + bcosθ = −1. 
When this simplified equation is combined with the third equation, we obtain: 
€ 
a = −sinθ and 
€ 
b = −cosθ . Thus, 
€ 
∂eϕ ∂ϕ = −er sinθ − eθ cosθ . 
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling 
b) Start by constructing the expressions for ex, ey, and ez in terms of 
€ 
er, 
€ 
eθ , and 
€ 
eϕ . This can be 
done my inverting the linear system 
€ 
cosϕ sinθ sinϕ sinθ cosθ
cosϕ cosθ sinϕ cosθ −sinθ
−sinϕ cosϕ 0
% 
& 
' 
( 
' 
) 
* 
' 
+ 
' 
ex
ey
ez
% 
& 
' 
( 
' 
) 
* 
' 
+ 
' 
=
er
eθ
eϕ
% 
& 
' 
( 
' 
) 
* 
' 
+ 
' 
 to find 
€ 
ex = er cosϕ sinθ + eθ cosϕ cosθ − eϕ sinϕ 
€ 
ey = er sinϕ sinθ + eθ sinϕ cosθ + eϕ cosϕ (1,2,3) 
€ 
ez = er cosθ − eθ sinθ . 
The next step is to use the coordinate definitions: 
€ 
r = x 2 + y 2 + z2 , 
€ 
θ = tan−1 x 2 + y 2 z( ) , and 
€ 
ϕ = tan−1 y x( ) (4,5,6) 
to transform the Cartesian partial derivatives. 
€ 
∂
∂x
=
∂r
∂x
∂
∂r
+
∂θ
∂x
∂
∂θ
+
∂ϕ
∂x
∂
∂ϕ
=
x
r
∂
∂r
+
1
1+ x 2 + y 2 z( )
2
2x
2z x 2 + y 2
∂
∂θ
−
y
x 2 + y 2
∂
∂ϕ
 
 
€ 
= cosϕ sinθ ∂
∂r
+
cosϕ cosθ
r
∂
∂θ
−
sinϕ
rsinθ
∂
∂ϕ
 
€ 
∂
∂y
=
∂r
∂y
∂
∂r
+
∂θ
∂y
∂
∂θ
+
∂ϕ
∂y
∂
∂ϕ
=
y
r
∂
∂r
+
1
1+ x 2 + y 2 z( )
2
2y
2z x 2 + y 2
∂
∂θ
+
x
x 2 + y 2
∂
∂ϕ
 
 
€ 
= sinϕ sinθ ∂
∂r
+
sinϕ cosθ
r
∂
∂θ
+
cosϕ
rsinθ
∂
∂ϕ
 
€ 
∂
∂z
=
∂r
∂z
∂
∂r
+
∂θ
∂z
∂
∂θ
+
∂ϕ
∂z
∂
∂ϕ
=
z
r
∂
∂r
+
1
1+ x 2 + y 2 z( )
2
− x 2 + y 2
z2
∂
∂θ
= cosθ ∂
∂r
−
sinθ
r
∂
∂θ
 
Now reassemble the gradient operator 
€ 
∇ = ex
∂
∂x
+ ey
∂
∂y
+ ez
∂
∂z
using the spherical coordinate unit 
vectors and differentiation definitions: 
€ 
∇ = er cosϕ sinθ + eθ cosϕ cosθ − eϕ sinϕ( ) cosϕ sinθ
∂
∂r
+
cosϕ cosθ
r
∂
∂θ
−
sinϕ
rsinθ
∂
∂ϕ
' 
( 
) 
* 
+ 
, 
 
€ 
+ er sinϕ sinθ + eθ sinϕ cosθ + eϕ cosϕ( ) sinϕ sinθ
∂
∂r
+
sinϕ cosθ
r
∂
∂θ
+
cosϕ
rsinθ
∂
∂ϕ
% 
& 
' 
( 
) 
* 
 
€ 
+ er cosθ − eθ sinθ( ) cosθ
∂
∂r
−
sinθ
r
∂
∂θ
% 
& 
' 
( 
) 
* . 
Collect all of the terms with like unit vectors and differential operators together: 
€ 
∇ = er cos
2ϕ sin2θ + sin2ϕ sin2θ + cos2θ( ) ∂
∂r
 
 
€ 
+er cos
2ϕ cosθ sinθ + sin2ϕ cosθ sinθ − cosθ sinθ( )1r
∂
∂θ
 
 
€ 
+er −cosϕ sinϕ sinθ + sinϕ cosϕ sinθ( )
1
rsinθ
∂
∂ϕ
 
 
€ 
+eθ cos
2ϕ cosθ sinθ + sin2ϕ cosθ sinθ − sinθ cosθ( ) ∂
∂r
 
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling 
 
€ 
+eθ cos
2ϕ cos2θ + sin2ϕ cos2θ + sin2θ( )1r
∂
∂θ
 
 
€ 
+eθ −cosϕ sinϕ cosθ + sinϕ cosϕ cosθ( )
1
rsinθ
∂
∂ϕ
 
 
€ 
+eϕ −sinϕ cosϕ sinθ + cosϕ sinϕ sinθ( )
∂
∂r
 
 
€ 
+eϕ −sinϕ cosϕ cosθ + cosϕ sinϕ cosθ( )
1
r
∂
∂θ
 
 
€ 
+eϕ sin
2ϕ + cos2ϕ( ) 1rsinθ
∂
∂ϕ
 
The terms in (,)-parentheses are either +1 or 0. When evaluated they produce: 
€ 
∇ = er
∂
∂r
+ eθ
1
r
∂
∂θ
+ eϕ
1
rsinθ
∂
∂ϕ
. 
c) In spherical coordinates, the divergence of the velocity is: 
€ 
∇ ⋅u = er
∂
∂r
+ eθ
1
r
∂
∂θ
+ eϕ
1
rsinθ
∂
∂ϕ
' 
( 
) 
* 
+ 
, ⋅ erur + eθ uθ + eϕuϕ( ). 
Further simplification requires that the unit vectors and the u's be differentiated. Completing this 
task term by term produces: 
€ 
er
∂
∂r
⋅ erur( ) = er ⋅ er
∂ur
∂r
+ urer ⋅
∂er
∂r
=
∂ur
∂r
, 
€ 
er
∂
∂r
⋅ eθ uθ( ) = er ⋅ eθ
∂uθ
∂r
+ uθer ⋅
∂eθ
∂r
= 0 , 
€ 
er
∂
∂r
⋅ eϕuϕ( ) = er ⋅ eϕ
∂uϕ
∂r
+ uϕer ⋅
∂eϕ
∂r
= 0 ,
€ 
eθ
r
∂
∂θ
⋅ erur( ) =
eθ ⋅ er
r
∂ur
∂θ
+
ur
r
eθ ⋅
∂er
∂θ
= 0 + ur
r
eθ ⋅ eθ =
ur
r
, 
€ 
eθ
r
∂
∂θ
⋅ eθ uθ( ) =
eθ ⋅ eθ
r
∂uθ
∂θ
+
uθ
r
eθ ⋅
∂eθ
∂θ
=
1
r
∂uθ
∂θ
−
uθ
r
eθ ⋅ er =
1
r
∂uθ
∂θ
, 
€ 
eθ
r
∂
∂θ
⋅ eϕuϕ( ) =
eθ ⋅ eϕ
r
∂uϕ
∂θ
+
uϕ
r
eθ ⋅
∂eϕ
∂θ
= 0 
€ 
eϕ
rsinθ
∂
∂ϕ
⋅ erur( ) =
eϕ ⋅ er
rsinθ
∂ur
∂ϕ
+
ur
rsinθ
eϕ ⋅
∂er
∂ϕ
=
ur
rsinθ
eϕ ⋅ eϕ sinθ( ) =
ur
r
, 
€ 
eϕ
rsinθ
∂
∂ϕ
⋅ eθ uθ( ) =
eϕ ⋅ eθ
rsinθ
∂uθ
∂ϕ
+
uθ
rsinθ
eϕ ⋅
∂eθ
∂ϕ
=
uθ
rsinθ
eϕ ⋅ eϕ cosθ =
uθ
r tanθ
, 
€ 
eϕ
rsinθ
∂
∂ϕ
⋅ eϕuϕ( ) =
1
rsinθ
eϕ ⋅ eϕ
∂uϕ
∂ϕ
+ uϕeϕ ⋅
∂eϕ
∂ϕ
& 
' 
( 
) 
* 
+ 
 
€ 
=
1
rsinθ
∂uϕ
∂ϕ
− uϕeϕ ⋅ er sinθ + eθ cosθ( )
' 
( 
) 
* 
+ 
, =
1
rsinθ
∂uϕ
∂ϕ
 
Reassembling the equation produces: 
€ 
∇ ⋅u = ∂ur
∂r
+
2ur
r
+
1
r
∂uθ
∂θ
+
uθ
r tanθ
+
1
rsinθ
∂uϕ
∂ϕ
 
The 1st & 2nd terms, and the 3trd and 4th terms on the right side are commonly combined to yield: 
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling 
€ 
∇ ⋅u = 1
r2
∂
∂r
r2ur( ) + 1rsinθ
∂
∂θ
sinθ uθ( ) +
1
rsinθ
∂uϕ
∂ϕ
. (10) 
d) The Laplacian operator is 
€ 
∇2 ≡ ∇ ⋅∇, and its form in spherical polar coordinates can be found 
by evaluating the dot product. Fortunately, the results of part c) can be used via the following 
replacements for the second gradient operator of the dot product: 
€ 
ur ↔
∂
∂r
 , 
€ 
uθ ↔
1
r
∂
∂θ
 , and 
€ 
uϕ ↔
1
rsinθ
∂
∂ϕ
. (7,8,9) 
Inserting (7,8,9) into (10), the Laplacian then becomes: 
€ 
∇2 =
1
r2
∂
∂r
r2 ∂
∂r
$ 
% 
& 
' 
( 
) +
1
r2 sinθ
∂
∂θ
sinθ ∂
∂θ
 
$ 
% 
& 
' 
( 
) +
1
r2 sin2θ
∂ 2
∂ϕ 2
. 
e) Start with the answer to part b) and compute the first dot product to find: 
€ 
u ⋅ ∇ = ur
∂
∂r
+ uθ
1
r
∂
∂θ
+ uϕ
1
rsinθ
∂
∂ϕ
 
This is the scalar operator applied to 
€ 
u = urer + uθeθ + uϕeϕ to find the advective acceleration: 
€ 
u ⋅ ∇( )u = ur
∂
∂r
+ uθ
1
r
∂
∂θ
+ uϕ
1
rsinθ
∂
∂ϕ
' 
( 
) 
* 
+ 
, urer + uθeθ + uϕeϕ( ) . 
Here the components of u and the unit vectors depend on the angular coordinates. 
€ 
u ⋅ ∇( )u = er ur
∂ur
∂r
+ uθ
1
r
∂ur
∂θ
+ uϕ
1
rsinθ
∂ur
∂ϕ
' 
( 
) 
* 
+ 
, + uruθ
1
r
∂er
∂θ
+ uruϕ
1
rsinθ
∂er
∂ϕ
+ 
 
€ 
eθ ur
∂uθ
∂r
+ uθ
1
r
∂uθ
∂θ
+ uϕ
1
rsinθ
∂uθ
∂ϕ
% 
& 
' 
( 
) 
* + uθ
2 1
r
∂eθ
∂θ
+ uθ uϕ
1
rsinθ
∂eθ
∂ϕ
+ 
 
€ 
eϕ ur
∂uϕ
∂r
+ uθ
1
r
∂uϕ
∂θ
+ uϕ
1
rsinθ
∂uϕ
∂ϕ
% 
& 
' 
( 
) 
* + uϕuθ
1
r
∂eϕ
∂θ
+ uϕ
2 1
rsinθ
∂eϕ
∂ϕUse the results of part a) to evaluate the unit vector derivatives. 
€ 
u ⋅ ∇( )u = er ur
∂ur
∂r
+ uθ
1
r
∂ur
∂θ
+ uϕ
1
rsinθ
∂ur
∂ϕ
' 
( 
) 
* 
+ 
, + uruθ
eθ
r
+ uruϕ
eϕ sinθ
rsinθ
+ 
 
€ 
eθ ur
∂uθ
∂r
+ uθ
1
r
∂uθ
∂θ
+ uϕ
1
rsinθ
∂uθ
∂ϕ
% 
& 
' 
( 
) 
* − uθ
2 er
r
+ uθ uϕ
eϕ cosθ
rsinθ
 
 
€ 
eϕ ur
∂uϕ
∂r
+ uθ
1
r
∂uϕ
∂θ
+ uϕ
1
rsinθ
∂uϕ
∂ϕ
% 
& 
' 
( 
) 
* + 0 + uϕ
2 1
rsinθ
−er sinθ − eθ cosθ( ) 
Collect components 
€ 
u ⋅ ∇( )u = er ur
∂ur
∂r
+ uθ
1
r
∂ur
∂θ
+ uϕ
1
rsinθ
∂ur
∂ϕ
−
uθ
2 + uϕ
2
r
( 
) 
* 
+ 
, 
- + 
 
€ 
eθ ur
∂uθ
∂r
+ uθ
1
r
∂uθ
∂θ
+ uϕ
1
rsinθ
∂uθ
∂ϕ
+
uruθ
r
−
uϕ
2
r
cotθ
& 
' 
( 
) 
* 
+ 
 
€ 
eϕ ur
∂uϕ
∂r
+ uθ
1
r
∂uϕ
∂θ
+ uϕ
1
rsinθ
∂uϕ
∂ϕ
+
uruϕ
r
+
uθ uϕ
r
cotθ
% 
& 
' 
( 
) 
* 
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling 
Exercise 3.3. In a steady two-dimensional flow, Cartesian-component particle trajectories are 
given by: x(t) = ro cos γ (t − to )+θo( ) and y(t) = ro sin γ (t − to )+θo( ) where ro = xo2 + yo2 and 
θo = tan
−1 yo xo( ) . 
a) From these trajectories determine the Lagrangian particle velocity components u(t) = dx/dt and 
v(t) = dy/dt, and convert these to Eulerian velocity components u(x,y) and v(x,y). 
b) Compute Cartesian particle acceleration components, ax = d2x/dt2 and ay = d2y/dt2, and show 
that they are equal to D/Dt of the Eulerian velocity components u(x,y) and v(x,y). 
 
Solution 3.3. a) Differentiate as suggested to find: 
u(t) = dx(t)
dt
= −γro sin γ (t − to )+θo( ) and v(t) =
dy(t)
dt
= γro cos γ (t − to )+θo( ) . 
Now use the original trajectory equations to eliminate the trig-functions: 
u = −γ y and v = γ x . 
b) Again differentiate as suggested to find: 
ax (t) =
d 2x(t)
dt2
= −γ 2ro cos γ (t − to )+θo( ) = −γ 2x and 
ay (t) =
d 2y(t)
dt2
= −γ 2ro sin γ (t − to )+θo( ) = −γ 2y . 
Compute Du/Dt and Dv/Dt from the final two answers of part a): 
Du
Dt
=
∂u
∂t
+u∂u
∂x
+ v ∂u
∂y
= 0−γ y(0)+γ x(−γ ) = −γ 2x and 
Dv
Dt
=
∂v
∂t
+u ∂v
∂x
+ v ∂v
∂y
= 0−γ y(γ )+γ x(0) = −γ 2y . 
The final equalities match as appropriate: ax = Du/Dt, and ay = Dv/Dt. 
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling 
Exercise 3.4. In a steady two-dimensional flow, polar coordinate particle trajectories are given 
by: r(t) = ro and θ(t) = γ (t − to )+θo . 
a) From these trajectories determine the Lagrangian particle velocity components ur(t) = dr/dt 
and, uθ (t) = rdθ/dt, and convert these to Eulerian velocity components ur(r,θ) and uθ (r,θ ) . 
b) Compute polar-coordinate particle acceleration components, ar = d
2r dt2 − r dθ dt( )
2 and 
aθ = rd
2θ dt2 + 2 dr dt( ) dθ dt( ) , and show that they are equal to D/Dt of the Eulerian velocity 
with components ur(r,θ) and uθ (r,θ ) . 
 
Solution 3.4. a) Differentiate as suggested to find: 
ur (t) =
dr(t)
dt
= 0 and uθ (t) = r
dθ(t)
dt
= rγ . 
These equations are readily interpreted as Eulerian velocity components: 
ur = 0 and uθ = γr . 
b) Start with the particle accelerations: 
ar = d
2r dt2 − r dθ dt( )
2
= 0− r(γ )2 = −γ 2r and aθ = rd
2θ dt2 + 2 dr dt( ) dθ dt( ) = 0+ 2(0)γ = 0 . 
Compute Dur/Dt and Duθ/Dt from the forms given in Appendix B and the two answers of part a): 
Dur
Dt
=
∂ur
∂t
+ur
∂ur
∂r
+
uθ
r
∂ur
∂θ
−
uθ
2
r
= 0+ 0− (γr)
2
r
= −γ 2r and 
Duθ
Dt
=
∂uθ
∂t
+ur
∂uθ
∂r
+
uθ
r
∂uθ
∂θ
+
uruθ
r
= 0+ 0(γ )+γ (0)+ 0 = 0 . 
The final equalities match as appropriate: ar = Dur/Dt, and aθ = Duθ/Dt. 
 
 
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling 
Exercise 3.5. if ds = (dx, dy, dz) is an element of arc length along a streamline (Figure 3.5) and u 
= (u, v, w) is the local fluid velocity vector, show that if ds is everywhere tangent to u then 
€ 
dx u = dy v = dz w . 
 
Solution 3.5. If ds = (dx, dy, dz) and u are parallel, then they must have the same unit tangent 
vector t: 
€ 
t = ds
ds
=
(dx,dy,dz)
(dx)2 + (dy)2 + (dz)2
=
(u,v,w)
u2 + v 2 + w2
=
u
u
. 
The three components of this equation imply: 
€ 
dx
ds
=
u
u
, 
€ 
dy
ds
=
v
u
, and 
€ 
dz
ds
=
w
u
. 
But these can be rearranged to find: 
€ 
ds
u
=
dx
u
=
dy
v
=
dz
w
. 
 
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling 
Exercise 3.6. For the two-dimensional steady flow having velocity components u = Sy and v = 
Sx, determine the following when S is a positive real constant having units of 1/time. 
a) equations for the streamlines with a sketch of the flow pattern 
b) the components of the strain rate tensor 
c) the components of the rotation tensor 
d) the coordinate rotation that diagonalizes the strain rate tensor, and the principal strain rates. 
e) How is this flow field related to that in Example 3.5. 
 
Solution 3.6. a) For steady streamlines in two dimensions: 
€ 
dx
u
=
dy
v
 or 
€ 
dy
dx
=
v
u
=
Sx
Sy
=
x
y
 , which implies: 
€ 
ydy = xdx→ y 2 2 = x 2 2 + const. 
Solving for y(x) produces: 
€ 
y = ± x 2 + const . These 
are hyperbolae that asymptote to the lines y = ±x. 
b) Compute the strain rate tensor from its definition: 
€ 
Sij =
1
2
∂ui
∂x j
+
∂u j
∂xi
# 
$ 
% % 
& 
' 
( ( 
 
€ 
=
∂u ∂x 12 ∂u ∂y + ∂v ∂x( )
1
2 ∂v ∂x + ∂u ∂y( ) ∂v ∂y
# 
$ 
% 
& 
' 
( 
 
 
€ 
=
0 12 S + S( )
1
2 S + S( ) 0
" 
# 
$ 
% 
& 
' 
=
0 S
S 0
" 
# 
$ 
% 
& 
' 
 
c) Compute the rotation tensor from its definition: 
€ 
Rij =
∂ui
∂x j
−
∂u j
∂xi
=
0 ∂u ∂y −∂v ∂x
∂v ∂x −∂u ∂y 0
$ 
% 
& 
' 
( 
) 
=
0 S − S
S − S 0
$ 
% 
& 
' 
( 
) 
=
0 0
0 0
$ 
% 
& 
' 
( 
) 
 
d) From Example 2.4, a θ = 45° coordinate rotation diagonalizes the strain rate tensor. The 
direction cosine matrix is: 
€ 
Cij =
cosθ −sinθ
sinθ cosθ
$ 
% 
& 
' 
( 
) 
=
1
2
1 −1
1 1
$ 
% 
& 
' 
( 
) 
, and the rotated strain rate matrix 
S´ is: 
€ 
" S = CT ⋅S ⋅C = 1
2
1 1
−1 1
% 
& 
' 
( 
) 
* 
0 S
S 0
% 
& 
' 
( 
) 
* 
1
2
1 −1
1 1
% 
& 
' 
( 
) 
* 
=
1
2
S S
S −S
% 
& 
' 
( 
) 
* 
1 −1
1 1
% 
& 
' 
( 
) 
* 
=
S 0
0 −S
% 
& 
' 
( 
) 
* 
. 
e) When this flow field is rotated 45° in the clockwise direction, it is the same as the flow field in 
Example 3.5. 
x
y
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling 
 
Exercise 3.7. At the instant shown in Figure 3.2b, the (u,v)-velocity field in Cartesian 
coordinates is 
€ 
u = A(y 2 − x 2) (x 2 + y 2)2 , and 
€ 
v = −2A xy (x 2 + y 2)2 where A is a positive 
constant. Determine the equations for the streamlines by rearranging the first equality in (3.7) to 
read
€ 
udy − vdx = 0 = ∂ψ ∂y( )dy + ∂ψ ∂x( )dx and then looking for a solution in the form ψ(x,y) = 
const. 
 
Solution 3.7. Rearrange the two-dimensional streamline condition, dx/u = dy/v, to obtain udy – 
vdx = 0 as the description of a streamline. Assume this differential equation is solved by the 
function ψ(x,y) = const, so that (∂ψ/∂x)dx + (∂ψ/∂y)dy = 0. Comparing the two equations 
requires: 
u = ∂ψ/∂y , and v = –∂ψ/∂x. 
Now use the given velocity field to find: 
€ 
∂ψ ∂y = A(y 2 − x 2) (x 2 + y 2)2 , and 
€ 
∂ψ ∂x = +2A xy (x 2 + y 2)2 . (a,b) 
Integrate (b) treating y as a constant: 
€ 
∂ψ
∂x
= Ay 2x
(x 2 + y 2)2
→ ψ −ψo = Ay
2xdx
(x 2 + y 2)2∫
= Ay −1
x 2 + y 2
' 
( 
) 
* 
+ 
, , 
where ψo may depend on y. Differentiate this result with respect to y to determine ψo: 
€ 
∂
∂y
ψ −ψo( ) =
∂
∂y
−Ay
x 2 + y 2
% 
& 
' 
( 
) 
* =
−A
x 2 + y 2
−
(−Ay)
(x 2 + y 2)2
(2y) = A(y
2 − x 2)
(x2 + y 2)2
= u. 
This result and equation (a) implies ∂ψo/∂y = 0, so it is at most a constant. Thus, the streamlines 
are given by: 
€ 
ψ(x,y) = const.= − Ay
x 2 + y 2
. 
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling 
Exercise 3.8. Determine the equivalent of the first equality in (3.7) for two dimensional (r,θ)-
polar coordinates, and then find the equation for the streamline that passes through (ro, θo) when 
u = (ur, uθ) = (A/r, B/r) where A and B are constants. 
 
Solution 3.8. The two-dimensional streamline condition in Cartesian coordinates is dx/u = dy/v, 
and is obtained from considering the streamline-tangent vector t: 
€ 
t = ds
ds
=
exdx + eydy
(dx)2 + (dy)2
=
exu + eyv
u2 + v 2
=
u
u
. 
In two-dimensional polar coordinates this becomes: 
€ 
t = ds
ds
=
erdr + eθ rdθ
(dr)2 + (rdθ)2
=
erur + eθ uθ
ur
2 + uθ
2
=
u
u
. 
Equating components produces two equations: 
€ 
dr
ds
=
ur
u
 and 
€ 
rdθ
ds
=
uθ
u
 , or 
€ 
ds
u
=
dr
ur
=
rdθ
uθ
. 
Thus, using the last equality and the given velocity field: 
€ 
1
r
dr
dθ
=
ur
uθ
=
A r
B r
=
A
B
→ ln(r) = A
B
θ + const. 
The initial condition allows the constant to be evaluated: 
€ 
ln(ro) =
A
B
θo + const., which leads to 
€ 
ln r
ro
" 
# 
$ 
% 
& 
' =
A
B
θ −θo( ) or 
€ 
r = ro exp
A
B
θ −θo( )
$ 
% 
& 
' 
( 
) . 
 
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling 
Exercise 3.9. Determine the streamline, path line, and streak line that pass through the origin of 
coordinates at t = t´ when u = Uo + ωξocos(ωt) and v = ωξosin(ωt) in two-dimensional Cartesian 
coordinates where Uo is a constant horizontal velocity. Compare your results those in Example. 
3.3 for 
€ 
Uo → 0. 
 
Solution 3.9. (i) For the streamline, time is a constant. Use the first equality of (3.7) to find: 
€ 
dy
dx
=
v
u
=
ωξo sin(ωt)
Uo +ωξo cos(ωt)
= m(t) , 
where m is the streamline slope. Since m does not depend on the spatial coordinate, this equation 
is readily integrated to find straight time-dependent streamlines: y = m(t)x + const. Thus, the 
streamline that passes through (0,0) at t = t´ is: 
€ 
y = ωξo sin(ω $ t )
Uo +ωξo cos(ω $ t )
x . 
(ii) For the path line, use both components of (3.8): 
€ 
dx
dt
=Uo +ωξo cos(ωt) and 
€ 
dy
dt
=ωξo sin(ωt) , 
and integrate in time to find: 
€ 
x − xo =Uot + ξo sin(ωt) and 
€ 
y − yo = –ξo cos(ωt) . 
Determine xo and yo by requiring the path line to pass through the origin at at t = t´: 
€ 
0 − xo = Uo # t + ξo sin(ω # t ) and 
€ 
0 − yo = –ξo cos(ω % t ) . 
The final component equations are: 
€ 
x = Uo(t − # t ) + ξo sin(ωt) − sin(ω # t )( ) and 
€ 
y = –ξo cos(ωt) − cos(ω % t )( ) . 
These two parametric equations for x(t) and y(t) can be combined to eliminate some of the t-
dependence: 
€ 
x −Uo(t − # t ) + ξo sin(ω # t )( )
2
+ y −ξo cos(ω # t )( )
2
= ξo
2 , 
which describes a moving circle with center located at 
€ 
Uo(t − # t ) −ξo sin(ω # t ),ξo cos(ω # t )( ) . 
(iii) For the streak line, use the path line results but this time evaluate the constants at t = to 
instead of at t = t´ to find: 
€ 
x =Uo(t − to) + ξo sin(ωt) − sin(ωto)( ) and 
€ 
y = –ξo cos(ωt) − cos(ωto)( ). 
Now evaluate these equations at t = t´ to produce two parametric equations for the streak line 
coordinates x(to) and y(to): 
€ 
x = Uo( " t − to) + ξo sin(ω " t ) − sin(ωto)( ) and 
€ 
y = –ξo cos(ω $ t ) − cos(ωto)( ) . 
Some of the to dependence can be eliminated by combining the equations: 
€ 
x −Uo( # t − to) −ξo sin(ω # t )( )
2
+ y + ξo cos(ω # t )( )
2
= ξo
2, 
which describes a circle with a to-dependent center located at 
€ 
Uo( " t − to) + ξo sin(ω " t ),−ξo cos(ω " t )( ). 
 These results differ from those in Example 3.3 by the uniform translation velocity Uo so 
they can be put into correspondence with a Galilean transformation x´ = x – Uo(t – t´). 
 
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling 
Exercise 3.10. Compute and compare the streamline, path line, and streak line that pass through 
(1,1,0) at t = 0 for the following Cartesian velocity field u = (x, –yt, 0). 
 
Solution 3.10. (i) For the streamline, time is a constant. Use the first equality of (3.7) to find: 
€ 
dy
dx
=
v
u
= −
yt
x
→
dy
y
= −t dx
x
→ ln y = −t ln x + const., or y = const.x–t. 
Evaluating at x = y = 1 and t = 0 requires the constant to be unity, so the streamline is: y = 1. 
(ii) For the path line, use both components of (3.8): 
€ 
dx
dt
= x and 
€ 
dy
dt
= −yt , 
and integrate these in time to find: 
€ 
x = C1e
t and 
€ 
y = C2 exp −t
2 2{ }, 
where C1 and C2 are constants. Evaluating at x = y = 1 and t = 0 requires C1 = C2 = 1. Eliminate t 
from the y-equation using t = ln(x) to find the path line as: 
€ 
y = exp −(ln x)2 2{ } . 
(iii) For the streak line, use the path line results but this time evaluate the constants at t = to 
instead of at t = 0 to find: 
€ 
x = exp t − to{ } and 
€ 
y = exp (to
2 − t 2) 2{ } . 
Evaluate at t = 0, and eliminate to from the resulting equations to find: 
€ 
y = exp +(ln x)2 2{ }. 
 
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling 
Exercise 3.11. Consider a time-dependent flow field in two-dimensional Cartesian coordinates 
where 
€ 
u = τ t 2 , 
€ 
v = xy τ , and 
€ 
 and τ are constant length and length time scales, respectively. 
a) Use dimensional analysis to determine the functional form of the streamline through x´ at time 
t´. 
b) Find the equation for the streamline through x´ at time t´ and put your answer in 
dimensionless form. 
c) Repeat b) for the path line through x´ at time t´. 
d) Repeat b) for the streak line through x´ at time t´. 
 
Solution 3.11. a) The streamline y(x) will depend on x, t, t´, x´= (x´,y´), 
€ 
 , and τ. There are eight 
parameters and two dimensions, thus there are six dimensionless groups: 
 
€ 
y

= Ψ
x

, # x 

, # y 

, t
τ
, # t 
τ
% 
& 
' 
( 
) 
* . 
Here there are too many variables and parameters for dimensional analysis to be really useful. 
However, this effort provides a reminder to check units throughout the remainder of the solution. 
i) For the streamline, time is a constant. Use the first equality of (3.7) to find: 
 
€ 
dy
dx
=
v
u
=
xy τ
τ t 2
→
dy
y
=
t 2
τ 2
xdx
2
→ ln y = t
2
τ 2
x 2
22
+ const. 
The initial condition requires, x = x´ and y = y´ at t = t´, and this allows the constant to be 
determined, yielding: 
 
€ 
ln y
" y 
# 
$ 
% 
& 
' 
( =
(t 2x 2 − " t 2 " x 2)
22τ 2
. 
(ii) For the path line, use both components of (3.8): 
 
€ 
dx
dt
=
τ
t 2
 and 
 
€ 
dy
dt
=
xy
τ
, 
and integrate the first of these in time and use the initial condition to find: 
 
€ 
x = −τ t + const. or 
€ 
x − # x = −τ t−1 − # t −1( ) . 
Use this result for x(t) in the second equation for y(t): 
 
€ 
dy
dt
=
y
τ
−τ
1
t
−
1
$ t 
% 
& 
' 
( 
) 
* + $ x 
% 
& 
' 
( 
) 
* or 
 
€ 
dy
y
=
1
" t 
−
1
t
+
1
τ
" x 
% 
& 
' 
( 
) 
* dt . 
The last expression can be integrated to find: 
 
€ 
ln y
" y 
# 
$ 
% 
& 
' 
( =
t
" t 
−1+ " x 
τ
− ln t
" t 
# 
$ 
% 
& 
' 
( =
" x 
τ
+
1
" t 
# 
$ 
% 
& 
' 
( (t − " t ) − ln
t
" t 
# 
$ 
% 
& 
' 
( or 
 
€ 
y
" y 
=
t
" t 
# 
$ 
% 
& 
' 
( 
−1
exp " x " t 
τ
+1
# 
$ 
% 
& 
' 
( 
t
" t 
−1
# 
$ 
% 
& 
' 
( 
+ 
, 
- 
. 
/ 
0 . 
Now use the final equations for x(t) and y(t) to eliminate t. The equation for x(t) can be 
rearranged to find: 
 
€ 
" t 
t
=1− (x − " x ) " t 
τ
, 
sothe equation for y becomes: 
 
€ 
y
" y 
= 1− (x − " x ) " t 
τ
% 
& 
' 
( 
) 
* exp
" x " t 
τ
+1
% 
& 
' 
( 
) 
* 1−
(x − " x ) " t 
τ
% 
& 
' 
( 
) 
* 
−1
−1
% 
& 
' ' 
( 
) 
* * 
+ 
, 
- 
- 
. 
/ 
0 
0 
. 
(iii) For the streak line, use the path line results but this time evaluate the integration constants at 
t = to instead of at t = t´ to find: 
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling 
 
€ 
x − # x = −τ t−1 − to
−1( ) , and 
 
€ 
y
" y 
=
t
to
# 
$ 
% 
& 
' 
( 
−1
exp " x to
τ
+1
# 
$ 
% 
& 
' 
( 
t
to
−1
# 
$ 
% 
& 
' 
( 
+ 
, 
- 
. 
/ 
0 . 
Now eliminate to find: 
 
€ 
y
" y 
= 1+ (x − " x )t
τ
% 
& 
' 
( 
) 
* 
−1
exp " x 
τ
1
t
+
(x − " x )
τ
% 
& 
' 
( 
) 
* 
−1
+1
% 
& 
' ' 
( 
) 
* * 
(x − " x )t
τ
% 
& 
' 
( 
) 
* 
+ 
, 
- 
- 
. 
/ 
0 
0 
. 
And evaluate at t = t´ to reach: 
 
€ 
y
" y 
= 1+ (x − " x ) " t 
τ
% 
& 
' 
( 
) 
* 
−1
exp " x 
τ
1
" t 
+
(x − " x )
τ
% 
& 
' 
( 
) 
* 
−1
+1
% 
& 
' ' 
( 
) 
* * 
(x − " x ) " t 
τ
% 
& 
' 
( 
) 
* 
+ 
, 
- 
- 
. 
/ 
0 
0 
. 
 
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling 
Exercise 3.12. The velocity components in an unsteady plane flow are given by 
€ 
u = x (1+ t) and 
€ 
v = 2y (2 + t). Determine equations for the streamlines and path lines subject to x = x0 at t = 0. 
 
Solution 3.12. i) For the streamline, time is a constant. Use the first equality of (3.7) to find: 
€ 
dy
dx
=
v
u
=
2y (2 + t)
x (1+ t)
→
dy
y
=
2(1+ t)
(2 + t)
dx
x
→ ln y = 2(1+ t)
(2 + t)
ln x + const. 
Use of the initial condition produces: 
€ 
ln y0 =
2(1+ 0)
(2 + 0)
ln x0 + const., 
so the final answer is: 
€ 
ln y
y0
" 
# 
$ 
% 
& 
' =
2(1+ t)
(2 + t)
ln x
x0
" 
# 
$ 
% 
& 
' or 
€ 
y
y0
=
x
x0
" 
# 
$ 
% 
& 
' 
2(1+ t )
(2+ t )
. 
(ii) For the path line, use both components of (3.8): 
€ 
dx
dt
=
x
1+ t
 and 
€ 
dy
dt
=
2y
2 + t
, 
and integrate the these in time to find: 
€ 
ln x = ln(1+ t) + const. and 
€ 
ln y = 2ln(2 + t) + const. 
Use the initial condition to determine the two constants, and exponentiate both equations: 
€ 
x = x0(1+ t) and 
€ 
y = y0 1+ t 2( )
2 . 
To determine the path line, eliminate t to find: 
€ 
y = y0 1+ (x − x0) 2x0( )
2 . 
 
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling 
Exercise 3.13. Using the geometry and notation of Fig. 3.8, prove (3.9). 
 
Solution 3.13. Before starting this problem, it is worthwhile to note that the acceleration of a 
fluid particle is invariant under the specified Galilean transformation so the components of U 
cannot be part of the final answer. Thus, transformation errors can be readily detected if terms 
are missing in the final results or extra ones have appeared. 
 Figure 3.8 supports the following vector addition formula: 
€ 
x = Ut + " x o + " x . Thus, the 
Cartesian-coordinate transformations in this case are given by: 
€ 
" x = x − ex ⋅U( )t − " x o , 
€ 
" y = y − ey ⋅U( )t − " y o , 
€ 
" z = z − ez ⋅U( )t − " z o , and 
€ 
" t = t . The transformation of the spatial 
derivatives between the stationary frame of reference, Oxyz, and the steadily moving frame, 
O´x´y´z´ is straightforward mathematics: 
 
€ 
∂
∂x
=
∂ # x 
∂x
∂
∂ # x 
+
∂ # y 
∂x
∂
∂ # y 
+
∂ # z 
∂x
∂
∂ # z 
+
∂ # t 
∂x
∂
∂ # t 
=
∂
∂ # x 
, 
 
€ 
∂
∂y
=
∂ # x 
∂y
∂
∂ # x 
+
∂ # y 
∂y
∂
∂ # y 
+
∂ # z 
∂y
∂
∂ # z 
+
∂ # t 
∂y
∂
∂ # t 
=
∂
∂ # y 
, and 
 
€ 
∂
∂z
=
∂ # x 
∂z
∂
∂ # x 
+
∂ # y 
∂z
∂
∂ # y 
+
∂ # z 
∂z
∂
∂ # z 
+
∂ # t 
∂z
∂
∂ # t 
=
∂
∂ # z 
, 
where the final equality on each line follows from differentiating the definitions of the moving 
coordinate variables given above. The time derivative requires more effort 
€ 
∂
∂t
=
∂ # x 
∂t
∂
∂ # x 
+
∂ # y 
∂t
∂
∂ # y 
+
∂ # z 
∂t
∂
∂ # z 
+
∂ # t 
∂t
∂
∂ # t 
= −ex ⋅U
∂
∂ # x 
− ey ⋅U
∂
∂ # y 
− ez ⋅U
∂
∂ # z 
+
∂
∂ # t 
. 
The first three equations imply: 
€ 
∇ = # ∇ and the fourth implies: 
€ 
−U ⋅ $ ∇ + ∂
∂ $ t 
. The velocities will 
be related by: u = U + u´. Now use these results to assemble the fluid particle acceleration 
starting in the stationary coordinate system, and converting each velocity and differential 
operation to the moving coordinate system. 
€ 
∂u
∂t
+ u ⋅ ∇( )u = −U ⋅ & ∇ + ∂
∂ & t 
' 
( 
) 
* 
+ 
, U + & u [ ] + U + & u [ ] ⋅ & ∇ ( ) U + & u [ ] 
 
€ 
= −U ⋅ $ ∇ U + ∂U
∂ $ t 
−U ⋅ $ ∇ $ u + ∂ $ u 
∂ $ t 
+ U ⋅ $ ∇ ( )U + $ u ⋅ $ ∇ ( )U + U ⋅ $ ∇ ( ) $ u + $ u ⋅ $ ∇ ( ) $ u 
 
€ 
= −U ⋅ $ ∇ $ u + ∂ $ u 
∂ $ t 
+ U ⋅ $ ∇ ( ) $ u + $ u ⋅ $ ∇ ( ) $ u = ∂
$ u 
∂ $ t 
+ $ u ⋅ $ ∇ ( ) $ u 
Here most of the simplifications occur because all derivatives of U are zero; it is a constant. 
Thus, as expected, the form of the fluid particle acceleration is frame invariant for coordinate 
systems related by Galilean transformations. 
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling 
Exercise 3.14. Determine the unsteady, ∂u/∂t, and advective, (u⋅∇)u, fluid acceleration terms for 
the following flow fields specified in Cartesian coordinates. 
a) 
€ 
u = u(y,z,t),0,0( ) 
b) 
€ 
u =Ω× x where 
€ 
Ω = 0,0,Ωz (t)( ) 
c) 
€ 
u = A(t) x,−y,0( ) 
d) u = (Uo + uosin(kx – Ωt), 0, 0) where Uo, uo, k and Ω are positive constants 
 
Solution 3.14. a) Here there is only one component of the fluid velocity; thus 
 
€ 
∂u ∂t = ∂ ∂t( ) u(y,z,t),0,0( ) = ∂u ∂t ,0,0( ) , and 
 
€ 
u ⋅ ∇[ ]u = u(y,z,t),0,0( ) ⋅ ∂ ∂x ,∂ ∂y ,∂ ∂z( )[ ] u(y,z,t),0,0( ) = u(∂ ∂x)[ ] u(y,z,t),0,0( ) = 0 . 
b) Here the fluid velocity has two components: 
€ 
u =Ω× x = (–Ωzy, +Ωzx, 0), so 
 
€ 
∂u ∂t = ∂ ∂t( ) −Ωz y,Ωz x,0( ) = −y
dΩz
dt
,x dΩz
dt
,0
% 
& 
' 
( 
) 
* =
dΩz
dt
−y,x,0( ) , and 
 
€ 
u ⋅ ∇[ ]u = −Ωz y,Ωz x,0( ) ⋅ ∂ ∂x ,∂ ∂y ,∂ ∂z( )[ ] −Ωz y,Ωz x,0( ) 
 
€ 
= −Ωz y(∂ ∂x) +Ωz x(∂ ∂y)[ ] −Ωz y,Ωz x,0( ) = −Ωz2x,−Ωz2y,0( ) = −Ωz2 x,y,0( ) . 
c) Again the fluid velocity has two components: (Ax, –Ay, 0), so 
 
€ 
∂u ∂t = ∂ ∂t( ) Ax,−Ay,0( ) = x dA
dt
,−y dA
dt
,0
$ 
% 
& 
' 
( 
) =
dA
dt
x,−y,0( ) , and 
 
€ 
u ⋅ ∇[ ]u = Ax,−Ay,0( ) ⋅ ∂ ∂x ,∂ ∂y ,∂ ∂z( )[ ] Ax,−Ay,0( ) 
 
€ 
= Ax(∂ ∂x) − Ay(∂ ∂y)[ ] Ax,−Ay,0( ) = A2x,A2y,0( ) = A2 x,y,0( ). 
d) Here again there is only one component of the fluid velocity; thus 
 
€ 
∂u ∂t = ∂ ∂t( ) Uo + uo sin(kx −Ωt),0,0( ) = −Ωuo cos(kx −Ωt),0,0( ) , and 
 
€ 
u ⋅ ∇[ ]u = Uo + uo sin(kx −Ωt),0,0( ) ⋅ ∂ ∂x ,∂ ∂y ,∂ ∂z( )[ ] Uo + uo sin(kx −Ωt),0,0( ). 
 
€ 
= Uo + uo sin(kx −Ωt)( )(∂ ∂x)[ ] Uo + uo sin(kx −Ωt),0,0( ) 
 
€ 
= Uo + uo sin(kx −Ωt)( )kuo cos(kx −Ωt),0,0( ) 
 
€ 
= kUouo cos(kx −Ωt) +
1
2 kuo
2 sin 2(kx −Ωt)[ ],0,0( ). 
 
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling 
Exercise 3.15. Consider the following Cartesian velocity field 
€ 
u = A(t) f (x),g(y),h(z)( ) where 
A, f, g, and h are non-constant functions of only one independent variable. 
a) Determine ∂u/∂t, and (u⋅∇)u in terms of A, f, g, and h, and their derivatives. 
b) Determine A, f, g, and h when Du/Dt = 0, u = 0 at x = 0, and u is finite for t > 0. 
c) For the conditions in b), determine the equation for the path line that passes through xo at time 
to, and show directly that the acceleration a of the fluid particle that follows this path is zero. 
 
Solution 3.15. a) Herethere are three components of the fluid velocity; thus 
 
€ 
∂u
∂t
=
∂
∂t
Af ,Ag,Ah( ) = f dA
dt
,g dA
dt
,h dA
dt
# 
$ 
% 
& 
' 
( =
dA
dt
f ,g,h( ) , and 
 
€ 
u ⋅ ∇[ ]u = A f ,g,h( ) ⋅ ∂
∂x
, ∂
∂y
, ∂
∂z
% 
& 
' 
( 
) 
* 
+ 
, 
- 
. 
/ 
0 Af ,Ag,Ah( ) 
 
€ 
= A2 f ∂
∂x
+ g ∂
∂y
+ h ∂
∂z
# 
$ 
% 
& 
' 
( f ,g,h( ) = A2 f
df
dx
,g dg
dy
,h dh
dz
) 
* 
+ 
, 
- 
. , 
where the partial derivatives have become total derivatives because f, g, and h are only functions 
of one variable. 
b) For the given velocity field, Du/Dt = 0 implies: 
€ 
f dA
dt
+ A2 f df
dx
= 0 , 
€ 
g dA
dt
+ A2g dg
dy
= 0 , and 
€ 
h dA
dt
+ A2h dh
dz
= 0. 
Start with the first equation, assume A2f is not zero, and divide by it to find: 
€ 
1
A2
dA
dt
+
df
dx
= 0 . 
The first term in this equation only depends on t while the second one only depends on x, thus, 
each must be equal and opposite, and constant (= C). So, 
€ 
1
A2
dA
dt
= −C→− dA
A2
= Cdt→ 1
A
= C(t − to)→ A =
1
C(t − to)
, and 
€ 
df
dx
= C→ f = C(x − xo) , 
where to and xo are constants of integration. Similarly, for the other two component directions: 
€ 
g = C(y − yo) and 
€ 
h = C(z − zo). 
Here, we presume to ≤ 0 so that A is finite for t > 0. 
c) The x-component of the path line is defined by: 
€ 
dx
dt
= u = Af = C(x − xo)
C(t − to)
→
dx
x − xo
=
dt
t − to
→ ln(x − xo) = ln(t − to) + const.→ x − xo =U(t − to), 
where U is a constant. Similarly for the other Cartesian directions: 
€ 
y − yo =V (t − to), and 
€ 
z − zo =W (t − to) , where V and W are constants. So defining the constant vector U = (U, V, W), 
the path line of interest is: 
x – xo = U(t – to). 
Thus, x(t) is linear function of t, so dx(t)/dt = U, and a = d2x(t)/dt2 = 0. This exercise illustrates 
how the unsteady and advective acceleration terms may be equal and opposite. 
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling 
Exercise 3.16 If a velocity field is given by u = ay and v = 0, compute the circulation around a 
circle of radius ro that is centered on at the origin. Check the result by using Stokes’ theorem. 
 
Solution 3.16. In plane polar coordinates, the vector path-length element, ds, on a circle of radius 
ro is ds = 
€ 
tds = eθ rodθ . From Example 2.1, the radial and angular velocity component are: 
ur = ucosθ + vsinθ = arsinθcosθ + 0, and uθ = –usinθ + vcosθ = –arsin2θ , 
where u = ay = arsinθ and v = 0 has been used. Thus, circulation is: 
€ 
Γ = u ⋅ ds =∫ u ⋅ eθ rodθθ = 0
θ = 2π
∫ = uθ rodθθ = 0
θ = 2π
∫ = − aro2 sin2θdθθ = 0
θ = 2π
∫ = −aπro2 . 
 Now use Stokes' theorem to reach the same result. Start by computing the vorticity: 
€ 
∇ × u =
ex ey ez
∂ ∂x ∂ ∂y ∂ ∂z
ay 0 0
= −aez . 
Insert this into the Stokes' theorem area integral using n = ez, and dA = rdrdθ: 
€ 
Γ = (∇ × u) ⋅ndA
A
∫∫ = (−aez )r= 0
r= ro∫θ = 0
θ = 2π
∫ ⋅ ezrdrdθ = −aπro2 , 
and this matches the result of the prior circulation calculation. 
 
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling 
Exercise 3.17. Consider a plane Couette flow of a viscous fluid confined between two flat plates 
a distance b apart. At steady state the velocity distribution is u = Uy/b and v = w = 0, where the 
upper plate at y = b is moving parallel to itself at speed U, and the lower plate is held stationary. 
Find the rates of linear strain, the rate of shear strain, and vorticity in this flow. 
 
Solution 3.17. Here there is only one velocity component: u = Uy/b. The strain rate tensor is: 
€ 
Sij =
1
2
∂ui
∂x j
+
∂u j
∂xi
# 
$ 
% % 
& 
' 
( ( → Sij =
∂u ∂x 12 ∂u ∂y + ∂v ∂x( )
1
2 ∂v ∂x + ∂u ∂y( ) ∂v ∂y
* 
+ 
, 
- 
. 
/ 
=
0 U 2b
U 2b 0
* 
+ 
, 
- 
. 
/ 
. 
Thus, the linear strain rates are both zero, and the shear strain rate is U/2b. 
The vorticity vector has one non-zero component: 
€ 
ωz =
∂v
∂x
−
∂u
∂y
= −
U
b
. 
 
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling 
Exercise 3.18. The steady two-dimensional flow field inside a sloping passage is given in (x,y)-
Cartesian coordinates by u = (u,v) = 3q 4h( ) 1− y h( )2( ) 1, y h( ) dh dx( )( ) where q is the volume 
flow rate per unit length into the page, and h is the passage's half thickness. Determine the 
streamlines, vorticity, and strain rate tensor in this flow away from x = 0 when h = αx where α is 
a positive constant. Sample profiles of u(x,y) vs. y are shown at two x-locations in the figure. 
What are the equations of the streamlines along which the x- and y-axes are aligned with the 
principal axes of the flow? What is the fluid particle rotation rate along these streamlines? 
 
 
Solution 3.18. For planar flow in Cartesian coordinates, the streamlines are determined from: 
dy
dx
=
v
u
=
y h( ) dh dx( )
1
= y αx( )α = y x . 
Separate the differentials and integrate to find: ln(y) = ln(x) + const. Exponentiate to find: y = Cx, 
where C is a constant. Thus, the streamlines are straight lines through the origin of coordinates. 
 The vorticity ωz is determined from: 
ωz =
∂v
∂x
−
∂u
∂y
=
∂
∂x
3q
4h
#
$
%
&
'
( 1−
y2
h2
#
$
%
&
'
(
y
h
dh
dx
)
*
+
,
-
.−
∂
∂y
3q
4h
#
$
%
&
'
( 1−
y2
h2
#
$
%
&
'
(
)
*
+
,
-
.
 = 3q
4
∂
∂x
1
αx
1− y
2
α 2x2
#
$
%
&
'
(
y
αx
α
)
*
+
,
-
.−
3q
4
∂
∂y
1
αx
1− y
2
α 2x2
#
$
%
&
'
(
)
*
+
,
-
.
 = 3q
4
−
2y
αx3
+
4y3
α 3x5
)
*
+
,
-
.−
3q
4
−
2y
α 3x3
)
*+
,
-.
=
3qy
2αx3
−1+ 2y
2
α 2x2
+
1
α 2
)
*
+
,
-
.
 
 The strain-rate tensor Sij is computed from the following velocity derivatives: 
S11 =
∂u
∂x
=
∂
∂x
3q
4h
"
#
$
%
&
' 1−
y2
h2
"
#
$
%
&
'
)
*
+
,
-
.=
3q
4
∂
∂x
1
αx
1− y
2
α 2x2
"
#
$
%
&
'
)
*
+
,
-
.=
3q
4
−
1
αx2
+
3y2
α 3x4
)
*
+
,
-
.=
3q
4αx2
−1+ 3y
2
α 2x2
)
*
+
,
-
. 
S22 =
∂v
∂y
=
∂
∂y
3q
4h
"
#
$
%
&
' 1−
y2
h2
"
#
$
%
&
'
y
h
dh
dx
)
*
+
,
-
.=
3q
4
∂
∂y
1
αx
1− y
2
α 2x2
"
#
$
%
&
'
y
αx
α
)
*
+
,
-
.=
3q
4αx2
1− 3y
2
α 2x2
"
#
$
%
&
' , and 
S12 = S21 =
1
2
∂u
∂y
+
∂v
∂x
"
#
$
%
&
'
=
1
2
∂
∂y
3q
4h
(
)
*
+
,
- 1−
y2
h2
(
)
*
+
,
-
/
0
1
2
3
4+
1
2
∂
∂x
3q
4h
(
)
*
+
,
- 1−
y2
h2
(
)
*
+
,
-
y
h
dh
dx
/
0
1
2
3
4=
3qy
4αx3
−
1
α 2
−1+ 2y
2
α 2x2
/
0
1
2
3
4 , 
where the differentiation details for S12 are available above in the calculation of ωz. 
 The principle axes occur then the off-diagonal strain rate components are zero. This 
occurs when 
x!
y!
h(x)!
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling 
S12 = S21 =
3qy
4αx3
−
1
α 2
−1+ 2y
2
α 2x2
"
#
$
%
&
'= 0 . 
This equality is satisfied along the x-axis where y = 0, and when the contents of the [,]-brackets 
are zero: 
1
α 2
+1= 2y
2
α 2x2
 or y = ±x (1+α 2 ) 2 
These streamlines only occur inside flow wedge when α > 1. 
 The fluid particle rotation rate is ωz/2, so from the results above for ωz: 
ωz
2
!
"
#
$
%
&
y=0
= 0 , and ωz
2
!
"
#
$
%
&
y=±x (1+α2 ) 2
= ±
3q (1+α 2 ) 2
2α 3x2
. 
 
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling 
Exercise 3.19. For the flow field 
€ 
u =U+Ω× x , where U and Ω are constant linear- and angular-
velocity vectors, use Cartesian coordinates to a) show that Sij is zero, and b) determine Rij. 
 
Solution 3.19. Since no simplifications are given, all the components of U = (U1, U2, U3) and Ω 
= (Ω1, Ω2, Ω3) should be treated as being non-zero. In Cartesian coordinates, the velocity field is 
 
€ 
u =U+Ω× x =U1e1 +U2e2 +U3e3 +
e1 e2 e3Ω1 Ω2 Ω3
x1 x2 x3
 
 
€ 
= U1 +Ω2x3 −Ω3x2( )e1 + U2 +Ω3x1 −Ω1x3( )e2 + U3 +Ω1x2 −Ω2x1( )e3 
a) Use this velocity field result to compute the velocity gradient tensor, and its transpose 
(indicated with a superscript "T" below) to sort out which derivatives are zero and which ones 
are not. 
€ 
∂ui
∂x j
=
∂u1 ∂x1 ∂u1 ∂x2 ∂u1 ∂x3
∂u2 ∂x1 ∂u2 ∂x2 ∂u2 ∂x3
∂u3 ∂x1 ∂u3 ∂x2 ∂u3 ∂x3
# 
$ 
% 
& 
% 
' 
( 
% 
) 
% 
=
0 −Ω3 +Ω2
+Ω3 0 −Ω1
−Ω2 +Ω1 0
# 
$ 
% 
& 
% 
' 
( 
% 
) 
% 
 
€ 
∂u j
∂xi
=
∂ui
∂x j
# 
$ 
% % 
& 
' 
( ( 
T
=
∂u1 ∂x1 ∂u2 ∂x1 ∂u3 ∂x1
∂u1 ∂x2 ∂u2 ∂x2 ∂u3 ∂x2
∂u1 ∂x3 ∂u2 ∂x3 ∂u3 ∂x3
) 
* 
+ 
, 
+ 
- 
. 
+ 
/ 
+ 
=
0 +Ω3 −Ω2
−Ω3 0 +Ω1
+Ω2 −Ω1 0
) 
* 
+ 
, 
+ 
- 
. 
+ 
/ 
+ 
 
This result can be used to construct the strain rate tensor Sij: 
 
€ 
Sij =
1
2
∂ui
∂x j
+
∂u j
∂xi
# 
$ 
% % 
& 
' 
( ( =
1
2
∂ui
∂x j
+
∂ui
∂x j
# 
$ 
% % 
& 
' 
( ( 
T# 
$ 
% 
% 
& 
' 
( 
( 
=
0 12 −Ω3 +Ω3( )
1
2 +Ω2 −Ω2( )
1
2 +Ω3 −Ω3( ) 0
1
2 −Ω1 +Ω1( )
1
2 −Ω2 +Ω2( )
1
2 +Ω1 −Ω1( ) 0
+ 
, 
- 
. 
- 
/ 
0 
- 
1 
- 
 
 
€ 
=
0 0 0
0 0 0
0 0 0
" 
# 
$ 
% 
$ 
& 
' 
$ 
( 
$ 
. 
b) Similarly for the rotation tensor: 
 
€ 
Rij =
∂ui
∂x j
−
∂u j
∂xi
=
∂ui
∂x j
−
∂ui
∂x j
$ 
% 
& & 
' 
( 
) ) 
T
=
0 −Ω3 −Ω3 +Ω2 +Ω2
+Ω3 +Ω3 0 −Ω−Ω1
−Ω2 −Ω2 +Ω1 +Ω1 0
+ 
, 
- 
. 
- 
/ 
0 
- 
1 
- 
 
 
€ 
=
0 −2Ω3 +2Ω2
+2Ω3 0 −2Ω1
−2Ω2 +2Ω1 0
$ 
% 
& 
' 
& 
( 
) 
& 
* 
& 
. 
 
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling 
Exercise 3.20. Starting with a small rectangular volume element δV = δx1δx2δx3, prove (3.14). 
 
Solution 3.20. The volumetric strain rate for a fluid element is: 
€ 
1
δV
D
Dt
δV( ) = 1
δx1δx2δx3
D
Dt
δx1δx2δx3( ) =
1
δx1
D
Dt
δx1( ) +
1
δx2
D
Dt
δx2( ) +
1
δx3
D
Dt
δx3( ). 
From Section 3.4 in the text, the linear strain rate corresponding to elongation or contraction of a 
fluid element in the first direction is: 
€ 
1
δx1
D
Dt
δx1( ) = S11 =
∂u1
∂x1
, 
and this can be immediately extended to the other two directions, 
€ 
1
δx2
D
Dt
δx2( ) = S22 =
∂u2
∂x2
 and 
€ 
1
δx3
D
Dt
δx3( ) = S33 =
∂u3
∂x3
, 
because its geometric derivation (see Figure 3.10) did not rely on any special properties of the 
first direction. Substitution of these relationships into the final version of the volumetric strain 
rate given above produces: 
€ 
1
δV
D
Dt
δV( ) = S11 + S22 + S33 =
∂u1
∂x1
+
∂u2
∂x2
+
∂u3
∂x3
=
∂ui
∂xi
= Sii. 
 
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling 
Exercise 3.21. Let Oxyz be a stationary frame of reference, and let the z-axis be parallel with the 
fluid vorticity vector in the vicinity of O so that 
€ 
ω =∇ ×u =ωzez in this frame of reference. Now 
consider a second rotating frame of reference 
€ 
O " x " y " z having the same origin that rotates about 
the z-axis at angular rate Ωez. Starting from the kinematic relationship, 
€ 
u = (Ωez) × x + $ u , show 
that in the vicinity of O the vorticity 
€ 
" ω = " ∇ × " u in the rotating frame of reference can only be 
zero when 2Ω = ωz, where 
€ 
" ∇ is the gradient operator in the primed coordinates. The following 
unit vector transformation rules may be of use: 
€ 
" e x = ex cos(Ωt) + ey sin(Ωt) , 
€ 
" e y = −ex sin(Ωt) + ey cos(Ωt) , and 
€ 
" e z = ez . 
 
Solution 3.21. The approach here is to compute 
€ 
" ω = " ∇ × " u in the stationary frame of reference 
and then determine the parameter choice(s) necessary for ω´ to be zero. The vector x must have a 
consistent representation in either frame, so 
€ 
x = xex + yey = " x " e x + " y " e y = " x ex cos(Ωt) + ey sin(Ωt)( ) + " y −ex sin(Ωt) + ey cos(Ωt)( ). 
Equating components in the stationary frame of reference produces: 
€ 
x = " x cos(Ωt) − " y sin(Ωt) , and 
€ 
y = " x sin(Ωt) + " y cos(Ωt) . 
The remaining independent variables are the same in either frame: 
€ 
t = " t , and 
€ 
z = " z . Thus, spatial 
derivatives are related by: 
 
€ 
∂
∂ # x 
=
∂x
∂ # x 
∂
∂x
+
∂y
∂ # x 
∂
∂y
+
∂z
∂ # x 
∂
∂z
+
∂t
∂ # x 
∂
∂t
= cos(Ωt) ∂
∂x
+ sin(Ωt) ∂
∂y
, 
 
€ 
∂
∂ # y 
=
∂x
∂ # y 
∂
∂x
+
∂y
∂ # y 
∂
∂y
+
∂z
∂ # y 
∂
∂z
+
∂t
∂ # y 
∂
∂t
= −sin(Ωt) ∂
∂x
+ cos(Ωt) ∂
∂y
, and 
 
€ 
∂
∂ # z 
=
∂x
∂ # z 
∂
∂x
+
∂y
∂ # z 
∂
∂y
+
∂z
∂ # z 
∂
∂z
+
∂t
∂ # z 
∂
∂t
=
∂
∂z
. 
Using the above information, the gradient operator in the rotating coordinates can be rewritten in 
terms of the stationary frame coordinates and unit vectors: 
€ 
" ∇ = " e x
∂
∂ " x 
+ " e y
∂
∂ " y 
+ " e z
∂
∂ " z 
 
 
€ 
= ex cos(Ωt) + ey sin(Ωt)( ) cos(Ωt)
∂
∂x
+ sin(Ωt) ∂
∂y
$ 
% 
& 
' 
( 
) 
 
€ 
+ −ex sin(Ωt) + ey cos(Ωt)( ) −sin(Ωt)
∂
∂x
+ cos(Ωt) ∂
∂y
% 
& 
' 
( 
) 
* + ez
∂
∂z
 
€ 
" ∇ = ex cos
2(Ωt) ∂
∂x
+ ex sin(Ωt)cos(Ωt)
∂
∂y
+ ey sin(Ωt)cos(Ωt)
∂
∂x
+ ey sin
2(Ωt) ∂
∂y
& 
' 
( 
) 
* 
+ 
 
€ 
+ ex sin
2(Ωt) ∂
∂x
− ex sin(Ωt)cos(Ωt)
∂
∂y
− ey sin(Ωt)cos(Ωt)
∂
∂x
+ ey cos
2(Ωt) ∂
∂y
% 
& 
' 
( 
) 
* + ez
∂
∂z
 
€ 
" ∇ = ex cos
2(Ωt) ∂
∂x
+ ex sin(Ωt)cos(Ωt)
∂
∂y
+ ey sin(Ωt)cos(Ωt)
∂
∂x
+ ey sin
2(Ωt) ∂
∂y
& 
' 
( 
) 
* 
+ 
 
€ 
+ ex sin
2(Ωt) ∂
∂x
− ex sin(Ωt)cos(Ωt)
∂
∂y
− ey sin(Ωt)cos(Ωt)
∂
∂x
+ ey cos
2(Ωt) ∂
∂y
% 
& 
' 
( 
) 
* + ez
∂
∂z
 
€ 
" ∇ = ex
∂
∂x
+ ey
∂
∂y
+ ez
∂
∂z
. 
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling 
This result seems too simple, but should not be a surprising to a routine user of vector calculus. 
 Now resolve rotating-frame velocity components in the stationary frame of reference: 
€ 
" u = u− (Ωez) × x = uex + vey + wez +Ωyex −Ωxey . 
Using the stationary frame components and unit vectors, the vorticity in the rotating frame is: 
 
€ 
" ω = " ∇ × " u =
ex ey ez
∂ ∂x ∂ ∂y ∂ ∂z
u +Ωy v −Ωx w
=
ex ey ez
∂ ∂x ∂ ∂y ∂ ∂z
u v w
+
ex ey ez
∂ ∂x ∂ ∂y ∂ ∂z
Ωy −Ωx 0
 
 
€ 
=∇ ×u+
ex ey ez
∂ ∂x ∂ ∂y ∂ ∂z
Ωy −Ωx 0
=ωzez + ex (0) + ey (0) + ez −Ω−Ω( ) = ez ωz − 2Ω( ) 
Thus, ω´ will be zero when ωz = 2Ω. 
 
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling 
Exercise 3.22. Consider a plane-polar area element having dimensions dr and rdθ. For two-
dimensional flow in this plane, evaluate the right-hand side of Stokes’ theorem 
€ 
ω ⋅ndA∫ = u ⋅ ds∫ and thereby show that the expression for vorticity in plane-polar coordinates 
is: 
€ 
ωz =
1
r
∂
∂r
ruθ( ) −
1
r
∂ur
∂θ
. 
 
Solution 3.22. Using the element shown with angular width dθ, 
 
 
 
application of Stokes' theorem provides: 
€ 
ω ⋅ndA∫ = u ⋅ ds∫ →ωzrdθdr = urdr + uθ +
∂uθ
∂r
dr
( 
) 
* 
+ 
, 
- r + dr( )dθ − ur +
∂ur
∂θ
dθ
( 
) 
* 
+ 
, 
- dr − uθ rdθ , or 
€ 
ωzrdθdr = urdr + ruθ dθ + r
∂uθ
∂r
drdθ + uθ drdθ − urdr −
∂ur
∂θ
dθdr − uθ rdθ + ... 
 
€ 
= r ∂uθ
∂r
drdθ + uθ drdθ −
∂ur
∂θ
dθdr + ... 
where + ... indicates the presence of higher order terms in dr and dθ. Division by rdθdr and 
passing to the limit where dr and dθ go to zero produces: 
€ 
ωz =
∂uθ
∂r
+
1
r
uθ −
1
r
∂ur
∂θ
=
1
r
∂
∂θ
ruθ( ) −
∂ur
∂θ
& 
' 
( 
) 
* 
+ . 
 
ur
u!
u! + (!u!/!")dr
ur + (!ur/!!)d!
dr
!
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling 
Exercise 3.23. The velocity field of a certain flow is given by 
€ 
u = 2xy 2 + 2xz2 , 
€ 
v = x 2y , and 
€ 
w = x 2z. Consider the fluid region inside a spherical volume x2 + y2 + z2 = a2. Verify the validity 
of Gauss’ theorem 
€ 
∇ ⋅udV = u ⋅ndA
A
∫∫
V
∫∫∫ by integrating over thesphere. 
 
Solution 3.23. First compute the divergence of the velocity field. 
€ 
∇ ⋅u = ∂u
∂x
+
∂v
∂y
+
∂w
∂z
=
∂
∂x
2xy 2 + 2xz2( ) + ∂
∂y
x 2y( ) + ∂
∂z
x 2z( ) = 2y 2 + 2z2 + 2x 2 = 2r2 . 
The volume integral of 
€ 
∇ ⋅u is: 
€ 
∇ ⋅udV = 2r2( )4πr2dr =
r= 0
r= a
∫
V
∫∫∫ 85
πa5 . 
Now work on the surface integration using spherical coordinates (see Figure 3.3d, and Appendix 
B). Here, 
€ 
n = er = ex cosϕ sinθ + ey sinϕ sinθ + ex cosθ , so 
 
€ 
u ⋅n = ucosϕ sinθ + v sinϕ sinθ + wcosθ 
 
€ 
= 2xy 2 + 2xz2( )cosϕ sinθ + x 2y( )sinϕ sinθ + x 2z( )cosθ 
Unfortunately, this result is in mixed variables so convert everything to spherical polar 
coordinates using 
€ 
x = rcosϕ sinθ , 
€ 
y = rsinϕ sinθ , 
€ 
z = rcosθ . 
This conversion produces: 
€ 
u ⋅n = 2r3 cosϕ sin2ϕ sin3θ + cosϕ sinθ cos2θ( )cosϕ sinθ 
 
€ 
+r3 cos2ϕ sinϕ sin3θ( )sinϕ sinθ + r3 cos2ϕ sin2θ cosθ( )cosθ 
 
€ 
= 2r3 sin2ϕ sin2θ + cos2θ( )cos2ϕ sin2θ + r3 sin2ϕ sin2θ( )cos2ϕ sin2θ + r3 cos2θ( )cos2ϕ sin2θ 
 
€ 
= r3 3sin2ϕ sin2θ + 3cos2θ( )cos2ϕ sin2θ = r3 34 sin2(2ϕ) 1− cos2θ( )
2
+ 3cos2ϕ cos2θ 1− cos2θ( )( ) 
So, the surface integral produces: 
€ 
u ⋅ndA
A
∫∫ = a3 34 sin
2(2ϕ) 1− cos2θ( )
2
+ 3cos2ϕ cos2θ 1− cos2θ( )( )
θ = 0
θ = π
∫
ϕ= 0
ϕ= 2π
∫ a2 sinθdθdϕ 
 
€ 
= a5 34 sin
2(2ϕ)dϕ 1− cos2θ( )
2
θ = 0
θ = π
∫
ϕ= 0
ϕ= 2π
∫ sinθdθ + a5 3cos2ϕdϕ cos2θ 1− cos2θ( )
θ = 0
θ = π
∫
ϕ= 0
ϕ= 2π
∫ sinθdθ 
 
€ 
= a5 3
4
π
# 
$ 
% 
& 
' 
( 1−β 2( )
2
−1
+1
∫ dβ + a5 3π( ) β 2 1−β 2( )
−1
+1
∫ dβ 
 
€ 
= a5 3
4
π
# 
$ 
% 
& 
' 
( 
16
15
+ a5 3π( ) 4
15
=
12
15
+
12
15
# 
$ 
% 
& 
' 
( πa5 =
8
5
πa5, 
which is the same as the result of the volume integration. 
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling 
Exercise 3.24. A flow field on the xy-plane has the velocity components u = 3x + y and v = 2x – 
3y. Show that the circulation around the circle (x − 1)2 + (y − 6)2 = 4 is 4π. 
 
Solution 3.24. The circle is centered at (1,6) and its radius is 2. So, if (x,y) is a point on the circle 
then x = 1 + 2cosθ, and y = 6 + 2sinθ, where θ is the angle from the horizontal. The velocity 
component tangent to the circle will be 
€ 
uθ = eθ ⋅u = −sinθ,cosθ( ) ⋅ (u,v) = −usinθ + v cosθ . 
Evaluate this velocity component. 
€ 
uθ = −(3x + y)sinθ + (2x − 3y)cosθ = (−3sinθ + 2cosθ)x − (sinθ + 3cosθ)y 
 
€ 
= (−3sinθ + 2cosθ)(1+ 2cosθ) − (sinθ + 3cosθ)(6 + 2sinθ)
= −3sinθ − 6sinθ cosθ + 2cosθ + 4cos2θ − 6sinθ − 2sin2θ −18cosθ − 6cosθ sinθ
= −9sinθ −12sinθ cosθ −16cosθ + 4cos2θ − 2sin2θ
 
Now compute the circulation: 
€ 
Γ = uθ
θ = 0
2π
∫ rdθ = −9sinθ −12sinθ cosθ −16cosθ + 4cos2θ − 2sin2θ( )
θ = 0
2π
∫ 2dθ 
 
€ 
= 0 + 0 + 0 + 4π − 2π( )2 = 4π 
 
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling 
Exercise 3.25. Consider solid-body rotation about the origin in two dimensions: ur = 0 and uθ = 
ω0r. Use a polar-coordinate element of dimension rdθ and dr, and verify that the circulation is 
vorticity times area. (In Section 5 this was verified for a circular element surrounding the origin.) 
 
Solution 3.25. Using the element shown with angular width dθ, 
 
 
 
the circulation is around the element is the sum or four terms: 
€ 
Γ = u ⋅ ds∫ = urdr + uθ +
∂uθ
∂r
dr
' 
( 
) 
* 
+ 
, r + dr( )dθ − ur +
∂ur
∂θ
dθ
' 
( 
) 
* 
+ 
, dr − uθ rdθ . 
Now substitute in the velocity field: ur = 0 and uθ = ω0r. 
€ 
Γ = ω0r +ω0dr( ) r + dr( )dθ −ω0r2dθ = 2ω0rdrdθ +ω0(dr)2dθ = (2ω0)rdrdθ , 
where the final equality holds in the limit as the differential elements become small and (dr)2dθ 
is negligible compared to rdrdθ. Thus, since the voriticity is 2ω0 and rdrdθ is the area of the 
element, the final relationship states: circulation = (vorticity) x (area). 
 
ur
u!
u! + (!u!/!")dr
ur + (!ur/!!)d!
dr
!
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling 
Exercise 3.26. Consider the steady Cartesian velocity field 
€ 
u = −Ay
x 2 + y 2( )β
, +Ax
x 2 + y 2( )β
,0
$ 
% 
& 
& 
' 
( 
) 
) 
. 
a) Determine the streamline that passes through 
€ 
x = (xo,yo,0) 
b) Compute Rij for this velocity field. 
c) For A > 0, explain the sense of rotation (i.e. clockwise or counter clockwise) for fluid elements 
for β < 1, β = 1, and β > 1. 
 
Solution 3.26. a) Use the definition of a streamline: 
€ 
dy
dx
=
v
u
=
Ax x 2 + y 2( )
β
−Ay x 2 + y 2( )β
= −
x
y
. Use the 
two ends of this extended equality to find: 
€ 
ydy = −xdx , and integrate the resulting differential 
relationship to get: y2 2 = −x2 2+ const . Evaluate the constant using the required condition: 
€ 
x 2 + y 2 = xo
2 + yo
2 . This is a circle with radius 
€ 
xo
2 + yo
2 . Therefore the streamlines are circles. 
b) The rotation tensor is: Rij =
∂ui
∂x j
−
∂uj
∂xi
=
0 ∂u ∂y−∂v ∂x 0
∂v ∂x −∂u ∂y 0 0
0 0 0
"
#
$
%
$
$
&
'
$
(
$
$
, where the 
second equality comes from putting the specified velocity field into the definition of Rij with 
u = (u,v,w) = (u1,u2,u3) = ui as usual. Evaluating the derivatives produces: 
Rij =
A
x2 + y2( )
β+1
0 −x2 − y2 + 2βy2 − x2 + y2 − 2βx2( ) 0
x2 + y2 − 2βx2 − −x2 − y2 + 2βy2( ) 0 0
0 0 0
"
#
$
$
%
$
$
&
'
$
$
(
$
$
 
 = 2A(1−β)
x2 + y2( )
β
0 −1 0
1 0 0
0 0 0
"
#
$
%
$
&
'
$
(
$
 
 
c) The following answers are based on A > 0. For β < 1, fluid particles rotate counter clockwise 
(positive ωz). For β = 1, fluid particles do not rotate. For β > 1, fluid particles rotate clockwise 
(negative ωz). Interestingly, the streamlines are the same (circular!) for all three possibilities. 
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling 
Exercise 3.27. Using indicial notation (and no vector identities) show that the acceleration a of a 
fluid particle is given by: 
€ 
a = ∂u ∂t +∇ 12 u
2( ) +ω ×u where ω is the vorticity. 
 
Solution 3.27. The acceleration of a fluid particle is a = Du Dt ≡ ∂u ∂t + u ⋅∇( )u . Thus, the task 
is to prove 
€ 
u ⋅ ∇( )u =∇ 12 u
2( ) +ω ×u. Start from the advective acceleration written in index 
notation and force the rotation tensor to appear: 
€ 
u j
∂ui
∂x j
= u j
∂ui
∂x j
−
∂u j
∂xi
$ 
% 
& & 
' 
( 
) ) − u j
∂u j
∂xi
= u jRij +
1
2
∂
∂xi
u j
2( ) . 
From (3.15), 
€ 
Rij = −εijkωk , so this becomes 
€ 
u j
∂ui
∂x j
= −εijku jωk +
1
2
∂
∂xi
u j
2( ) = εikjωku j + 12
∂
∂xi
u j
2( ) =ω ×u+∇ 12 u
2( 
) 
* 
+ 
, 
- , 
where the final equality follows from the index notation definitions of the cross product (2.21), 
gradient (2.22), and vector magnitude (
€ 
u j
2 = u 2 ). 
 
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling 
Exercise 3.28. Starting from (3.29), show that the maximum uθ in a Gaussian vortex occurs 
when 
€ 
1+ 2(r2 σ 2) = exp(r2 σ 2) . Verify that this implies r ≈ 1.12091σ. 
 
Solution 3.28. Differentiate the uθ equation from (3.29) with respect to r and set this derivative 
equal to zero. 
€ 
d
dr
uθ (r)( ) =
Γ
2π
d
dr
1− exp −r2 σ 2( )
r
' 
( 
) 
) 
* 
+ 
, 
, =
Γ
2π
−
1− exp −r2 σ 2( )
r2
−
exp −r2 σ 2( )
r
−
2r
σ 2
' 
( 
) 
* 
+ 
, 
' 
( 
) 
) 
* 
+ 
, 
, = 0. 
Eliminate common factors assuming r ≠ 0. 
€ 
0 = −1+ exp −r2 σ 2( ) + 2r2 σ 2( )exp −r2 σ 2( ) = −1+ 1+ 2r2 σ 2( )exp −r2 σ 2( ). 
This can be rearranged to: 
€ 
exp r2 σ 2( ) =1+ 2r2 σ 2 , 
which is the desired result. When r/σ ≈ 1.12091, then 
€ 
exp r2 σ 2( ) = 3.51289 and 
€1+ 2r2 σ 2 = 3.51288 , 
which is suitable numerical agreement. 
 
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling 
Exercise 3.29. Using (3.35) in two dimensions with F = 1, show that the time-rate-of-change of 
the area of the parallelogram shown is hl(dθ/dt)cosθ when θ depends on time while h and l are 
constants. 
 
 
Solution 3.29. With F = 1, the volume integral of ∂F/∂t on the right side of (3.35) is zero, and 
the integrands are simplified, so (3.35) simplifies to: 
d
dt
dV
V*(t )∫ = + b ⋅ndAA*(t )∫ . 
(Here the volume integration is two dimensional, and produces the parallelogram's area, hlsinθ, 
which can be time differentiated, d(hlsinθ)/dt, to reach the desired result. However, this is not the 
intended solution path for this problem.) 
 When θ is time-dependent, the parallelogram has three moving sides (a left side of length 
h, a right side of length h, and a top side of length l). Thus, the simplified version of (3.35) 
reduces to: 
d
dt
dV
V*(t )∫ = +left side∫ +right side∫ top side∫{ }b ⋅ndA . 
Here we note that the left and right sides move identically, so b will be the same. However, n 
points in opposite directions on these two sides, so the contributions from these two sides cancel. 
The x-coordinates of points on the parallelogram's top side are hcosθ ≤ x(t) ≤ l + hcosθ. The y-
coordinate of the parallelogram's top side is y(t) = hsinθ. Time differentiate the location of any 
point on the parallelogram's top side to find b: 
b = (dx/dt, dy/dt) = (–hsinθ, hcosθ)(dθ/dt). 
And, on the parallelogram's top side, n is ey, so 
b ⋅n = hcosθ( ) dθ dt( ) , 
and in two dimensions dA is just dx. Thus, the simplified version of (3.35) is: 
d
dt
dV
V*(t )∫ = b ⋅ntop side∫ dA = hcosθ
hcosθ
l+hcosθ
∫ dθdt dx = hl cosθ
dθ
dt
, 
and this is the desired result. 
 
x!
y!
l!
h!θ(t)!
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling 
Exercise 3.30. Using (3.35) in two dimensions with F = 1, show that the time-rate-of-change of 
the area of the triangle shown is 12 b dh dt( ) when h depends on time and b is constant. 
 
 
Solution 3.30. With F = 1, the volume integral of ∂F/∂t on the right side of (3.35) is zero, and 
the integrands are simplified, so (3.35) simplifies to: 
d
dt
dV
V*(t )∫ = + b ⋅ndAA*(t )∫ . 
(Here the volume integration is two dimensional, and produces the triangle's area, hb/2, which 
can be time differentiated, (d/dt)(hb/2) to reach the desired result. However, this is not the 
intended solution path for this problem.) 
 When h is time-dependent, the triangle has two moving sides (a top side of length b, and 
a hypotenuse of length [h2 + b2]1/2). Thus, the simplified version of (3.35) reduces to: 
d
dt
dV
V*(t )∫ = +top side∫ hypotenuse∫{ }b ⋅ndA . 
The x-coordinates of the triangle's top side are 0 ≤ x ≤ b. The y-coordinate of the triangle's top 
side is y(t) = h(t). Time differentiate the location of any point on the triangles's top side to find b: 
b = (dx/dt, dy/dt) = (0, dh/dt). 
And, on the triangle's top side, n is ey, so 
b ⋅n = dh dt , 
 The x-y coordinates of the triangle's hypotenuse fall on the line y = (h/b)x for 0 ≤ x ≤ b. 
The motion of points on the hypotenuse can be described by a purely vertical velocity. So, for 
any constant x-location, differentiate the equation of this line to find b: 
b = (dx/dt, dy/dt) = (0, (x/b)dh/dt). 
Tangent and normal vectors to an x-y curve are (1, dy/dx) and (dy/dx, –1), respectively. Thus, the 
outward unit normal on the hypotenuse of the triangle is: 
n =
h b,−1( )
(h / b)2 +1
=
(h,−b)
h2 + b2
, so b ⋅n = −
x dh dt( )
h2 + b2
. 
 Thus, the simplified version of (3.35) can be written: 
d
dt
dV
V*(t )∫ = b ⋅n
top side
∫ dA+ b ⋅n
hypotenuse
∫ dA = dhdt0
b
∫ dx −
x dh dt( )
h2 + b20
b
∫ 1+ (h / b)2dx$%
&
' , 
where, in two dimensions, dA is just dx on the top side and dA is a path length element along the 
hypotenuse, ds = [1 + (dy/dx)2]1/2dx. The factor in [,]-brackets is this path length element in terms 
of dx. Perform the integrations to find: 
d
dt
dV
V*(t )∫ = b
dh
dt
−
1
b
dh
dt
x2
2
#
$
%
&
'
(
0
b
=
b
2
dh
dt
 , 
and this is the desired result. 
x!
y!
b!
h(t)!
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling 
Exercise 3.31. Using (3.35) in two dimensions with F = 1, show that the time-rate-of-change of 
the area of the ellipse shown is πb da dt( ) when a depends on time and b is constant. 
 
 
 
Solution 3.31. With F = 1, the volume integral of ∂F/∂t on the right side of (3.35) is zero, and 
the integrands are simplified, so (3.35) simplifies to: 
d
dt
dV
V*(t )∫ = + b ⋅ndAA*(t )∫ . 
(Here the volume integration is two dimensional, and produces the ellipse's area, πab, which can 
be time differentiated, (d/dt)( πab) to reach the desired result. However, this is not the intended 
solution path for this problem.) 
 When a is time-dependent, the contour that defines the ellipse, (y/b)2 + (x/a)2 = 1, is also 
time dependent. However, the symmetry of the ellipse allows the analysis to completed in the 
first quadrant alone and then multiplied by 4. Thus, the simplified version of (3.35) reduces to: 
d
dt
dV
V*(t )∫ = 4 b ⋅n
first quadrant
∫ dA . 
The equation for the ellipse in the first quadrant is: 
x = +a 1− y b( )2 or y = +b 1− x a( )2 . 
The motion of points on this curve can be described by a purely horizontal velocity when a 
varies but b is constant. So, for any constant y-location, differentiate the first equation to find b: 
b = dx
dt
, 0
!
"
#
$
%
&=
da
dt
1− y b( )2 , 0
!
"
#
$
%
&=
x
a
da
dt
, 0
!
"
#
$
%
& , 
where the final equality comes from changing the independent variable from y to x. 
 Tangent and normal vectors to an x-y curve are (1, dy/dx) and (–dy/dx, 1), respectively. 
Thus, the outward unit normal on the ellipse in the first quadrant is: 
n =
−dy / dx,1( )
(dy / dx)2 +1
, so b ⋅n = x(−dy / dx)
a (dy / dx)2 +1
da
dt
. 
 Thus, the simplified version of (3.35) can be written: 
d
dt
dV
V*(t )∫ = 4 b ⋅n
first quadrant
∫ dA = 4 x(−dy / dx)
a (dy / dx)2 +1
da
dt0
a
∫ 1+ (dy / dx)2dx$%
&
' , 
where, in two dimensions, dA is a path length element, ds = [1 + (dy/dx)2]1/2dx, along the first-
quadrant portion of the ellipse. This path length element is the factor in [,]-brackets above. 
x!
y!
b!
a(t)!
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling 
Simplify the integrand, insert –dy/dx = (bx/a2)/[1 – (x/a)2]1/2, and perform the integration using 
the substitution x = asinθ to find: 
d
dt
dV
V*(t )∫ = 4
x(−dy / dx)
a
da
dt0
a
∫ dx = 4a
da
dt
xb(x / a2 )
1− (x / a)20
a
∫ dx
 = 4b da
dt
sin2θ
1− sin2θ0
π 2
∫ cosθdθ = 4b dadt sin
2θ
0
π 2
∫ dθ = 4b dadt
π
4
= πb da
dt
,
 
and this is the desired result. 
 
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling 
Exercise 3.32. For the following time-dependent volumes V*(t) and smooth single-valued 
integrand functions F, choose an appropriate coordinate system and show that 
€ 
d dt( ) FdVV *( t )∫ 
obtained from (3.30) is equal to that obtained from (3.35). 
a) V*(t) = L1(t)L2L3 is a rectangular solid defined by 0 ≤ xi ≤ Li, where L1 depends on time while 
L2 and L3 are constants, and the integrand function F(x1,t) depends only on the first coordinate 
and time. 
b) V*(t) = (π/4)d2(t)L is a cylinder defined by 0 ≤ R ≤ d(t)/2 and 0 ≤ z ≤ L, where the cylinder’s 
diameter d depends on time while its length L is constant, and the integrand function F(R,t) 
depends only on the distancefrom the cylinder’s axis and time. 
c) V*(t) = (π/6)D3(t) is a sphere defined by 0 ≤ r ≤ D(t)/2 where the sphere’s diameter D depends 
on time, and the integrand function F(r,t) depends only on the radial distance from the center of 
the sphere and time. 
 
Solution 3.32. The two equations are (3.30) 
€ 
d
dt
F(x, t)dx
x=a( t )
x=b( t )
∫ = ∂F
∂t
dx
a
b
∫ + dbdt
F b,t( ) − da
dt
F a,t( ), 
and (3.35) 
€ 
d
dt
F(x,t)dV
V *(t )
∫ = ∂F(x,t)
∂t
dV
V *(t )
∫ + F(x,t)b ⋅ndA
A*(t )
∫ . 
a) Use Cartesian coordinates with the origin at xi = 0. The cross sectional area of the rectangular 
solid, L2L3, is constant, so dV = L2L3dx1. The volume integral proceeds from x1 = 0 (= a) to x1 = 
L1(t) (= b) so (3.30) implies: 
€ 
d
dt
F(x,t)dV
V *(t )
∫ = ddt
F(x1,t)L2L3dx1
x1 =0
x=L1 ( t )
∫ = ∂F
∂t
L2L3dx1
0
L1
∫ + dL1dt
F L1,t( )L2L3 . (a1) 
Now start from (3.35) using a control volume that encloses the rectangular solid. In this case the 
only control surface that moves lies at x1 = L1, has outward normal n = e1, and moves with 
velocity b = (dL1/dt)ex. First evaluate the left side term from (3.35). 
€ 
d
dt
F(x,t)dV
V *(t )
∫ = ddt
F(x1,t)
x3= 0
L3
∫
x2= 0
L2
∫ dx1dx2dx3
x1= 0
L1 ( t )
∫ = L2L3
d
dt
F(x1,t)dx1
x1= 0
L1 ( t )
∫ . 
Now evaluate the right side terms from (3.35). 
€ 
d
dt
F(x,t)dV
V *(t )
∫ = ∂F(x1,t)
∂tx3 = 0
L3
∫
x2 = 0
L2
∫ dx1dx2dx3
x1 = 0
L1 ( t )
∫ + F(L1,t)
x3 = 0
L3
∫
x2 = 0
L2
∫ dL1dt
e1
$ 
% 
& 
' 
( 
) ⋅ e1dx2dx3
 = L2L3
∂F(x1,t)
∂t
dx1
x1 = 0
L1 (t )
∫ + L2L3
dL1
dt
F(L1,t)
 
Setting the left and right side terms equal produces 
€ 
L2L3
d
dt
F(x1,t)dx1
x1= 0
L1 ( t )
∫ = L2L3
∂F(x1,t)
∂t
dx1
x1= 0
L1 ( t )
∫ + L2L3
dL1
dt
F(L1,t) , (a2) 
which is identical to (a1). 
b) Use cylindrical coordinates with the origin at R = z = 0. The length area of the cylinder, L, is 
constant, so dV = L(2πRdR). The volume integral proceeds from R = 0 (= a) to R = d(t)/2 (= b) so 
(3.30) implies: 
€ 
d
dt
F(x,t)dV
V *(t )
∫ = 2πL ddt
F(R,t)RdR
R =0
d ( t ) / 2
∫ = 2πL ∂F
∂t
RdR
0
d / 2
∫ + 2πL
d d 2( )
dt
F d /2,t( ) d
2
. (b1) 
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling 
Now start from (3.35) using a control volume that encloses the cylinder. In this case the only 
control surface that moves lies at R = d/2, has outward normal n = eR, and moves with velocity b 
= (d(d/2)/dt)eR. First evaluate the left side term from (3.35). 
€ 
d
dt
F(x,t)dV
V *(t )
∫ = ddt
F(R,t)
ϕ= 0
2π
∫
R= 0
d / 2
∫ RdϕdRdz
z= 0
L
∫ = 2πL ddt
F(R,t)RdR
0
d / 2
∫ . 
Now evaluate the right side terms from (3.35). 
€ 
d
dt
F(x,t)dV
V *(t )
∫ = ∂F(R,t)
∂tϕ= 0
2π
∫
R= 0
d / 2
∫ RdϕdRdz
z= 0
L
∫ + F(d/2,t)
ϕ= 0
2π
∫
z= 0
L
∫ d(d /2)dt
eR
& 
' 
( 
) 
* 
+ ⋅ eR
d
2
dϕdz
 = 2πL ∂F(R,t)
∂t
RdR
R= 0
d / 2
∫ + 2πL d(d /2)dt
F(d/2,t) d
2
 
Setting the left and right side terms equal produces 
€ 
2πL d
dt
F(R,t)RdR
0
d / 2
∫ = 2πL ∂F(R,t)
∂t
RdR
R= 0
d / 2
∫ + 2πL d(d /2)dt
F(d/2,t) d
2
 , (b2) 
which is identical to (b1). 
c) Use spherical coordinates with the origin at r = 0. The sphere expands symmetrically so the 
volume element is dV = 4πr2dr. The volume integral proceeds from r = 0 (= a) to r = D(t)/2 (= b) 
so (3.30) implies: 
€ 
d
dt
F(x,t)dV
V *(t )
∫ = 4π ddt
F(r,t)r2dr
r=0
D(t ) / 2
∫ = 4π ∂F
∂t
r2dr
0
D / 2
∫ + 4π
d D 2( )
dt
F D /2,t( ) D
2
% 
& 
' 
( 
) 
* 
2
. (c1) 
Now start from (3.35) using a control volume that encloses the sphere. In this case the moving 
control surface lies at r = D/2, has outward normal n = er, and moves with velocity b = 
(d(D/2)/dt)er. First evaluate the left side term from (3.35). 
€ 
d
dt
F(x,t)dV
V *(t )
∫ = ddt
F(R,t)
ϕ= 0
2π
∫
θ = 0
π
∫ r2drsinθdθdϕ
r= 0
D / 2
∫ = 4π ddt
F(r,t)r2dr
0
D / 2
∫ . 
Now evaluate the right side terms from (3.35). 
€ 
d
dt
F(x,t)dV
V *(t )
∫ = ∂F(r,t)
∂tϕ= 0
2π
∫
θ = 0
π
∫ r2drsinθdθdϕ
r= 0
D / 2
∫ + F(d/2,t)
ϕ= 0
2π
∫
θ = 0
π
∫ d(D /2)dt
eR
' 
( 
) 
* 
+ 
, ⋅ eR
D
2
' 
( 
) 
* 
+ 
, 
2
sinθdθdϕ
 = 4π ∂F(r,t)
∂t
r2dr
r= 0
D / 2
∫ + 4π d(D /2)dt
F(D/2,t) D
2
' 
( 
) 
* 
+ 
, 
2
 
Setting the left and right side terms equal produces 
€ 
4π d
dt
F(r,t)r2dr
0
D / 2
∫ = 4π ∂F(r,t)
∂t
r2dr
r= 0
D / 2
∫ + 4π d(D /2)dt
F(D/2,t) D
2
% 
& 
' 
( 
) 
* 
2
, (c2) 
which is identical to (c1). 
 
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling 
Exercise 3.33. Starting from (3.35), set F = 1 and derive (3.14) when b = u and V*(t) = δV
€ 
→ 0. 
 
Solution 3.33. With F = 1, b = u, and V*(t) = δV with surface δA, (3.35) becomes: 
€ 
d
dt
dV
δV
∫ = 0 + u ⋅ndA
δA
∫ . 
The first integral is merely δV. Use Gauss' divergence theorem on the second term to convert it 
to volume integral. 
€ 
d
dt
δV( ) = ∇ ⋅udV
δV
∫ . 
As δV
€ 
→ 0 the integral reduces to a product of δV and the integrand evaluated at the center point 
of δV. Divide both sides of the last equation by δV and take the limit as δV
€ 
→ 0: 
€ 
lim
δV→0
1
δV
d
dt
δV( ) = lim
δV→0
1
δV
∇ ⋅udV
δV
∫ = lim
δV→0
1
δV
∇ ⋅u( )δV + ...[ ] =∇ ⋅u = Sii, 
and this is (3.14). 
 
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling 
Exercise 3.34. For a smooth single valued function F(x) that only depends on space and an 
arbitrarily-shaped control volume that moves with velocity b(t) that only depends on time, show 
that 
€ 
d dt( ) F(x)dVV *( t )∫ = b ⋅ ∇F(x)dVV *( t )∫( ) . 
 
Solution 3.34. Start from Reynolds transport theorem (3.35): 
€ 
d
dt
F(x,t)dV
V *(t )
∫ = ∂F(x,t)
∂t
dV
V *(t )
∫ + F(x,t)b ⋅ndA
A*(t )
∫ . 
When F does not depend on time, the first term on the right drops out. 
€ 
d
dt
F(x,t)dV
V *(t )
∫ = F(x,t)b ⋅ndA
A*(t )
∫ . 
When b does not depend on location, it can be taken outside the surface integral. 
€ 
d
dt
F(x,t)dV
V *(t )
∫ = b ⋅ F(x,t)ndA
A*( t )
∫
$ 
% 
& 
' 
( 
) . 
Apply Gauss' theorem to the integral in large parentheses, to reach the desired form: 
€ 
d
dt
F(x,t)dV
V *(t )
∫ = b ⋅ ∇F(x,t)dA
V *( t )
∫
% 
& 
' 
( 
) 
* . 
 
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling 
3.35. Show that (3.35) reduces to (3.5) when V*(t) = δV
€ 
→ 0 and the control surface velocity b is 
equal to the fluid velocity u(x,t). 
 
Solution 3.35. When V*(t) = δV with surface δA, δV is small, and b = u, δV represents a fluid 
particle. Under these conditions (3.35) becomes: 
€ 
d
dt
F(x,t)dV
δV
∫ = ∂F(x,t)
∂t
dV
δV
∫ + F(x,t)u ⋅ndA
δA
∫ , 
and the time derivative is evaluated following δV. Use Gauss' divergence theorem on the final 
term to convert it to a volume integral, 
€ 
F(x,t)u ⋅ndA
δA
∫ = ∇ ⋅ F(x,t)u( )
δV
∫ dV , 
so that (3.35) becomes: 
€ 
d
dt
F(x,t)dV
δV
∫ = ∂F(x,t)
∂t
+∇ ⋅ F(x,t)u( )
' 
( ) 
* 
+ , 
dV
δV
∫ = ∂F(x,t)
∂t
+ F(x,t)∇ ⋅u+ u ⋅ ∇( )F(x,t)
' 
( ) 
* 
+ , 
dV
δV
∫ , 
where the second equality follows from expanding the divergence of the product Fu. 
 As δV
€ 
→ 0 the various integrals reduce to a product of δV and the integrand evaluated at 
the center point of δV. Divide both sides of the prior equation by δV and take the limit as δV
€ 
→ 0
to find: 
€ 
lim
δV→0
1
δV
d
dt
F(x,t)dV
δV
∫ = lim
δV→0
1
δV
∂F(x,t)
∂t
+ F(x,t)∇ ⋅u+ u ⋅ ∇( )F(x,t)
( 
) * 
+ 
, - 
dV
δV
∫ , 
€ 
lim
δV→0
1
δV
d
dt
F(x,t)δV + ...[ ] = lim
δV→0
1
δV
∂F(x,t)
∂t
+ F(x,t)∇ ⋅u+ u ⋅ ∇( )F(x,t)
' 
( 
) 
* 
+ 
, δV + ...
- 
. 
/ 
0 
1 
2 , or 
€ 
d
dtF(x,t) + F(x,t) lim
δV→0
1
δV
d
dt
δV( ) = ∂F(x,t)
∂t
+ F(x,t)∇ ⋅u+ u ⋅ ∇( )F(x,t) , 
where the product rule for derivative has been used on product FδV in [,]-braces on the left. 
From (3.14) or Exercise 3.33: 
€ 
lim
δV→0
1
δV
d
dt
δV( ) =∇ ⋅u , so the second terms on both sides of the 
last equation are equal and may be subtracted out leaving: 
€ 
d
dt
F(x,t) = ∂F(x,t)
∂t
+ u ⋅ ∇( )F(x,t) , 
and this is (3.5) when the identification 
€ 
D Dt ≡ d dt is made.