Respostas do livro de Cálculo 3

Prévia do material em texto

f (x,y)=xy A=4
V !
3
i=1
!
2
j=1
f x
i
,y
j( ) A
= f 2,2( ) A+ f 2,4( ) A+ f 4,2( ) A+ f 4,4( ) A+ f 6,2( ) A+ f 6,4( ) A
=4 4( )+8 4( )+8 4( )+16 4( )+12 4( )+24 4( )=288
V !
3
i=1
!
2
j=1
f x
i
,y
j( ) A
= f 1,1( ) A+ f 1,3( ) A+ f 3,1( ) A+ f 3,3( ) A+ f 5,1( ) A+ f 5,3( ) A
=1 4( )+3 4( )+3 4( )+9 4( )+5 4( )+15 4( )=144
 A=1
""
R
y
2
#2x
2( )dA !4
i=1
!
2
j=1
f x
*
ij
,y
*
ij
 A
= f #1,1( ) A+ f #1,2( ) A+ f 0,1( ) A+ f 0,2( ) A
+ f 1,1( ) A+ f 1,2( ) A+ f 2,1( ) A+ f 2,2( ) A
=#1 1( )+2 1( )+1 1( )+4 1( ) #1 1( )+2 1( )#7 1( ) #4 1( )=#4
 A=!
2
/4
""
R
sin (x+y)dA !
2
i=1
!
2
j=1
f x
*
ij
,y
*
ij
 A
= f (0,0) A+ f 0,
!
2
 A+ f
!
2
,0 A+ f
!
2
,
!
2
 A
1. (a) The subrectangles are shown in the figure.
The surface is the graph of and ,
so we estimate
(b)
2. The subrectangles are shown in the figure.
Since , we estimate
3. (a) The subrectangles are shown in the figure. Since , we estimate
 1
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.1 Double Integrals over Rectangles
=0
!
2
4
+1
!
2
4
+1
!
2
4
+0
!
2
4
=
!
2
2
 4.935
""
R
sin (x+y)dA !
2
i=1
!
2
j=1
f(x
i
,y
j
) A
=f
!
4
,
!
4
 A+f
!
4
,
3!
4
 A
+f
3!
4
,
!
4
 A+f
3!
4
,
3!
4
 A
=1
!
2
4
+0
!
2
4
+0
!
2
4
+(#1)
!
2
4
=0
f (x,y)=x+2y
2
 A=2
V =""
R
(x+2y
2
)dA !
2
i=1
!
2
j=1
f x
*
ij
,y
*
ij
 A
= f (1,0) A+ f (1,2) A+ f (2,0) A+ f (2,2) A
=1(2)+9(2)+2(2)+10(2)=44
(b)
4. (a) The subrectangles are shown in the figure.
The surface is the graph of and , so we estimate
 2
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.1 Double Integrals over Rectangles
V =""
R
(x+2y
2
)dA !
2
i=1
!
2
j=1
f (x
i
,y
j
) A
= f
1
2
,1 A+ f
1
2
,3 A+ f
3
2
,1 A+ f
3
2
,3 A
=
5
2
(2)+
37
2
(2)+
7
2
(2)+
39
2
(2)=88
 A=2 f
""
R
f (x,y)dA !
2
i=1
!
2
j=1
f x
i
,y
j( ) A
= f 1.5,1( ) A+ f 1.5,3( ) A
+ f 2.5,1( ) A+ f 2.5,3( ) A
=1 2( )+ #8( ) 2( )+5 2( )+ #1( ) 2( )=#6
(b)
5. (a) Each subrectangle and its midpoint are shown in the figure. The area of each subrectangle is
 , so we evaluate at each midpoint and estimate
 3
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.1 Double Integrals over Rectangles
 A=
1
2
""
R
f x,y( ) dA !
4
i=1
!
4
j=1
f x
i
,y
j( ) A
= f 1.5,1( ) A+ f 1.5,2( ) A+ f 1.5,3( ) A+ f 1.5,4( ) A
+ f 2,1( ) A+ f 2,2( ) A+ f 2,3( ) A+ f 2,4( ) A
+ f 2.5,1( ) A+ f 2.5,2( ) A+ f 2.5,3( ) A+ f 2.5,4( ) A
+ f 3,1( ) A+ f 3,2( ) A+ f 3,3( ) A+ f 3,4( ) A
=1
1
2
+ #4( )
1
2
+ #8( )
1
2
+ #6( )
1
2
+3
1
2
+0
1
2
+ #5( )
1
2
+ #8( )
1
2
+5
1
2
+3
1
2
+ #1( )
1
2
+ #4( )
1
2
+8
1
2
+6
1
2
+3
1
2
+0
1
2
=#3.5
R
R x y
f (x,y) x,y( )
R= 0,20 $ 0,30 f (x,y)
m=2 n=3 A=100
(b) The subrectangles are shown in the figure. In each subrectangle, the sample point farthest from
the origin is the upper right corner, and the area of each subrectangle is . Thus we estimate
6. To approximate the volume, let be the planar region corresponding to the surface of the water in
the pool, and place on coordinate axes so that and correspond to the dimensions given. Then we
define to be the depth of the water at , so the volume of water in the pool is the volume
of the solid that lies above the rectangle and below the graph of . We can
estimate this volume using the Midpoint Rule with and , so . Each subrectangle
with its midpoint is shown in the figure. Then
 4
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.1 Double Integrals over Rectangles
V !
2
i=1
!
3
j=1
f x
i
,y
j( ) A
= A f 5,5( )+ f 5,15( )+ f 5,25( )+ f 15,5( )+ f 15,15( )+ f 15,25( )
= 100 3+7+10+3+5+8( )=3600
3600
m=4 n=6
 A=25
V !
4
i=1
!
6
j=1
f x
i
,y
j( ) A
= 25 3+4+7+8+10+8+4+6+8+10+12+10+3+4
+5+6+8+7+2+2+2+3+4+4.
= 25(140)=3500
3500
3
f (x,y)= 52#x
2
#y
2
U<V<L R
f
1
2
,
1
2
 11
""
R
f x,y( )dA 1% f
1
2
,
1
2
 11 R
""
R
f x,y( )dA 
1
4
f
1
4
,
1
4
+ f
1
4
,
3
4
+ f
3
4
,
1
4
+ f
3
4
,
3
4
 
1
4
(11+13+9.5+11) 11
""
R
f x,y( )dA 11
Thus, we estimate that the pool contains cubic feet of water.
Alternatively, we can approximate the volume with a Riemann sum where , and the sample
points are taken to be, for example, the upper right corner of each subrectangle. Then and
So we estimate that the pool contains ft of water.
7. The values of get smaller as we move farther from the origin, so on any of the
subrectangles in the problem, the function will have its largest value at the lower left corner of the
subrectangle and its smallest value at the upper right corner, and any other value will lie between
these two. So using these subrectangles we have . (Note that this is true no matter how is
divided into subrectangles.)
8. From the level curves we see that . So, using the Midpoint Rule with only one
subrectangle, we get . Dividing into four squares of equal size,
we get
.
Using sixteen squares we get the same result. So .
 5
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.1 Double Integrals over Rectangles
m=n=2 A=4 f
""
R
f x,y( ) dA !
2
i=1
!
2
j=1
f x
i
,y
j( ) A= A f 1,1( )+ f 1,3( )+ f 3,1( )+ f 3,3( )
 4(27+4+14+17)=248
f
ave
=
1
A R( )
""
R
f (x,y)dA 
1
16
(248)=15.5
R= 0,388 $ 0,276 f (x,y)
x,y( )
f
ave
=
1
A(R)
""
R
f (x,y)dA=
1
388 % 276
""
R
f (x,y)dA
m=n=4
R 16
 A=
388
4
%
276
4
=6693
""
R
f (x,y)dA !
4
i=1
!
4
j=1
f x
i
,y
j( ) A
 A
+72.0+74.9+68.4+63.7+73.2+72.3+70.3+67.7]
=6693(1111.5)
9. (a)With , we have . Using the contour map to estimate the value of at the center
of each subrectangle, we have
(b)
10. As in Example 4, we place the origin at the southwest corner of the state. Then
 (in miles) is the rectangle corresponding to Colorado and we define to be
the temperature at the location . The average temperature is given by
We can use the Midpoint Rule with to give a reasonable estimate of the value of the double
integral.
Thus, we divide into regions of equal size, as shown in the figure, with the center of each
subrectangle indicated. The area of each subrectangle is , so using the contour
map to estimate the function values at each midpoint, we have
Therefore,
 6
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.1 Double Integrals over Rectangles
f
ave
 
6693% 1111.5
388% 276
 69.5
69.5&
m=n=2
 A=
388
2
%
276
2
=26
772
""
R
f (x,y)dA !
2
i=1
!
2
j=1
f x
i
,y
j( ) A 26,772[70.0+66.5+74.3+68.5]
= 26,772% 279.3
f
ave
 
26,772% 279.3
388% 276
 69.8&
z=3>0 S z=3
#2,2 $ 1,6 S ""
R
3dA=4% 5% 3=60
z=5#x' 0 0( x( 5 S
z=5#x 0,5 $ 0,3 S
3 area of triangle( )=3
1
2
% 5% 5 =37.5
""
R
(5#x)dA=37.5
z= f (x,y)=4#2y' 0 0( y( 1
0,1 $ 0,1 $ 0,4 z=4#2y
""
R
(4#2y)dA=(1)(1)(2)+
1
2
(1)(1)(2)=3
 , so the average temperature in Colorado on May 1, 1996, was
approximately .
Alternatively, we can use the Midpoint Rule with which is easier computationally but will
most likely be less accuratesince we have fewer subrectangles. In this case, ,
 and we can use the same grid to estimate the function values at the midpoints of the four
subrectangles. Then
and .
11. , so we can interpret the integral as the volume of the solid that lies below the plane 
and above the rectangle . is a rectangular solid, thus .
12. for , so we can interpret the integral as the volume of the solid that lies
below the plane and above the rectangle . is a triangular cylinder whose volume
is . Thus,
13. for . Thus the integral represents the volume of that part of the
rectangular solid which lies below the plane .
So
 7
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.1 Double Integrals over Rectangles
z= 9#y
2
z
2
+y
2
=9 z' 0
z
2
+y
2
=9 0,4 $ 0,2
f (x,y)=e
#x
2
#y
2
n=m
2
n
1 0.6065
4 0.5694
16 0.5606
64 0.5585
256 0.5579
1024 0.5578
n
1 0.9922
14. Here , so , . Thus the integral represents the volume of the top half of the
part of the circular cylinder that lies above the rectangle .
15. To calculate the estimates using a programmable calculator, we can use an algorithm similar to
that of Exercise 5.1.7 [ET 5.1.7]. In Maple, we can define the function (calling it f),
load the student package, and then use the command
middlesum(middlesum(f,x=0..1,m),
y=0..1,m);
to get the estimate with squares of equal size. Mathematica has no special Riemann sum
command, but we can define f and then use nested Sum commands to calculate the estimates.
estimate
16.
estimate
 8
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.1 Double Integrals over Rectangles
4 0.9262
16 0.8797
n
64 0.8660
256 0.8625
1024 0.8616
R mn ""
R
kdA !
m
i=1
!
n
j=1
f x
*
ij
,y
*
ij
 A
x
*
ij
,y
*
ij
f x
*
ij
,y
*
ij
=k !
m
i=1
!
n
j=1
 A= R=(b#a)(d#c)
!
m
i=1
!
n
j=1
f x
*
ij
,y
*
ij
 A=k!
m
i=1
!
n
j=1
 A=k(b#a)(d#c)
""
R
kdA = lim
m,n)*
!
m
i=1
!
n
j=1
f x
*
ij
,y
*
ij
 A= lim
m,n)*
k!
m
i=1
!
n
j=1
 A
=
lim
m,n)*
k(b#a)(d#c)=k(b#a)(d#c)
R 0( x+y( 2<! sin"' 0 0("(! f (x,y)=sin (x+y)' 0 (x,y)+ R
0( sin (x+y)( 1 ""
R
0dA(""
R
sin (x+y)dA(""
R
1dA
0(""
R
sin (x+y)dA( 1
estimate
17. If we divide into subrectangles, for any choice of sample
points . But always and area of . Thus, no matter
how we choose the sample points, and so
18. On , and for . Thus for all .
Since , Property (9) gives , so by Exercise 17 we
have .
 9
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.1 Double Integrals over Rectangles
 
0
3
2x+3x
2
y( )dx= x2+x3y
x=3
x=0
= 9+27y( )! 0+0( )=9+27y
 
0
4
2x+3x
2
y( )dy= 2xy+3x2 y
2
2
y=4
y=0
= 8x+3x
2
"
16
2
! 0+0( )=8x+24x
2
 
0
3
y
x+2
dx= yln x+2. x=3
x=0
=yln 5!yln 2=yln
5
2
 
0
4
y
x+2
dy=
1
x+2
y
2
2
y=4
y=0
=
1
x+2
16
2
!0 =
8
x+2
 
1
3
 
0
1
(1+4xy)dxdy= 
1
3
x+2x
2
y
x=1
x=0
dy= 
1
3
(1+2y)dy= y+y
2 3
1
= 3+9( )! 1+1( )=10
 
2
4
 
!1
1
x
2
+y
2( )dydx = 
2
4
x
2
y+
1
3
y
3 y=1
y=!1
dx= 
2
4
x
2
+
1
3
! !x
2
!
1
3
dx
= 
2
4
2x
2
+
2
3
dx=
2
3
x
3
+
2
3
x
4
2
=
128
3
+
8
3
!
16
3
+
4
3
=
116
3
 
0
2
 
0
 /2
xsin ydydx= 
0
2
xdx 
0
 /2
sin ydy =
x
2
2
2
0
!cos y
 /2
0
=(2!0)(0+1)=2
 
1
4
 
0
2
x+ y( )dxdy = 
1
4
1
2
x
2
+x y
x=2
x=0
dy= 
1
4
2+2 y( )dy
= 2y+2"
2
3
y
3/2 4
1
= 8+
4
3
" 8 ! 2+
4
3
=
46
3
 
0
2
 
0
1
(2x+y)
8
dxdy = 
0
2
1
2
(2x+y)
9
9
x=1
x=0
dy
1. ,
2. ,
3.
4.
5. [ as in Example 5] .
6.
7.
 1
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.2 Iterated Integrals
=
1
18
 
0
2
[(2+y)
9
!(0+y)
9
]dy=
1
18
(2+y)
10
10
!
y
10
10
2
0
=
1
180
[(4
10
!2
10
)!(2
10
!0
10
)]=
1,046,528
180
=
261,632
45
 
0
1
 
1
2
xe
x
y
dydx = 
0
1
xe
x
dx 
1
2
1
y
dy
= xe
x
!e
x 1
0
ln y
2
1
=[(e!e)!(0!1)](ln 2!0)=ln 2
 
1
4
 
1
2
x
y
+
y
x
dydx = 
1
4
xln y +
1
x
"
1
2
y
2 y=2
y=1
dx= 
1
4
xln 2+
3
2x
dx
=
1
2
x
2
ln 2+
3
2
ln x
4
1
=8ln 2+
3
2
ln 4!
1
2
ln 2
=
15
2
ln 2+3ln 4
1/2
=
21
2
ln 2
 
1
2
 
0
1
x+y( )
!2
dxdy = 
1
2
!(x+y)
!1 x=1
x=0
dy= 
1
2
y
!1
!(1+y)
!1
dy
= ln y!ln 1+y( )
2
1
=ln 2!ln 3!0+ln 2=ln
4
3
 
0
ln 2
 
0
ln 5
e
2x! y
dxdy = 
0
ln 5
e
2x
dx 
0
ln 2
e
!y
dy =
1
2
e
2x ln 5
0
!e
!y ln 2
0
=
25
2
!
1
2
!
1
2
+1 =6
8.
 [ as in Example 5]
 [ by integrating by parts]
9.
10.
11.
12.
 2
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.2 Iterated Integrals
 
0
1
 
0
1
xy
x
2
+y
2
+1
dydx = 
0
1
x x
2
+y
2
+1
y=1
y=0
dx= 
0
1
x x
2
+2 ! x
2
+1( )dx
=
1
3
(x
2
+2)
3/2
!(x
2
+1)
3/2 1
0
=
1
3
(3
3/2
!2
3/2
)!(2
3/2
!1)
=
1
3
3 3!4 2+1( )
 
R
(6x
2
y
3
!5y
4
)dA = 
0
3
 
0
1
(6x
2
y
3
!5y
4
)dydx= 
0
3
3
2
x
2
y
4
!y
5 y=1
y=0
dx
= 
0
3
3
2
x
2
!1 dx=
1
2
x
3
!x
3
0
=
27
2
!3=
21
2
 
R
cos (x+2y)dA
= 
0
 
 
0
 /2
cos (x+2y)dydx
= 
0
 
1
2
sin (x+2y)
y= /2
y=0
dx=
1
2
 
0
 
(sin (x+ )!sin x)dx
=
1
2
!cos (x+ )+cos x
 
0
=
1
2
!cos 2 +cos !(!cos +cos 0)
=
1
2
(!1!1!(1+1))=!2
 
R
xy
2
x
2
+1
dA = 
0
1
 
!3
3
xy
2
x
2
+1
dydx= 
0
1
x
x
2
+1
dx 
!3
3
y
2
dy
=
1
2
ln (x
2
+1)
1
0
1
3
y
3 3
!3
=
1
2
(ln 2!ln 1)"
1
3
(27+27)=9ln 2
 
R
1+x
2
1+y
2
dA = 
0
1
 
0
1
1+x
2
1+y
2
dydx= 
0
1
(1+x
2
)dx 
0
1
1
1+y
2
dy
13.
14.
15.
16.
 3
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.2 Iterated Integrals
= x+
1
3
x
3 1
0
tan
!1
y
1
0
= 1+
1
3
!0
 
4
!0 =
 
3
 
0
 /6
 
0
 /3
xsin (x+y)dydx
= 
0
 /6
!xcos (x+y)
y= /3
y=0
dx= 
0
 /6
xcos x!xcos x+
 
3
dx
=x sin x!sin x+
 
3
 /6
0
! 
0
 /6
sin x!sin x+
 
3
dx
=
 
6
1
2
!1 ! !cos x+cos x+
 
3
 /6
0
=!
 
12
! !
3
2
+0! !1+
1
2
=
3!1
2
!
 
12
 
R
x
1+xy
dA= 
0
1
 
0
1
x
1+xy
dydx
= 
0
1
ln (1+xy)
y=1
y=0
dx= 
0
1
ln (1+x)!ln 1 dx
= 
0
1
ln (1+x)dx= (1+x)ln (1+x)!x
1
0
=(2ln 2!1)!(ln 1!0)=2ln 2!1
 
R
xye
x
2
y
dA = 
0
2
 
0
1
xye
x
2
y
dxdy= 
0
2
1
2
e
x
2
y
x=1
x=0
dy=
1
2
 
0
2
(e
y
!1)dy
=
1
2
e
y
!y
2
0
=
1
2
[(e
2
!2)!(1!0)]=
1
2
(e
2
!3)
 
0
1
 
1
2
x
x
2
+y
2
dxdy= 
0
1
1
2
ln (x2
+y
2
)
x=2
x=1
dy=
1
2
 
0
1
ln (4+y
2
)!ln (1+y
2
) dy
u=ln (4+y
2
)#
17.
[by integrating by parts separately for each term]
18.
[by integrating by parts]
19.
20.
To evaluate the first term, we integrate by parts with 
 4
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.2 Iterated Integrals
du=
2y
4+y
2
dy
dv=dy# v=y
 ln (4+y
2
)dy = yln (4+y
2
)! 2y
2
4+y
2
dy=yln (4+y
2
)! 2! 8
4+y
2
dy
= yln (4+y
2
)!2y+8"
1
2
tan
!1 y
2
=yln (4+y
2
)!2y+4tan
!1 y
2
 ln (1+y
2
)dy=yln (1+y
2
)!2y+2tan
!1
y
 
0
1
 
1
2
x
x
2
+y
2
dxdy =
1
2
 
0
1
ln (4+y
2
)!ln (1+y
2
) dy
=
1
2
yln (4+y
2
)!2y+4tan
!1 y
2
!yln (1+y
2
)+2y!2tan
!1
y
1
0
=
1
2
ln 5+4tan
!1 1
2
!ln 2!2tan
!1
1 !0
=
1
2
ln 5!ln 2+4tan
!1 1
2
!2
 
4
=
1
2
ln
5
2
+2tan
!1 1
2
!
 
4
z= f (x,y)=4!x!2y$ 0 0% x% 1 0% y% 1
z=4!x!2y
0,1 & 0,1
z=2!x
2
!y
2
$ 0 0% x% 1 0% y% 1
z=2!x
2
!y
2
0,1 & 0,1
 and
 . Then
Similarly, . Thus,
21. for and .
So the solid is the region in the first octant which lies below the plane and above
 .
22. for and . So the solid is the region in the first octant which lies
below the circular paraboloid and above .
 5
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.2 Iterated Integrals
V = 
R
(12!3x!2y)dA= 
!2
3
 
0
1
(12!3x!2y)dxdy= 
!2
3
12x!
3
2
x
2
!2xy
x=1
x=0
dy
= 
!2
3
21
2
!2y dy=
21
2
y!y
2 3
!2
=
95
2
V = 
R
(4+x
2
!y
2
)dA= 
!1
1
 
0
2
(4+x
2
!y
2
)dydx= 
!1
1
4y+x
2
y!
1
3
y
3 y=2
y=0
dx
= 
!1
1
2x
2
+
16
3
dx=
2
3
x
3
+
16
3
x
1
!1
=
2
3
+
16
3
+
2
3
+
16
3
=12
V = 
!2
2
 
!1
1
1!
1
4
x
2
!
1
9
y
2
dxdy=4 
0
2
 
0
1
1!
1
4
x
2
!
1
9
y
2
dxdy
=4 
0
2
x!
1
12
x
3
!
1
9
y
2
x
x=1
x=0
dy=4 
0
2
11
12
!
1
9
y
2
dy=4
11
12
y!
1
27
y
3 2
0
=4"
83
54
=
166
27
V = 
!1
1
 
0
 
(1+e
x
sin y)dydx= 
!1
1
y!e
x
cos y
y= 
y=0
dx= 
!1
1
( +e
x
!0+e
x
)dx
= 
!1
1
( +2e
x
)dx= x+2e
x 1
!1
=2 +2e!
2
e
23.
24.
25.
26.
 6
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.2 Iterated Integrals
z=x x
2
+y
R= 0,1 & 0,1 xy !
V 
0
1
 
0
1
x x
2
+y dxdy= 
0
1
1
3
(x
2
+y)
3/2 x=1
x=0
dy=
1
3
 
0
1
(1+y)
3/2
!y
3/2
dy
=
1
3
"
2
5
(1+y)
5/2
!y
5/2 1
0
=
4
15
2 2 !1( )
z=1+(x!1)
2
+4y
2
R= 0,3 & 0,2 xy !
V = 
0
3
 
0
2
1+(x!1)
2
+4y
2
dydx= 
0
3
y+(x!1)
2
y+
4
3
y
3 y=2
y=0
dx
= 
0
3
2+2(x!1)
2
+
32
3
dx=
38
3
x+
2
3
(x!1)
3 3
0
=44
z$ 0# y% 3
V= 
0
3
 
0
2
(9!y
2
)dxdy= 
0
3
9x!y
2
x
x=2
x=0
dy= 
0
3
(18!2y
2
)dy= 18y!
2
3
y
3 3
0
=36
z=6!xy
R= !2,2 & 0,3 xy !
V = 
!2
2
 
0
3
(6!xy)dydx
= 
!2
2
6y!
1
2
xy
2 y=3
y=0
dx
= 
!2
2
18!
9
2
x dx
= 18x!
9
4
x
2 2
!2
=72
27. Here we need the volume of the solid lying under the surface and above the square
 in the plane.
28. Here we need the volume of the solid lying under the surface and above the
rectangle in the plane.
29. In the first octant, , so
30. (a) Here we need the volume of the solid lying under the surface and above the rectangle
 in the plane.
(b) The solid occupies the region between the two surfaces shown.
 7
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.2 Iterated Integrals
 
R
x
5
y
3
e
xy
dA=21e!57'0.0839
! ! !
! ! !
 
R
(2!x
2
!y
2
)! e
!x
2
cos (x
2
+y
2
)( ) dA'3.0271
R !1,1 & 0,5 A(R)=2" 5=10
31. In Maple, we can calculate the integral by defining the integrand as f and then using the command
int(int(f,x=0..1),y=0..1);.
In Mathematica, we can use the command Integrate[Integrate[f,{x,0,1}], {y,0,1}]. We find that
 . We can use plot3d (in Maple) or Plot3d (in Mathematica) to graph
the function.
32. In Maple, we can calculate the integral by defining f:=exp( x^2)*cos(x^2+y^2); and g:=2 x^2 y
^2; and then using the command evalf(int(int(g f,x= 1..1),y= 1..1),5);. In Mathematica, we can use
the command N[Integrate[Integrate[f,{x,0,1}],{y,0,1}],5].
In each of these commands, the 5 indicates that we want only five significant digits; this speeds up the
calculation considerably. We find that . We can use the
plot3d command (in Maple) or Plot3d (in Mathematica) to graph both functions on the same screen.
33. is the rectangle . Thus, and
 8
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.2 Iterated Integrals
f
ave =
1
A(R)
 
R
f (x,y)dA=
1
10
 
0
5
 
!1
1
x
2
ydxdy=
1
10
 
0
5
1
3
x
3
y
x=1
x=!1
dy=
1
10
 
0
5
2
3
ydy
=
1
10
1
3
y
2 5
0
=
5
6
A(R)=4" 1=4
f
ave =
1
A(R)
 
R
f (x,y)dA=
1
4
 
0
4
 
0
1
e
y
x+e
y
dydx=
1
4
 
0
4
2
3
(x+e
y
)
3/2 y=1
y=0
dx
=
1
4
"
2
3
 
0
4
(x+e)
3/2
!(x+1)
3/2
dx=
1
6
2
5
(x+e)
5/2
!
2
5
(x+1)
5/2 4
0
=
1
6
"
2
5
[(4+e)
5/2
!5
5/2
!e
5/2
+1]=
1
15
[(4+e)
5/2
!e
5/2
!5
5/2
+1]'3.327
f (x,y)=
x!y
(x+y)
3
 
0
1
 
0
1
f (x,y)dydx=
1
2
 
0
1
 
0
1
f (x,y)dxdy=!
1
2
f
0,0( )
!
g
xy
y !
g
x
=
d
dx
g(x,y)=
d
dx
 
a
x
 
c
y
f (s,t)dt ds= 
c
y
f (x,t)dt
g
xy
=
d
dy
 
c
y
f (x,t)dt= f (x,y)
g
yx
 
a
x
 
c
y
f (s,t)dtds= 
c
y
 
a
x
f (s,t)dtds
g
yx
= f (x,y) g
xy
=g
yx
= f (x,y)
34. , so
35. Let . Then a CAS gives and .
To explain the seeming violation of Fubini’s Theorem, note that has an infinite discontinuity at
 and thus does not satisfy the conditions of Fubini’s Theorem. In fact, both iterated integrals
involve improper integrals which diverge at their lower limits of integration.
36. (a) Loosely speaking, Fubini’s Theorem says that the order of integration of a function of two
variables does not affect the value of the double integral, while Clairaut’s Theorem says that the order
of differentiation of such a function does not affect the value of the second order derivative. Also,
both theorems require continuity (though Fubini’s allows a finite number of smooth curves to contain
discontinuities).
(b) To find , we first hold constant and use the single variable Fundamental Theorem of
Calculus, Part 1: . Now we use the Fundamental
Theorem again: .
To find , we first use Fubini’s Theorem to find that , and then use the
Fundamental Theorem twice, as above, to get . So .
 9
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.2 Iterated Integrals
 
0
1
 
0
x
2
(x+2y)dydx = 
0
1
xy+y
2 y=x
2
y=0
dx= 
0
1
x(x
2
)+(x
2
)
2
!0!0 dx
= 
0
1
(x
3
+x
4
)dx=
1
4
x
4
+
1
5
x
5 1
0
=
9
20
 
1
2
 
y
2
xydxdy = 
1
2
1
2
x
2
yx=2
x=y
dy= 
1
2
1
2
y(4!y
2
)dy=
1
2
 
1
2
(4y!y
3
)dy
=
1
2
2y
2
!
1
4
y
4 2
1
=
1
2
8!4!2+
1
4
=
9
8
 
0
1
 
y
e
y
x dxdy = 
0
1
2
3
x
3/2 x=e
y
x=y
dy=
2
3
 
0
1
(e
3y/2
!y
3/2
)dy=
2
3
2
3
e
3y/2
!
2
5
y
5/2 1
0
=
2
3
2
3
e
3/2
!
2
5
!
2
3
e
0
+0 =
4
9
e
3/2
!
32
45
 
0
1
 
x
2!x
(x
2
!y)dydx = 
0
1
x
2
y!
1
2
y
2 y=2!x
y=x
dx= 
0
1
x
2
(2!x)!
1
2
(2!x)
2
!x
2
(x)+
1
2
x
2
dx
= 
0
1
(!2x
3
+2x
2
+2x!2)dx= !
1
2
x
4
+
2
3
x
3
+x
2
!2x
1
0
=!
5
6
 
0
 /2
 
0
cos!
e
sin!
dr d! = 
0
 /2
re
sin! r=cos!
r=0
d! = 
0
 /2
(cos! )e
sin!
d! = e
sin! /2
0
=e
sin ( /2)
!e
0
=e!1
1.
2.
3.
4.
5.
6.
 1
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.3 Double Integrals over General Regions
 
0
1
 
0
v
1!v
2
dudv = 
0
1
u 1!v
2
u=v
u=0
dv= 
0
1
v 1!v
2
dv= !
1
3
(1!v
2
)
3/2 1
0
=!
1
3
(0!1)=
1
3
 
D
x
3
y
2
dA = 
0
2
 
!x
x
x
3
y
2
dydx= 
0
2
1
3
x
3
y
3 y=x
y=!x
dx=
1
3
 
0
2
2x
6
dx
=
2
3
1
7
x
7 2
0
=
2
21
2
7
!0 =
256
21
 
D
4y
x
3
+2
dA = 
1
2
 
0
2x
4y
x
3
+2
dydx= 
1
2
2y
2
x
3
+2
y=2x
y=0
dx= 
1
2
8x
2
x
3
+2
dx
=
8
3
ln x
3
+2
2
1
=
8
3
(ln 10!ln 3)=
8
3
ln
10
3
 
0
1
 
0
x
2y
x
2
+1
dydx = 
0
1
y
2
x
2
+1
y= x
y=0
dx= 
0
1
x
x
2
+1
dx
=
1
2
ln x
2
+1
1
0
=
1
2
(ln 2!ln 1)=
1
2
ln 2
 
0
1
 
0
y
e
y
2
dxdy= 
0
1
xe
y
2 x=y
x=0
dy= 
0
1
ye
y
2
dy=
1
2
e
y
2 1
0
=
1
2
(e!1)
 
1
2
 
y
y
3
e
x/y
dxdy= 
1
2
ye
x/y x= y
3
x= y
dy= 
1
2
ye
y
2
!ey( )dy= 1
2
e
y
2
!
1
2
ey
2
2
1
=
1
2
(e
4
!4e)
 
0
1
 
0
y
x y
2
!x
2
dxdy= 
0
1
!
1
3
(y
2
!x
2
)
3/2 x=y
x=0
dy=
1
3
 
0
1
y
3
dy=
1
3
"
1
4
y
4 1
0
=
1
12
7.
8.
9.
10.
11.
12.
 2
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.3 Double Integrals over General Regions
 
0
1
 
0
x
2
xcos ydydx= 
0
1
xsin y
y=x
2
y=0
dx= 
0
1
xsin x
2
dx= !
1
2
cos x
2 1
0
=
1
2
(1!cos 1)
 
0
1
 
x
2
x
(x+y)dydx = 
0
1
xy+
1
2
y
2 y= x
y=x
2
dx
= 
0
1
x
3/2
+
1
2
x!x
3
!
1
2
x
4
dx
=
2
5
x
5/2
+
1
4
x
2
!
1
4
x
4
!
1
10
x
5 1
0
=
3
10
 
1
2
 
2!y
2y!1
y
3
dxdy = 
1
2
xy
3 x=2y!1
x=2!y
dy
= 
1
2
(2y!1)!(2!y) y
3
dy
= 
1
2
(3y
4
!3y
3
)dy=
3
5
y
5
!
3
4
y
4 2
1
=
96
5
!12!
3
5
+
3
4
=
147
20
13.
14.
15.
16.
 3
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.3 Double Integrals over General Regions
 
D
xy
2
dA
= 
!1
1
 
0
1!y
2
xy
2
dxdy
= 
!1
1
y
2 1
2
x
2 x= 1!y
2
x=0
dy=
1
2
 
!1
1
y
2
(1!y
2
)dy
=
1
2
 
!1
1
(y
2
!y
4
)dy=
1
2
1
3
y
3
!
1
5
y
5 1
!1
=
1
2
1
3
!
1
5
+
1
3
!
1
5
=
2
15
 
!2
2
 
! 4!x
2
4!x
2
(2x!y)dydx
= 
!2
2
2xy!
1
2
y
2 y= 4!x
2
y=! 4!x
2
dx
= 
!2
2
2x 4!x
2
!
1
2
4!x
2( )+2x 4!x2 + 1
2
4!x
2( ) dx
= 
!2
2
4x 4!x
2
dx= !
4
3
4!x
2( ) 3/2
2
!2
=0
4x 4!x
2
17.
(Or, note that is an odd function, so
 4
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.3 Double Integrals over General Regions
 
!2
2
4x 4!x
2
dx=0
 
D
xydA
= 
0
1
 
2x
3!x
2xydydx= 
0
1
xy
2 y=3!x
y=2x
dx
= 
0
1
x[(3!x)
2
!(2x)
2
]dx
= 
0
1
(!3x
3
!6x
2
+9x)dx
= !
3
4
x
4
!2x
3
+
9
2
x
2 1
0
=!
3
4
!2+
9
2
=
7
4
V = 
0
1
 
x
4
x
(x+2y)dydx
= 
0
1
xy+y
2 y=x
y=x
4
dx= 
0
1
(2x
2
!x
5
!x
8
)dx
=
2
3
x
3
!
1
6
x
6
!
1
9
x
9 1
0
=
2
3
!
1
6
!
1
9
=
7
18
 .)
18.
19.
 5
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.3 Double Integrals over General Regions
V = 
0
1
 
y
3
y
2
(2x+y
2
)dxdy
= 
0
1
x
2
+xy
2 x=y
2
x=y
3
dy= 
0
1
(2y
4
!y
6
!y
5
)dy
=
2
5
y
5
!
1
7
y
7
!
1
6
y
6 1
0
=
19
210
V = 
1
2
 
1
7!3y
xydxdy= 
1
2
1
2
x
2
y
x=7!3y
x=1
dy
=
1
2
 
1
2
(48y!42y
2
+9y
3
)dy
=
1
2
24y
2
!14y
3
+
9
4
y
4 2
1
=
31
8
20.
21.
22.
 6
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.3 Double Integrals over General Regions
V = 
0
1
 
x
1
(x
2
+3y
2
)dydx
= 
0
1
x
2
y+y
3 y=1
y=x
dx= 
0
1
(x
2
+1!2x
3
)dx
=
1
3
x
3
+x!
1
2
x
4 1
0
=
5
6
V = 
0
1
 
0
1!x
1!x!y( ) dydx
= 
0
1
y!xy!
y
2
2
y=1!x
y=0
dx
= 
0
1
(1!x)
2
!
1
2
(1!x)
2
dx
= 
0
1
1
2
(1!x)
2
dx= !
1
6
(1!x)
3 1
0
=
1
6
23.
24.
 7
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.3 Double Integrals over General Regions
V = 
0
1
 
x
2!x
xdydx
= 
0
1
x y
y=2!x
y=x
dx= 
0
1
(2x!2x
2
)dx
= x
2
!
2
3
x
3 1
0
=
1
3
V = 
!2
2
 
x
2
4
x
2
dydx
= 
!2
2
x
2
y
y=4
y=x
2
dx= 
!2
2
(4x
2
!x
4
)dx
=
4
3
x
3
!
1
5
x
5 2
!2
=
32
3
!
32
5
+
32
3
!
32
5
=
128
15
25.
26.
 8
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.3 Double Integrals over General Regions
V = 
0
2
 
0
2y
4!y
2
dxdy
= 
0
2
x 4!y
2
x=2y
x=0
dy= 
0
2
2y 4!y
2
dy
= !
2
3
4!y
2( ) 3/2
2
0
=0+
16
3
=
16
3
V = 
0
1
 
0
1!x
2
ydydx= 
0
1
y
2
2
y= 1!x
2
y=0
dx
= 
0
1
1!x
2
2
dx=
1
2
x!
1
3
x
3 1
0
=
1
3
V 8 V
1
V
1 = 
0
r
 
0
r
2
! y
2
r
2
!y
2
dxdy
= 
0
r
x r
2
!y
2 x= r
2
! y
2
x=0
dy
27.
28.
By symmetry, the desired volume is times the volume in the first octant. Now
 9
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.3 Double Integrals over General Regions
= 
0
r
(r
2
!y
2
)dy= r
2
y!
1
3
y
3 r
0
=
2
3
r
3
V=
16
3
r
3
x=0 x#1.213
 
D
xdA # 
0
1.213
 
x
4
3x!x
2
xdydx= 
0
1.213
xy
y=3x!x
2
y=x
4
dx
= 
0
1.213
(3x
2
!x
3
!x
5
)dx= x
3
!
1
4
x
4
!
1
6
x
6 1.213
0
# 0.713
y=cos xy=x x#0.7391
V # 
0
0.7391
 
x
cos x
zdydx= 
0
0.7391
 
x
cos x
xdydx
Thus .
29.
From the graph, it appears that the two curves intersect at and at . Thus the desired
integral is
30.
The desired solid is shown in the first graph. From the second graph, we estimate that 
intersects at . Therefore the volume of the solid is
 10
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.3 Double Integrals over General Regions
= 
0
0.7391
xy
y=cos x
y=x
dx= 
0
0.7391
(xcos x!x
2
)dx
= cos x+xsin x!
1
3
x
3 0.7391
0
#0.1024
y=0
V# 
0
0.7391
 
0
x
xdydx+ 
0.7391
 /2
 
0
cos x
xdydx#0.4684
y=1!x
2
y=x
2
!1 $1,0( ) 1!x
2
% x
2
!1 !1,1
z=2x+2y+10 z=2!x!y
V = 
!1
1
 
x
2
!1
1!x
2
(2x+2y+10)dydx! 
!1
1
 
x
2
!1
1!x
2
(2!x!y)dydx
= 
!1
1
 
x
2
!1
1!x
2
(2x+2y+10!(2!x!y))dydx= 
!1
1
 
x
2
!1
1!x
2
(3x+3y+8)dydx
= 
!1
1
3xy+
3
2
y
2
+8y
y=1!x
2
y=x
2
!1
dx
= 
!1
1
3x(1!x
2
)+
3
2
(1!x
2
)
2
+8(1!x
2
)!3x(x
2
!1)!
3
2
(x
2
!1)
2
!8(x
2
!1) dx
= 
!1
1
(!6x
3
!16x
2
+6x+16)dx= !
3
2
x
4
!
16
3
x
3
+3x
2
+16x
1
!1
= !
3
2
!
16
3
+3+16+
3
2
!
16
3
!3+16=
64
3
y=1 z=3
y=x
2
y=1 2+y% 3y 0& y& 1
z=2+y z=3y
Note: There is a different solid which can also be construed to satisfy the conditions stated in the
exercise. This is the solid bounded by all of the given surfaces, as well as the plane . In case you
calculated the volume of this solid and want to check your work, its volume is
 .
31. The two bounding curves and intersect at with on .
Within this region, the plane is above the plane , so
32. The two planes intersect in the line , , so the region of integration is the plane region
enclosed by the parabola and the line . We have for , so the solid region is
bounded above by and bounded below by .
 11
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.3 Double Integrals over General Regions
V = 
!1
1
 
x
2
1
(2+y)dydx! 
!1
1
 
x
2
1
(3y)dydx= 
!1
1
 
x
2
1
(2+y!3y)dydx= 
!1
1
 
x
2
1
(2!2y)dydx
= 
!1
1
2y!y
2 y=1
y=x
2
dx= 
!1
1
(1!2x
2
+x
4
)dx= x!
2
3
x
3
+
1
5
x
5 1
!1
=
16
15
y=x
3
!x y=x
2
+x x=2 x
2
+x>x
3
!x
0,2( )
V= 
0
2
 
x
3
!x
x
2
+x
zdydx= 
0
2
 
x
3
!x
x
2
+x
(x
3
y
4
+xy
2
)dydx=
13,984,735,616
14,549,535
x & 1 y & 1 2x
2
+y
2
<8!x
2
!2y
2
!1& x& 1 ! 1!x
2
& y& 1!x
2
V= 
!1
1
 
! 1!x
2
1!x
2
(8!x
2
!2y
2
)!(2x
2
+y
2
) dydx=
13 
2
x
2
+y
2
=1 z=0 D
x
2
+y
2
& 1 
D
(1!x
2
!y
2
)dA= 
!1
1
 
! 1!x
2
1!x
2
(1!x
2
!y
2
)dydx=
 
2
xy ! x
2
+y
2
=2y'
x
2
+y
2
!2y=0' x
2
+(y!1)
2
=1 !1& x& 1
33. The two bounding curves and intersect at the origin and at , with 
on . Using a CAS, we find that the volume is
34. For and , . Also, the cylinder is described by the inequalities
 , . So the volume is given by
[using a CAS]
35. The two surfaces intersect in the circle , and the region of integration is the disk :
 . Using a CAS, the volume is .
36. The projection onto the plane of the intersection of the two surfaces is the circle 
 , so the region of integration is given by ,
 12
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.3 Double Integrals over General Regions
1! 1!x
2
& y& 1+ 1!x
2
2y% x
2
+y
2
V= 
!1
1
 
1! 1!x
2
1+ 1!x
2
[2y!(x
2
+y
2
)]dydx=
 
2
D = (x,y) |0& y& x ,0& x& 4{ }
= (x,y) | y
2
& x& 4,0& y& 2{ }
 
0
4
 
0
x
f (x,y)dydx= 
D
f (x,y)dA= 
0
2
 
y
2
4
f (x,y)dxdy
D = x,y( ) |4x& y& 4,0& x& 1{ }
= x,y( ) |0& x&
y
4
,0& y& 4{ }
 
0
1
 
4x
4
f (x,y)dydx
= 
D
f (x,y)dA
 . In this region, so, using a CAS, the volume is
 .
37.
Because the region of integration is
we have
 .
38.
Because the region of integration is
we have
 13
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.3 Double Integrals over General Regions
= 
0
4
 
0
y/4
f (x,y)dxdy
D = (x,y) |! 9!y
2
& x& 9!y
2
,0& y& 3{ }
= (x,y) |0& y& 9!x
2
,!3& x& 3{ }
 
0
3
 
! 9!y
2
9!y
2
f(x,y)dxdy= 
D
f(x,y)dA= 
!3
3
 
0
9!x
2
f(x,y)dydx
D = (x,y) |0& x& 9!y ,0& y& 3{ }
= (x,y) |0& y& 3,0& x& 6{ }( x,y( ) |0& y& 9!x
2
, 6& x& 3{ }
 
0
3
 
0
9!y
f (x,y)dxdy
= 
D
f (x,y)dA
39.
Because the region of integration is
we have
40.
To reverse the order, we must break the region into two separate type I regions. Because the region of
integration is
we have
 14
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.3 Double Integrals over General Regions
= 
0
6
 
0
3
f (x,y)dydx+ 
6
3
 
0
9!x
2
f (x,y)dydx
D = (x,y) |0& y& ln x,1& x& 2{ }
= (x,y) |e
y
& x& 2,0& y& ln 2{ }
 
1
2
 
0
ln x
f (x,y)dydx =
 
D
f (x,y)dA
= 
0
ln 2
 
e
y
2
f (x,y)dxdy
D = (x,y) |arctanx& y&
 
4
,0& x& 1{ }
= (x,y) |0& x& tan y,0& y&
 
4{ }
 
0
1
 
arctanx
 /4
f (x,y)dydx =
 
D
f (x,y)dA
= 
0
 /4
 
0
tan y
f (x,y)dxdy
41.
Because the region of integration is
we have
42.
Because the region of integration is
we have
 15
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.3 Double Integrals over General Regions
 
0
1
 
3y
3
e
x
2
dxdy = 
0
3
 
0
x/3
e
x
2
dydx
= 
0
3
e
x
2
y
y=x/3
y=0
dx= 
0
3
x
3
e
x
2
dx
=
1
6
e
x
2 3
0
=
e
9
!1
6
 
0
1
 
y
1
x
3
+1 dxdy = 
0
1
 
0
x
2
x
3
+1 dydx
= 
0
1
x
3
+1 y
y=x
2
y=0
dx= 
0
1
x
2
x
3
+1 dx
=
2
9
(x
3
+1)
3/2 1
0
=
2
9
2
3/2
!1( )
 
0
3
 
y
2
9
ycos x
2
dxdy = 
0
9
 
0
x
ycos x
2
dydx
43.
44.
45.
 16
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.3 Double Integrals over General Regions
= 
0
9
cos x
2 y
2
2
y= x
y=0
dx= 
0
9
1
2
xcos x
2
dx
=
1
4
sin x
2 9
0
=
1
4
sin 81
 
0
1
 
x
2
1
x
3
sin (y
3
)dydx = 
0
1
 
0
y
x
3
sin (y
3
)dxdy
= 
0
1
x
4
4
sin (y
3
)
x= y
x=0
dy
= 
0
1
1
4
y
2
sin (y
3
)dy
= !
1
12
cos (y
3
)
1
0
=
1
12
(1!cos 1)
 
0
1
 
arcsiny
 /2
cos x 1+cos
2
x dxdy
= 
0
 /2
 
0
sin x
cos x 1+cos
2
x dydx
= 
0
 /2
cos x 1+cos
2
x y
y=sin x
y=0
dx
46.
47.
 17
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.3 Double Integrals over General Regions
= 
0
 /2
cos x 1+cos
2
x sin xdx
Let u=cos x,du=!sin xdx,dx=du/(!sin x)
= 
1
0
!u 1+u
2
du= !
1
3
1+u
2( ) 3/2
0
1
=
1
3
8 !1( )=
1
3
2 2 !1( )
 
0
8
 
3
y
2
e
x
4
dxdy = 
0
2
 
0
x
3
e
x
4
dydx
= 
0
2
e
x
4
y
y=x
3
y=0
dx= 
0
2
x
3
e
x
4
dx
=
1
4
e
x
4 2
0=
1
4
(e
16
!1)
D = (x,y) |0& x& 1,!x+1& y& 1{ }( (x,y) |!1& x& 0,x+1& y& 1{ }
( (x,y) |0& x& 1,!1& y& x!1{ }( (x,y) |!1& x& 0,!1& y&!x!1{ }
 
D
x
2
dA = 
0
1
 
1!x
1
x
2
dydx+ 
!1
0
 
x+1
1
x
2
dydx+ 
0
1
 
!1
x!1
x
2
dydx+ 
!1
0
 
!1
!x!1
x
2
dydx
= 4 
0
1
 
1!x
1
x
2
dydx[ by symmetry of the regions and because f (x,y)=x
2
% 0]
= 4 
0
1
x
3
dx=4
1
4
x
4 1
0
=1
48.
49.
 ,
all type I.
 18
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.3 Double Integrals over General Regions
D = (x,y) |!1& x& 0,!1& y& 1+x
2{ } ( (x,y) |0& x& 1, x& y& 1+x2{ }
( (x,y) |0& x& 1,!1& y&! x{ }
 
D
xydA = 
!1
0
 
!1
1+x
2
xydydx+ 
0
1
 
x
1+x
2
xydydx+ 
0
1
 
!1
! x
xydydx
= 
!1
0
1
2
xy
2 y=1+x
2
y=!1
dx+ 
0
1
1
2
xy
2 y=1+x
2
y= x
dx+ 
0
1
1
2
xy
2 y=! x
y=!1
dx
= 
!1
0
x
3
+
1
2
x
5
dx+ 
0
1
1
2
(x
5
+2x
3
!x
2
+x)dx+ 
0
1
1
2
(x
2
!x)dx
=
1
4
x
4
+
1
12
x
6 0
!1
+
1
2
1
6
x
6
+
1
2
x
4
!
1
3
x
3
+
1
2
x
2 1
0
+
1
2
1
3
x
3
!
1
2
x
2 1
0
= !
1
3
+
5
12
!
1
12
=0
D= 0,1 ) 0,1 0& x
3
+y
3
& 2 A(D)=1 0& 
D
x
3
+y
3
dA& 2
D= (x,y) |x
2
+y
2
&
1
4{ } 1=e
0
& e
x
2
+y
2
& e
1/4
A(D)=
 
4
 
4
& 
D
e
x
2
+y
2
dA& (e
1/4
)
 
4
m& f (x,y)&M 
D
mdA& 
D
f (x,y)dA& 
D
MdA '
m 
D
1dA& 
D
f (x,y)dA&M 
D
1dA 'mA(D)& 
D
f (x,y)dA&MA(D)
50.
 ,
all type I.
51. For , and , so .
52. Since , and , so 
.
53. Since , by (8) 
 by (7) by (10).
54.
 19
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.3 Double Integrals over General Regions
 
D
f (x,y)dA
= 
0
1
 
0
2y
f (x,y)dxdy+ 
1
3
 
0
3! y
f (x,y)dxdy
= 
0
2
 
x/2
3!x
f (x,y)dydx
 
D
(x
2
tan x+y
3
+4) dA= 
D
x
2
tan xdA+ 
D
y
3
dA+ 
D
4dA x
2
tan x x D
y ! 
D
x
2
tan xdA=0 y
3
y D
x ! 
D
y
3
dA=0
 
D
(x
2
tan x+y
3
+4)dA=4 
D
dA=4(area of D)=4" 2( )
2
=8 
D y !
3x x 
D
3xdA=0 4y y D
x ! 
D
4ydA=0
 
D
2!3x+4y( ) dA = 
D
2dA=2 
D
dA
= 2(area of D)=2(50)
= 100
1!x
2
!y
2
% 0 
D
1!x
2
!y
2
dA
55. . But is an odd function of and is
symmetric with respect to the axis, so . Similarly, is an odd function of and 
is symmetric with respect to the axis, so . Thus
56. First, 0in0.19in ] The region , shown in the figure, is symmetric with respect to the axis and
 is an odd function of , so . Similarly, is an odd function of and is symmetric
with respect to the axis, so . Then
57. Since , we can interpret as the volume of the solid that lies below
 20
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.3 Double Integrals over General Regions
z= 1!x
2
!y
2
D xy ! z= 1!x
2
!y
2
x
2
+y
2
+z
2
=1 z% 0 xy ! x
2
+y
2
=1 D
1
1
2
4
3
 (1)
3
=
2
3
 
z !
! ! ! !
! ! ! !
y=
1$ 13+4x!4x
2
2
13+4x!4x
2
=0
x=
1$ 14
2
V= 
1! 14( ) /2
1+ 14( ) /2
 
1! 13+4x!4x
2( ) /2
1+ 13+4x!4x
2( ) /2
[(4!x
2
!y
2
)!(1!x!y)]dydx=
49 
8
the graph of and above the region in the plane. is equivalent to
 , which meets the plane in the circle , the boundary of . Thus, the
solid is an upper hemisphere of radius which has volume .
58. To find the equations of the boundary curves, we require that the values of the two surfaces be
the same. In Maple, we use the command solve(4 x^2 y^2=1 x y,y);
and in Mathematica, we use Solve[4 x^2 y^ 2==1 x y,y] . We find that the curves have equations
 .
To find the two points of intersection of these curves, we use the CAS to solve , finding
that . So, using the CAS to evaluate the integral, the volume of intersection is
 .
 21
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.3 Double Integrals over General Regions
R R= (r, ) |0 r 2,0 2!{ }
!!
R
f (x,y)dA= !
0
2!
!
0
2
f (rcos ,rsin )r dr d 
R R= (x,y) |0 x 2,0 y 2"x{ }
!!
R
f (x,y)dA=!
0
2
!
0
2"x
f (x,y)dydx
R R= (x,y) |"2 x 2,x y 2{ }
!!
R
f (x,y)dA=!
"2
2
!
x
2
f (x,y)dydx
R R= (r, ) |1 r 3,0 
!
2{ }
!!
R
f (x,y)dA= !
0
! /2
!
1
3
f (rcos ,rsin )r dr d 
R R= (r, ) |2 r 5,0 2!{ }
!!
R
f (x,y)dA= !
0
2!
!
2
5
f (rcos ,rsin )r dr d 
R
R= (r, ) |0 r 2 2,
!
4
 
5!
4{ } !!
R
f (x,y)dA= !
! /4
5! /4
!
0
2 2
f (rcos ,rsin )r dr d 
!
!
2!
!
4
7
r dr d R= (r, ) |4 r 7,! 2!{ }
!
!
2!
!
4
7
r dr d = !
!
2!
d !
4
7
r dr
= 
2!
!
1
2
r
2 7
4
=! #
1
2
49"16( )=
33!
2
1. The region is more easily described by polar coordinates: . Thus
 .
2. The region is more easily described by rectangular coordinates: .
Thus .
3. The region is more easily described by rectangular coordinates: .
Thus .
4. The region is more easily described by polar coordinates: .
Thus .
5. The region is more easily described by polar coordinates: . Thus
 .
6. The region is more easily described by polar coordinates:
 . Thus .
7. The integral represents the area of the region , the lower
half of a ring.
 1
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.4 Double Integrals in Polar Coordinates
!
0
! /2
!
0
4cos 
r dr d R= (r, ) |0 r 4cos ,0 ! /2{ }
r=4cos $r
2
=4rcos $x
2
+y
2
=4x$(x"2)
2
+y
2
=4 R
2,0( )
!
0
! /2
!
0
4cos 
r dr d = !
0
! /2
1
2
r
2 r=4cos 
r=0
d = !
0
! /2
8cos
2
 d 
= !
0
! /2
4(1+cos 2 )d =4 +
1
2
sin 2 
! /2
0
=2!
D D= (r, ) |0 r 3,0 2!{ }
!!
D
xydA
= !
0
2!
!
0
3
(rcos )(rsin )r dr d = !
0
2!
sin cos d !
0
3
r
3
dr
=
1
2
sin
2
 
2!
0
1
4
r
4 3
0
=0
!!
R
(x+y)dA
= !
! /2
3! /2
!
1
2
(rcos +rsin )r dr d = !
! /2
3! /2
!
1
2
r
2
(cos +sin dr d 
= !
! /2
3! /2
(cos +sin d !
1
2
r
2
dr = sin "cos 
3! /2
! /2
1
3
r
3 2
1
=("1"0"1+0)
8
3
"
1
3
="
14
3
8. The integral represents the area of the region .
Since , is the portion in the first quadrant of a
circle of radius 2 with center .
9. The disk can be described in polar coordinates as . Then
10.
 2
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.4 Double Integrals in Polar Coordinates
!!
R
cos (x
2
+y
2
)dA =!
0
!
!
0
3
cos (r
2
)r dr d = !
0
!
d !
0
3
rcos (r
2
)dr
= 
!
0
1
2
sin (r
2
)
3
0
=! #
1
2
(sin 9"sin 0)=
!
2
sin 9
!!
R
4"x
2
"y
2
dA = !
"! /2
! /2
!
0
2
4"r
2
r dr d = !
"! /2
! /2
d !
0
2
r 4"r
2
dr
= 
! /2
"! /2
"
1
2
#
2
3
(4"r
2
)
3/2 2
0
=
!
2
+
!
2
"
1
3
(0"4
3/2
) =
8
3
!
!!
D
e
"x
2
"y
2
dA = !
"! /2
! /2
!
0
2
e
"r
2
r dr d = !
"! /2
! /2
d !
0
2
re
"r
2
dr
= 
! /2
"! /2
"
1
2
e
"r
2 2
0
=! "
1
2
(e
"4
"e
0
)=
!
2(1"e
"4
)
!!
R
ye
x
dA= !
0
! /2
!
0
5
(rsin )e
rcos 
r dr d =!
0
5
!
0
! /2
r
2
sin e
rcos 
d dr
!
0
! /2
r
2
sin e
rcos 
d u=rcos %du="rsin d 
!
0
! /2
r
2
sin e
rcos 
d = !
u=r
u=0
"r e
u
du="r[e
0
"e
r
]=re
r
"r
!
0
5
!
0
! /2
r
2
sin e
rcos 
d dr=!
0
5
(re
r
"r)dr= re
r
"e
r
"
1
2
r
2 5
0
=4e
5
"
23
2
R R= (r, ) |0 ! /4,1 r 2{ }
!!
R
arctan(y/x)dA= !
0
! /4
!
1
2
arctan(tan )r dr d 
y/x=tan (tan )= 0 ! /4
11.
12.
13.
14. . First we integrate
 : Let , and
 . Then
 , where we integrated by parts in the
first term.
15. is the region shown in the figure, and can be described by .
Thus
since . Also, arctan for ,
so the integral becomes
 3
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.4 Double Integrals in Polar Coordinates
!
0
! /4
!
1
2
 r dr d = !
0
! /4
 d !
1
2
r dr=
1
2
 
2 ! /4
0
1
2
r
2 2
1
=
!
2
32
#
3
2
=
3
64
!
2
!!
D
xdA
= !!
x
2
+ y
2
 4
x& 0,y& 0
xdA" !!
x"1( )
2
+ y
2
 1
y& 0
xdA
= !
0
! /2
!
0
2
r
2
cos dr d " !
0
! /2
!
0
2cos 
r
2
cos dr d 
= !
0
! /2
1
3
(8cos )d " !
0
! /2
1
3
(8cos
4
 )d 
=
8
3
"
8
12
cos
3
 sin +
3
2
( +sin cos )
! /2
0
=
8
3
"
2
3
0+
3
2
!
2
=
16"3!
6
D= (r, ) "! /6 ! /6,0 r 3 { }
!!
D
dA
= !
"! /6
! /6
!
0
cos 3 
r dr d = !
"! /6
! /6
1
2
r
2 r=cos 3 
r=0
d 
= !
"! /6
! /6
1
2
cos
2
3 d =2 !
0
! /6
1
2
1+cos 6 
2
d 
 .
16.
17. One loop is given by the region
 , so the area is
 4
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.4 Double Integrals in Polar Coordinates
=
1
2
 +
1
6
sin 6 
! /6
0
=
!
12
D= (r, ) |0 2! ,0 r 4+3cos { }
A(D) =!!
D
dA= !
0
2!
!
0
4+3cos 
r dr d = !
0
2!
1
2
r
2 r=4+3cos 
r=0
d =
1
2
!
0
2!
(4+3cos )
2
d 
=
1
2
!
0
2!
(16+24cos +9cos
2
 )d =
1
2
!
0
2!
16+24cos +9#
1+cos 2 
2
d 
=
1
2
16 +24sin +
9
2
 +
9
4
sin 2 
2!
0
=
41
2
!
A =2 !
0
! /4
!
0
sin 
r dr d =2 !
0
! /4
1
2
r
2 r=sin 
r=0
d 
= !
0
! /4
sin
2
 d = !
0
! /4
1
2
(1"cos 2 )d 
=
1
2
 "
1
2
sin 2 
! /4
0
=
1
2
!
4
"
1
2
sin
!
2
"0+
1
2
sin 0 =
1
8
! "2( )
18. , so
19. By symmetry,
 5
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.4 Double Integrals in Polar Coordinates
2=4sin =
!
6
5!
6
A = !
! /6
5! /6
!
2
4sin 
r dr d = !
! /6
5! /6
1
2
r
2 r=4sin 
r=2
d = !
! /6
5! /6
(8sin
2
 "2)d 
= !
! /6
5! /6
[4(1"cos 2 )"2]d = 2 "2sin 2 
5! /6
! /6
=
4!
3
+2 3
V= !!
x
2
+ y
2
 9
(x
2
+y
2
)dA= !
0
2!
!
0
3
(r
2
)r dr d = !
0
2!
d !
0
3
r
3
dr= 
2!
0
1
4
r
4 3
0
=2!
81
4
=
81!
2
x
2
+y
2
+z
2
=16 xy " x
2
+y
2
=16
V = 2 !!
4 x
2
+y
2
 16
16"x
2
"y
2
dA [by symmetry]
= 2 !
0
2!
!
2
4
16"r
2
r dr d =2 !
0
2!
d !
2
4
r(16"r
2
)
1/2
dr
= 2 
2!
0
"
1
3
(16"r
2
)
3/2 4
2
="
2
3
(2! )(0"12
3/2
)=
4!
3
12 12( )=32 3!
V =2 !!
x
2
+ y
2
 a
2
a
2
"x
2
"y
2
dA=2 !
0
2!
!
0
a
a
2
"r
2
r dr d =2 !
0
2!
d !
0
a
r a
2
"r
2
dr
=2 
2!
0
"
1
3
(a
2
"r
2
)
3/2 a
0
=2(2! ) 0+
1
3
a
3
=
4!
3
a
3
z=10"3x
2
"3y
2
z=4 4=10"3x
2
"3y
2
x
2
+y
2
=2
20. implies that or , so
 .
21.
22. The sphere intersects the plane in the circle , so
23. By symmetry,
24. The paraboloid intersects the plane when or . So
 6
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.4 Double Integrals in Polar Coordinates
V = !!
x
2
+ y
2
 2
(10"3x
2
"3y
2
)"4 dA= !
0
2!
!
0
2
(6"3r
2
)r dr d 
= !
0
2!
d !
0
2
(6r"3r
3
)dr= 
2!
0
3r
2
"
3
4
r
4 2
0
=6!
z= x
2
+y
2
x
2
+y
2
+z
2
=1 x
2
+y
2
+ x
2
+y
2( )
2
=1 x
2
+y
2
=
1
2
V = !!
x
2
+ y
2
 1/2
1"x
2
"y
2
" x
2
+y
2( )dA= !
0
2!
!
0
1/ 2
1"r
2
"r( ) r dr d 
= !
0
2!
d !
0
1/ 2
r 1"r
2
"r
2( )dr= 2!
0
"
1
3
(1"r
2
)
3/2
"
1
3
r
3 1/ 2
0
= 2! "
1
3
1
2
"1 =
!
3
2" 2( )
3x
2
+3y
2
=4"x
2
"y
2
x
2
+y
2
=1
V = !!
x
2
+ y
2
 1
[(4"x
2
"y
2
)"3(x
2
+y
2
)]dA= !
0
2!
!
0
1
4(1"r
2
)r dr d 
= !
0
2!
d !
0
1
(4r"4r
3
)dr= 
2!
0
2r
2
"r
4 1
0
=2!
x
2
+y
2
=4 z= 64"4x
2
"4y
2
z=" 64"4x
2
"4y
2
V = !!
x
2
+ y
2
 4
64"4x
2
"4y
2
" " 64"4x
2
"4y
2( ) dA
= !!
x
2
+y
2
 4
2 64"4x
2
"4y
2
dA=4 !
0
2!
!
0
2
16"r
2
r dr d 
25. The cone intersects the sphere when or .
So
26. The two paraboloids intersect when or . So
27. The given solid is the region inside the cylinder between the surfaces 
and . So
 7
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.4 Double Integrals in Polar Coordinates
= 4 !
0
2!
d !
0
2
r 16"r
2
dr=4 
2!
0
"
1
3
(16"r
2
)
3/2 2
0
= 8! "
1
3
(12
3/2
"16
2/3
)=
8!
3
64"24 3( )
xy " r
2
1
 x
2
+y
2
 r
2
2
xy "
V = 2 !!
r
2
1
 x
2
+ y
2
 r
2
2
r
2
2
"x
2
"y
2
dA=2 !
0
2!
!
r
1
r
2
r
2
2
"r
2
r dr d 
= 2 !
0
2!
d !
r
1
r
2
r
2
2
"r
2
r dr=
4!
3
"(r
2
2
"r
2
)
3/2 r2
r
1
=
4!
3
(r
2
2
"r
2
1
)
3/2
" r
2
2
=
1
2
h
2
+r
2
1
1
4
h
2
=r
2
2
"r
2
1
h V=
4!
3
1
4
h
2 3/2
=
!
6
h
3
!
0
1
!
0
1"x
2
e
x
2
+y
2
dydx = !
0
! /2
!
0
1
e
r
2
r dr d 
28. (a) Here the region in the plane is the annular region and the desired volume is
twice that above the plane. Hence
(b) A cross sectional cut is shown in the figure. So or .
Thus the volume in terms of is .
29.
 8
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.4 Double Integrals in Polar Coordinates
= !
0
! /2
d !
0
1
re
r
2
dr
= 
! /2
0
1
2
e
r
2 1
0
=
1
4
! (e"1)
!
"! /2
! /2
!
0
a
(r
2
)
3/2
r dr d = !
"! /2
! /2
d !
0
a
r
4
dr
= 
! /2
"! /2
1
5
r
5 a
0
=
1
5
! a
5
!
0
!
!
0
2
(rcos )
2
(rsin )
2
r dr d =!
0
!
(sin cos )
2
d !
0
2
r
5
dr
=!
0
!
1
2
sin 2 
2
d !
0
2
r
5
dr
=
1
4
1
2
 "
1
8
sin 4 
!
0
1
6
r
6 2
0
=
1
4
!
2
64
6
=
4!
3
30.
31.
32.
 9
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.4 Double Integrals in Polar Coordinates
!
0
! /2
!
0
2cos 
r
2
dr d = !
0
! /2
1
3
r
3 r=2cosr=0
d 
= !
0
! /2
8
3
cos
3
 d 
=
8
3
sin "
1
3
sin
3
 
! /2
0
=
16
9
D 20 D
D f (x,y)
x,y( )
D= (x,y) |x
2
+y
2
 400{ } f (x,y) y "
x "
y " f (0,"20)=2 f (0,20)=7 yz "
(0,"20,2) (0,20,7)
7"2
20" "20( )
=
1
8
z"7=
1
8
(y"20)% z=
1
8
y+
9
2
f (x,y) x f (x,y)=
1
8
y+
9
2
!!
D
f (x,y)dA
D= (r, ) |0 r 20,0 2!{ } x=rcos y=rsin 
!
0
2!
!
0
20
1
8
rsin +
9
2
r dr d = !
0
2!
1
24
r
3
sin +
9
4
r
2 r=20
r=0
d = !
0
2!
1000
3
sin +900 d 
= "
1000
3
cos +900 
2!
0
=1800!
1800!'5655
3
R
V = !
0
2!
!
0
R
e
"r
r dr d = !
0
2!
d !
0
R
re
"r
dr= 
2!
0
"re
"r
"e
"r R
0
= 2! ["Re
"R
"e
"R
+0+1]=2! (1"Re
"R
"e
"R
)ft
3
33. The surface of the water in the pool is a circular disk with radius ft. If we place on
coordinate axes with the origin at the center of and define to be the depth of the water at
 , then the volume of water in the pool is the volume of the solid that lies above
 and below the graph of . We can associate north with the positive 
direction, so we are given that the depth is constant in the direction and the depth increases linearly
in the direction from to . The trace in the plane is a line segment from
 to . The slope of this line is , so an equation of the line is
 . Since is independent of , . Thus the volume is
given by , which is most conveniently evaluated using polar coordinates. Then
 and substituting , the integral becomes
Thus the pool contains ft of water.
34. (a) The total amount of water supplied each hour to the region within feet of the sprinkler is
 10
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.4 Double Integrals in Polar Coordinates
R
V
areaofregion
=
V
! R
2
=
2 1"Re
"R
"e
"R( )
R
2
3
!
1/ 2
1
!
1"x
2
x
xydydx+ !
1
2
!
0
x
xydydx+ !
2
2
!
0
4"x
2
xydydx
= !
0
! /4
!
1
2
r
3
cos sin dr d = !
0
! /4
r
4
4
cos sin 
r=2
r=1
d 
=
15
4
!
0
! /4
sin cos d =
15
4
sin
2
 
2
! /4
0
=
15
16
!!
D
a
e
"(x
2
+y
2
)
dA= !
0
2!
!
0
a
re
"r
2
dr d =2! "
1
2
e
"r
2 a
0
=! 1"e
"a
2
( ) a
lim
a()
! 1"e
"a
2
( )=! e"a
2
(0 a() !
")
)
!
")
)
e
"(x
2
+y
2
)
dA=!
!!
S
a
e
"(x
2
+y
2
)
dA=!
"a
a
!
"a
a
e
"x
2
e
"y
2
dxdy= !
"a
a
e
"x
2
dx !
"a
a
e
"y
2
dy a
! =!!
R
2
"(x
2
+y
2
)dA
! =lim
a()
!!
S
a
e
"(x
2
+y
2
)
dA=lim
a()
!
"a
a
e
"x
2
dx !
"a
a
e
"y
2
dy = !
")
)
e
"x
2
dx !
")
)
e
"y
2
dy .
lim
a()
!
"a
a
e
"x
2
dx !
"a
a
e
"y
2
dy
(b) The average amount of water per hour per square foot supplied to the region within feet of the
sprinkler is ft (per hour per square foot). See the definition
of the average value of a function on page 1022 [ET 986].
35.
36. (a) for each . Then
 since as . Hence .
(b) for each .
Then, from (a), , so
To evaluate , we are using the fact that these integrals are bounded.
 11
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.4 Double Integrals in Polar Coordinates
"1,1 0<e
"x
2
 1 ") ,"1( ) 0<e
"x
2
 e
x
1,)( ) 0<e
"x
2
<e
"x
0 !
")
)
e
"x
2
dx !
")
"1
e
x
dx+!
"1
1
dx+!
1
)
e
"x
dx=2(e
"1
+1)
!
")
)
e
"x
2
dx !
")
)
e
"y
2
dy =! y x !
")
)
e
"x
2
dx
2
=!
!
")
)
e
"x
2
dx=* ! e
"x
2
& 0 x !
")
)
e
"x
2
dx= !
t= 2 x !
")
)
e
"x
2
dx= !
")
)
1
2
e
"t
2
/2( )dt ! = 1
2
!
")
)
e
"t
2
/2
dt !
")
)
e
"t
2
/2
dt= 2!
u=x dv=xe
"x
2
dx du=dx v="
1
2
e
"x
2
!
0
)
x
2
e
"x
2
dx = lim
t()
!
0
t
x
2
e
"x
2
dx=lim
t()
"
1
2
xe
"x
2 t
0
+!
0
t
1
2
e
"x
2
dx
= lim
t()
"
1
2
te
"t
2
+
1
2
!
0
)
e
"x
2
dx=0+
1
2
!
0
)
e
"x
2
dx [by l’Hospital’s Rule]
=
1
4
!
")
)
e
"x
2
dx [since e
"x
2
is an even function]
=
1
4
! [by Exercise 36(c)]
u= x u
2
=x%dx=2udu%
!
0
)
x e
"x
dx = lim
t()
!
0
t
x e
"x
dx=lim
t()
!
0
t
ue
"u
2
2udu=2!
0
)
u
2
e
"u
2
du
= 2
1
4
! [by part(a)]=
1
2
!
This is true since on , while on , and on , .
Hence .
(c) Since and can be replaced by , implies that
 . But for all , so .
(d) Letting , , so that or .
37. (a)We integrate by parts with and . Then and , so
(b) Let . Then 
 12
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.4 Double Integrals in Polar Coordinates
Q = 
D
 (x,y)dA= 
1
3
 
0
2
(2xy+y
2
)dydx= 
1
3
xy
2
+
1
3
y
3 y=2
y=0
dx
= 
1
3
4x+
8
3
dx= 2x
2
+
8
3
x
3
1
=16+
16
3
=
64
3
C
Q = 
D
 (x,y)dA= 
D
(x+y+x
2
+y
2
)dA= 
0
2!
 
0
2
(rcos" +rsin" +r
2
)r dr d"
= 
0
2!
 
0
2
dr d" = 
0
2!
1
3
r
3
(cos" +sin" )+
1
4
r
4 r=2
r=0
d"
= 
0
2!
8
3
(cos" +sin" )+4 d" =
8
3
(sin" !cos" )+4"
2!
0
=8!C
m= 
D
# (x,y)dA= 
0
2
 
!1
1
xy
2
dydx= 
0
2
xdx 
!1
1
y
2
dy=
1
2
x
2 2
0
1
3
y
3 1
!1
=2"
2
3
=
4
3
x=
1
m
 
D
x# (x,y)dA=
3
4
 
0
2
 
!1
1
x
2
y
2
dydx=
3
4
 
0
2
x
2
dx 
!1
1
y
2
dy=
3
4
1
3
x
3 2
0
1
3
y
3 1
!1
=
3
4
"
8
3
"
2
3
=
4
3
y=
1
m
 
D
y# (x,y)dA=
3
4
 
0
2
 
!1
1
xy
3
dydx=
3
4
 
0
2
xdx 
!1
1
y
3
dy=
3
4
1
2
x
2 2
0
1
4
y
4 1
!1
=
3
4
" 2" 0=0
x,y( )=
4
3
,0
m= 
D
# (x,y)dA= 
0
a
 
0
b
cxydydx=c 
0
a
xdx 
0
b
ydy=c
1
2
x
2 a
0
1
2
y
2 b
0
=
1
4
a
2
b
2
c
M
y
= 
D
x# (x,y)dA= 
0
a
 
0
b
cx
2
ydydx=c 
0
a
x
2
dx 
0
b
ydy=c
1
3
x
3 a
0
1
2
y
2 b
0
=
1
6
a
3
b
2
c
M
x
= 
D
y# (x,y)dA= 
0
a
 
0
b
cxy
2
dydx=c 
0
a
xdx 
0
b
y
2
dy=c
1
2
x
2 a
0
1
3
y
3 b
0
=
1
6
a
2
b
3
c
x,y( )=
M
y
m
,
M
x
m
=
2
3
a,
2
3
b
1.
2.
3. ,
 ,
 .
Hence, .
4. ,
 , and
 .
Hence, .
 1
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.5 Applications of Double Integrals
m = 
0
2
 
x/2
3!x
(x+y)dydx= 
0
2
xy+
1
2
y
2 y=3!x
y=x/2
dx= 
0
2
x 3!
3
2
x +
1
2
(3!x)
2
!
1
8
x
2
dx
= 
0
2
!
9
8
x
2
+
9
2
dx= !
9
8
1
3
x
3
+
9
2
x
2
0
=6
M
y
= 
0
2
 
x/2
3!x
(x
2
+xy)dydx= 
0
2
x
2
y+
1
2
xy
2 y=3!x
y=x/2
dx= 
0
2
9
2
x!
9
8
x
3
dx=
9
2
M
x
= 
0
2
 
x/2
3!y
(xy+y
2
)dydx= 
0
2
1
2
xy
2
+
1
3
y
3 y=3!x
y=x/2
dx=0
2
9!
9
2
x dx=9 m=6
x,y( )=
M
y
m
,
M
x
m
=
3
4
,
3
2
m= 
0
1
 
y
4!3y
xdxdy= 
0
1
1
2
(4!3y)
2
!
1
2
y
2
dy= !
1
18
(4!3y)
3
!
1
6
y
3 1
0
=
10
3
M
y
= 
0
1
 
y
4!3y
x
2
dxdy= 
0
1
1
3
(4!3y)
3
!
1
3
y
3
dy= !
1
36
(4!3y)
4
!
1
12
y
4 1
0
=7
M
x
= 
0
1
 
y
4!3y
xydxdy= 
0
1
1
2
y(4!3y)
2
!
1
2
y
3
dy= 
0
1
(8y!12y
2
+4y
3
)dy=1
m=
10
3
x,y( )= 2.1,0.3( )
m= 
0
1
 
0
e
x
ydydx= 
0
1
1
2
y
2 y=e
x
y=0
dx=
1
2
 
0
1
e
2x
dx=
1
4
e
2x 1
0
=
1
4
(e
2
!1)
M
y
= 
0
1
 
0
e
x
xydydx=
1
2
 
0
1
xe
2x
dx=
1
2
1
2
xe
2x
!
1
4
e
2x 1
0
=
1
8
(e
2
+1)
M
x
= 
0
1
 
0
e
x
y
2
dydx= 
0
1
1
3
y
3 y=e
x
y=0
dx=
1
3
 
0
1
e
3x
dx=
1
3
1
3
e
3x 1
0
=
1
9
(e
3
!1)
m=
1
4
(e
2
!1) x,y( )=
1
8
(e
2
+1)
1
4
(e
2
!1)
,
1
9
(e
3
!1)
1
4
(e
2
!1)
=
e
2
+1
2(e
2
!1)
,
4(e
3
!1)
9(e
2
!1)
5.
 ,
 , and
 . Hence ,
 .
6. ,
 ,
 .
Hence , .
7. ,
 , and
 .
Hence , .
8.
 2
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.5 Applications of Double Integrals
m= 
0
1
 
0
x
xdydx= 
0
1
x y
y= x
y=0
dx= 
0
1
x
3/2
dx=
2
5
x
5/2 1
0
=
2
5
M
y
= 
0
1
 
0
x
x
2
dydx= 
0
1
x y
y= x
y=0
dx= 
0
1
x
5/2
dx=
2
7
x
7/2 1
0
=
2
7
M
x
= 
0
1
 
0
x
yxdydx= 
0
1
x
1
2
y
2 y= x
y=0
dx=
1
2
 
0
1
x
2
dx=
1
2
1
3
x
3 1
0
=
1
6
m=
2
5
x,y( )=
2/7
2/5
,
1/6
2/5
=
5
7
,
5
12
m= 
!1
2
 
y
2
y+2
3dxdy= 
!1
2
(3y+6!3y
2
)dy=
27
2
M
y = 
!1
2
 
y
2
y+2
3xdxdy= 
!1
2
3
2
(y+2)
2
!y
4
dy
=
1
2
(y+2)
3
!
3
10
y
5 2
!1
=
108
5
M
x = 
!1
2
 
y
2
y+2
3ydxdy= 
!1
2
(3y
2
+6y!3y
3
)dy
= y
3
+3y
2
!
3
4
y
4 2
!1
=
27
4
m=
27
2
x,y( )=
8
5
,
1
2
m= 
0
! /2
 
0
cos x
xdydx= 
0
! /2
xcos xdx= xsin x+cos x
! /2
0
=
!
2
!1
M
y
= 
0
! /2
 
0
cos x
x
2
dydx= 
0
! /2
x
2
cos xdx= x
2
sin x+2xcos x!2sin x
! /2
0
=
!
2
4
!2
 ,
 , and
 .
Hence , .
9.
 ,
and
Hence , .
10. ,
 , and
 3
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.5 Applications of Double Integrals
M
x
= 
0
! /2
 
0
cos x
xydydx= 
0
! /2
1
2
xcos
2
xdx=
1
2
1
4
x
2
+
1
4
xsin 2x+
1
8
cos 2x
! /2
0
=
!
2
32
!
1
8
m=
! !2
2
x,y( )=
!
2
!8
2 ! !2( )
,
! +2
16
# (x,y)=ky=krsin" m= 
0
! /2
 
0
1
kr
2
sin" dr d" =
1
3
k 
0
! /2
sin" d" =
1
3
k !cos"
! /2
0
=
1
3
k
M
y
= 
0
! /2
 
0
1
kr
3
sin" cos" dr d" =
1
4
k 
0
! /2
sin" cos" d" =
1
8
k !cos 2"
! /2
0
=
1
8
k
M
x
= 
0
! /2
 
0
1
kr
3
sin
2
" dr d" =
1
4
k 
0
! /2
sin
2
" d" =
1
8
k " +sin 2"
! /2
0
=
!
16
k
x,y( )=
3
8
,
3!
16
# (x,y)=k(x
2
+y
2
)=kr
2
m= 
0
! /2
 
0
1
kr
3
dr d" =
!
8
k
M
y
= 
0
! /2
 
0
1
kr
4
cos" dr d" =
1
5
k 
0
! /2
cos" d" =
1
5
k sin"
! /2
0
=
1
5
k
M
x
= 
0
! /2
 
0
1
kr
4
sin" dr d" =
1
5
k 
0
! /2
sin" d" =
1
5
k !cos"
! /2
0
=
1
5
k
x,y( )=
8
5!
,
8
5!
0,0( ) # (x,y)=k(x
2
+y
2
)
m = 
0
a
 
0
a!x
k x
2
+y
2( )dydx=k 
0
a
ax
2
!x
3
+
1
3
a!x( )
3
dx
= k
1
3
ax
3
!
1
4
x
4
!
1
12
a!x( )
4 a
0
=
1
6
ka
4
M
y
=M
x = 
0
a
 
0
a!x
ky(x
2
+y
2
)dydx=k 
0
a
1
2
(a!x)
2
x
2
+
1
4
(a!x)
4
dx
=k
1
6
a
2
x
3
!
1
4
ax
4
+
1
10
x
5
!
1
20
(a!x)
5 a
0
=
1
15
ka
5
 .
Hence , .
11. , ,
 ,
 .
Hence .
12. , ,
 ,
 .
Hence .
13. Placing the vertex opposite the hypotenuse at , . Then
By symmetry,
 4
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.5 Applications of Double Integrals
x,y( )=
2
5
a,
2
5
a .
# (x,y)=k/ x
2
+y
2
=k/r
m = 
! /6
5! /6
 
1
2sin"
k
r
r dr d" =k 
! /6
5! /6
[(2sin" )!1]d"
=k !2cos" !"
5! /6
! /6
=2k 3!
!
3
D f x( )=x M
y
=0
M
x = 
! /6
5! /6
 
1
2sin"
krsin" dr d" =
1
2
k 
! /6
5! /6
(4sin
3
" !sin" )d"
=
1
2
k !3cos" +
4
3
cos
3
"
5! /6
! /6
= 3 k
x,y( )= 0,
3 3
2 3 3 !!( )
.
I
x = 
D
y
2
# x,y( ) dA= 
0
1
 
0
e
x
y
2
" y dydx= 
0
1
1
4
y
4 y=e
x
y=0
dx=
1
4
 
0
1
e
4x
dx
=
1
4
1
4
e
4x 1
0
=
1
16
(e
4
!1)
I
y = 
D
x
2
# (x,y)dA= 
0
1
 
0
e
x
x
2
ydydx= 
0
1
x
2 1
2
y
2 y=e
x
y=0
dx=
1
2
 
0
1
x
2
e
2x
dx
Hence
14.
 ,
By symmetry of and , , and
Hence
15.
 ,
 5
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.5 Applications of Double Integrals
=
1
2
1
2
x
2
!
1
2
x+
1
4
e
2x 1
0
=
1
8
(e
2
!1)
I
0
=I
x
+I
y
=
1
16
(e
4
!1)+
1
8
(e
2
!1)=
1
16
(e
4
+2e
2
!3)
I
x
= 
0
! /2
 
0
1
(r
2
sin
2
" )(kr
2
)r dr d" =
1
6
k 
0
! /2
sin
2
" d" =
1
6
k
1
4
(2" !sin 2" )
! /2
0
=
!
24
k
I
y
= 
0
! /2
 
0
1
(r
2
cos
2
" )(kr
2
)r dr d" =
1
6
k 
0
! /6
cos
2
" d" =
1
6
k
1
4
(2" +sin 2" )
! /2
0
=
!
24
k
I
0
=I
x
+I
y
=
!
12
k
I
x
= 
!1
2
 
y
2
y+2
3y
2
dxdy= 
!1
2
(3y
3
+6y
2
!3y
4
)dy=
3
4
y
4
+2y
3
!
3
5
y
5 2
!1
=
189
20
I
y
= 
!1
2
 
y
2
y+2
3x
2
dxdy= 
!1
2
(y+2)
3
!y
6
dy=
1
4
(y+2)
4
!
1
7
y
7 2
!1
=
1269
28
I
0
=I
x
+I
y
=
1917
35
x ! y !
x !
I
x = 
D
y
2
# (x,y)dA= 
0
2
 
0
2
y
2
(1+0.1x)dydx= 
0
2
(1+0.1x)
1
3
y
3 y=2
y=0
dx
=
8
3
 
0
2
(1+0.1x)dx=
8
3
x+0.1"
1
2
x
2 2
0
=
8
3
(2.2)#5.87
!
I
y = 
D
x
2
# (x,y)dA= 
0
2
 
0
2
x
2
(1+0.1x)dydx= 
0
2
x
2
(1+0.1x) y
y=2
y=0
dx
= 2 
0
2
(x
2
+0.1x
3
)dx=2
1
3
x
3
+0.1"
1
4
x
4 2
0
=2
8
3
+0.4 #6.13
 [ integrate by parts twice]
 ,
and .
16. ,
 , and
 .
17. ,
 , and .
18. If we find the moments of inertia about the and axes, we can determine in which direction
rotation will be more difficult. (See the explanation following Example 4.) The moment of inertiaabout the axis is given by
Similarly, the moment of inertia about the y axis is given by
Since
 6
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.5 Applications of Double Integrals
I
y
>I
x
y !
x=
1
m
 
D
x# (x,y)dA=
8
!
2
 
0
!
 
0
sin x
x
2
ydydx=
2!
3
!
1
!
y=
1
m
 
D
y# (x,y)dA=
8
!
2
 
0
!
 
0
sin x
xy
2
dydx=
16
9!
x,y( )=
2!
3
!
1
!
,
16
9!
I
x
= 
D
y
2
# (x,y)dA= 
0
!
 
0
sin x
xy
3
dydx=
3!
2
64
I
y
= 
D
x
2
# (x,y)dA= 
0
!
 
0
sin x
x
3
ydydx=
!
2
16
(!
2
!3) I
0
=I
x
+I
y
=
!
2
64
(4!
2
!9)
m= 
D
x
2
+y
2
dA= 
0
2!
 
0
1+cos"
r
2
dr d" =
5
3
!
x=
1
m
 
D
x x
2
+y
2
dA=
3
5!
 
0
2!
 
0
1+cos"
r
3
cos" dr d" =
21
20
y=
1
m
 
D
y x
2
+y
2
dA=
3
5!
 
0
2!
 
0
1+cos"
r
3
sin" dr d" =0 x,y( )=
21
20
,0
I
x
= 
D
y
2
x
2
+y
2
dA= 
0
2!
 
0
1+cos"
r
4
sin
2
" dr d" =
33
40
!
I
y
= 
D
x
2
x
2
+y
2
dA= 
0
2!
 
0
1+cos"
r
4
cos
2
" dr d" =
93
40
! I
0
=I
x
+I
y
=
63
20
!
I
x
= 
0
a
 
0
a
# y
2
dxdy=# 
0
a
dx 
0
a
y
2
dy=# x
a
0
1
3
y
3 a
0
=# a
1
3
a
3
=
1
3
# a
4
=I
y
m=# a
2
x
2
=
I
y
m
$ x=
1
3
# a
4
/(# a
2
)
1/2
=
1
3
a
y
2
=
I
x
m
$ y=
1
3
a
 , more force is required to rotate the fan blade about the axis.
19. Using a CAS, we find . Then and
 , so .
The moments of inertia are ,
 , and .
20. Using a CAS, we find ,
 and
 , so .
The moments of inertia are ,
 , and .
21. by symmetry, and 
since the lamina is homogeneous. Hence and
 .
22.
 7
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.5 Applications of Double Integrals
m = 
0
!
 
0
sin x
# dydx=# 
0
!
sin xdx=# !cos x
!
0
=2#
I
x = 
0
!
 
0
sin x
# y
2
dydx=
1
3
# 
0
!
sin
3
xdx=
1
3
# 
0
!
(1!cos
2
x)sin xdx
=
1
3
# !cos" +
1
3
cos
3
"
!
0
=
4
9
#
I
y = 
0
!
 
0
sin x
# x
2
dydx=# 
0
!
x
2
sin xdx
=# !x
2
cos x+2xsin x+2cos x
!
0
=# (!
2
!4)
y
2
=
I
x
m
=
2
9
y=
2
3
x
2
=
I
y
m
=
!
2
!4
2
x=
!
2
!4
2
f (x,y) 
R
2
f (x,y)dA=1 f (x,y)=0
0,1 % 0,2
 
R
2
f (x,y)dA
= 
!&
&
 
!&
&
f (x,y)dydx= 
0
1
 
0
2
Cx(1+y)dydx
= C 
0
1
x y+
1
2
y
2 y=2
y=0
dx=C 
0
1
4xdx=C 2x
2 1
0
=2C
2C=1$C=
1
2
P X' 1,Y' 1( ) = 
!&
1
 
!&
1
f (x,y)dydx= 
0
1
 
0
1
1
2
x(1+y)dydx
= 
0
1
1
2
x y+
1
2
y
2 y=1
y=0
dx= 
0
1
1
2
x
3
2
dx=
3
4
1
2
x
2 1
0
=
3
8
0.375
P X+Y' 1( )=P X,Y( )(D( ) D
 ,
 ,
 .
Then , so and , so .
23. (a) is a joint density function, so we know . Since outside the
rectangle , we can say
Then .
(b)
 or 
(c) where is the triangular region shown in the figure. Thus
 8
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.5 Applications of Double Integrals
P X+Y' 1( )= 
D
f (x,y)dA= 
0
1
 
0
1!x
1
2
x(1+y)dydx
= 
0
1
1
2
x y+
1
2
y
2 y=1!x
y=0
dx= 
0
1
1
2
x
1
2
x
2
!2x+
3
2
dx
=
1
4
 
0
1
x
3
!4x
2
+3x( )dx= 1
4
x
4
4
!4
x
3
3
+3
x
2
2
1
0
=
5
48
#0.1042
f (x,y)) 0 f 
R
2
f (x,y)dA=1 f (x,y)=0
0,1 % 0,1 
R
2
f (x,y)dA= 
0
1
 
0
1
4xydydx= 
0
1
2xy
2 y=1
y=0
dx= 
0
1
2xdx= x
2 1
0
=1
f (x,y)
Y
P X)
1
2
= 
1/2
&
 
!&
&
f (x,y)dydx= 
1/2
1
 
0
1
4xydydx
= 
1/2
1
2xy
2 y=1
y=0
dx= 
1/2
1
2xdx= x
2 1
1/2
=
3
4
Y
P X)
1
2
= 
1/2
&
 
!&
&
f (x,y)dydx= 
1/2
1
 
0
1
4xydydx
= 
1/2
1
2xy
2 y=1
y=0
dx= 
1/2
1
2xdx= x
2 1
1/2
=
3
4
24. (a) , so is a joint density function if . Here, outside the
square , so .
Thus, is a joint density function.
(b)
(a) No restriction is placed on , so
(b) No restriction is placed on , so
 9
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.5 Applications of Double Integrals
P X)
1
2
,Y'
1
2
= 
1/2
&
 
!&
1/2
f (x,y)dydx= 
1/2
1
 
0
1/2
4xydydx
= 
1/2
1
2xy
2 y=1/2
y=0
dx= 
1/2
1
1
2
xdx=
1
2
"
1
2
x
2 1
1/2
=
3
16
P X)
1
2
,Y'
1
2
= 
1/2
&
 
!&
1/2
f (x,y)dydx= 
1/2
1
 
0
1/2
4xydydx
= 
1/2
1
2xy
2 y=1/2
y=0
dx= 
1/2
1
1
2
xdx=
1
2
"
1
2
x
2 1
1/2
=
3
16
X $
1
= 
R
2
x f (x,y)dA= 
0
1
 
0
1
x(4xy)dydx= 
0
1
2x
2
y
2 y=1
y=0
dx
=2 
0
1
x
2
dx=2
1
3
x
3 1
0
=
2
3
Y $
2
= 
R
2
y f (x,y)dA= 
0
1
 
0
1
y(4xy)dydx= 
0
1
4x
1
3
y
3 y=1
y=0
dx
=
4
3
 
0
1
xdx=
4
3
1
2
x
2 1
0
=
2
3
f (x,y)) 0 f 
R
2
f (x,y)dA=1 f (x,y)=0
 
R
2
f (x,y)dA
= 
0
&
 
0
&
0.1e
!(0.5x+0.2y)
dydx=0.1 
0
&
 
0
&
e
!0.5x
e
!0.2y
dydx
= 0.1 
0
&
e
!0.5x
dx 
0
&
e
!0.2y
dy=0.1lim
t*&
 
0
t
e
!0.5x
dx lim
t*&
 
0
t
e
!0.2y
dy
= 0.1lim
t*&
!2e
!0.5x t
0
lim
t*&
!5e
!0.2y t
0
= 0.1lim
t*&
!2(e
!0.5t
!1) lim
t*&
!5(e
!0.2t
!1)
(c)
(d)
(c) The expected value of is given by 
The expected value of is 
25. (a) , so is a joint density function if . Here, outside the first
quadrant, so
 10
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.5 Applications of Double Integrals
= (0.1)" (!2)(0!1)" (!5)(0!1)=1
f (x,y)
X
P(Y ) 1)= 
!&
&
 
1
&
f (x,y)dydx= 
0
&
 
1
&
0.1e
!(0.5x+0.2y)
dydx
= 0.1 
0
&
e
!0.5x
dx 
1
&
e
!0.2y
dy=0.1lim
t*&
 
0
t
e
!0.5x
dx lim
t*&
 
1
t
e
!0.2y
dy
= 0.1lim
t*&
!2e
!0.5x t
0
lim
t*&
!5e
!0.2y t
1
= 0.1lim
t*&
!2(e
!0.5t
!1) lim
t*&
!5(e
!0.2t
!e
!0.2
)
= (0.1)" (!2)(0!1)" (!5)(0!e
!0.2
)=e
!0.2
#0.8187
X
P(Y ) 1)= 
!&
&
 
1
&
f (x,y)dydx= 
0
&
 
1
&
0.1e
!(0.5x+0.2y)
dydx
= 0.1 
0
&
e
!0.5x
dx 
1
&
e
!0.2y
dy=0.1lim
t*&
 
0
t
e
!0.5x
dx lim
t*&
 
1
t
e
!0.2y
dy
= 0.1lim
t*&
!2e
!0.5x t
0
lim
t*&
!5e
!0.2y t
1
= 0.1lim
t*&
!2(e
!0.5t
!1) lim
t*&
!5(e
!0.2t
!e
!0.2
)
= (0.1)" (!2)(0!1)" (!5)(0!e
!0.2
)=e
!0.2
#0.8187
P(X' 2,Y' 4) = 
!&
2
 
!&
4
f (x,y)dydx= 
0
2
 
0
4
0.1e
!(0.5x+0.2y)
dydx
=0.1 
0
2
e
!0.5x
dx 
0
4
e
!0.2y
dy=0.1 !2e
!0.5x 2
0
!5e
!0.2y 4
0
=(0.1)" (!2)(e
!1
!1)" (!5)(e
!0.8
!1)
=(e
!1
!1)(e
!0.8
!1)=1+e
!1.8
!e
!0.8
!e
!1
#0.3481
Thus is a joint density function.
(b)
(a) No restriction is placed on , so
(b)