calculus - Triple Integral $ _iiint_limits_S_frac1{_sqrt{_left(x-a_right)^{2}+_left(y-b_right)^{2}+_left(z-c_right)^{2}}}_;dx_;dy_;dz $ - Mathematics Stack Exchange

Prévia do material em texto

11/7/2019 calculus - Triple Integral $ \iiint\limits_S\frac1{\sqrt{\left(x-a\right)^{2}+\left(y-b\right)^{2}+\left(z-c\right)^{2}}}\;d…
https://math.stackexchange.com/questions/474441/triple-integral-iiint-limits-s-frac1-sqrt-leftx-a-right2-lefty-b-rig 1/3
Mathematics Stack Exchange is a question and answer
site for people studying math at any level and
professionals in related fields. Join them; it only takes a
minute:
Sign up
Here's how it works:
Anybody can ask a question
Anybody can answer
The best answers are voted up and rise to the top
Triple Integral ∭
S
1
√ ( x− a ) 2 + ( y− b ) 2 + ( z− c ) 2
dx dy dz
2
4
I used spherical coordinates and couldn't calculate this triple integral.
∭S
1
√(x − a)2 + (y − b)2 + (z − c)2
dx dy dz
where is a solid sphere of radius and center at the origin, and is a fixed point outside this sphere.S R (a, b, c)
 calculus integration multivariable-calculus definite-integrals
edited Oct 5 '16 at 16:05 asked Aug 23 '13 at 16:09
user91500
3,676 9 46 105
7 – That's a triple integral alright. Did you mean to ask a question about it? Henning Makholm Aug 23 '13 at 16:15
4
 – 
In case you have a physics background: this is essentially the electric potential at generated by the uniformly
charged ball .
(a, b, c)
S Start wearing purple Aug 23 '13 at 16:33
2 – Isn't this a sphere centered at ?(a, b, c) Ataraxia Aug 23 '13 at 16:53
2 – 
 
@Ataraxia Looks like the integrand is the reciprocal of the distance of to .(x, y, z) (a, b, c) rschwieb Aug 23 '13 at
17:12
2 – @rschwieb Ah ok. I didn't take notice of the minus sign. Ataraxia Aug 23 '13 at 17:21By using our site, you acknowledge that you have read and understand our , , and our
.
Cookie Policy Privacy Policy
Terms of Service
11/7/2019 calculus - Triple Integral $ \iiint\limits_S\frac1{\sqrt{\left(x-a\right)^{2}+\left(y-b\right)^{2}+\left(z-c\right)^{2}}}\;d…
https://math.stackexchange.com/questions/474441/triple-integral-iiint-limits-s-frac1-sqrt-leftx-a-right2-lefty-b-rig 2/3
1 Answer
5
Method 1
This integral is actually very easy to evaluate because for any , the function satisfies the
 away from .
→p ∈ R3 ϕ(→x) =
1
| →x − →p |
Laplace's equation →p
∇
2ϕ(→x) = 0 for x ∈ R3 ∖ {→p}
This sort of function is called and one of its most useful property is the so called Harmonic function mean
value property:
If is a function harmonic over a ball centered at with radius , thenϕ B(→q, R) →q R
ϕ(→q) =
1
4πR2 ∫∂B ( →q ,R )ϕ(
→x)dS(→x) =
3
4πR3 ∫B ( →q ,R )ϕ(
→x)dV(→x)
i.e. the average of over the surface and the body of the ball are both equal to the value of at
the center. Apply this to the integral at hand, one can read off the value of the integral as:
ϕ(→x) B(→q, R) ϕ(→x)
4πR3
3√a2 + b2 + c2
Method 2
If one insists to evaluate the integral using polar-coordinates, one way to do it is to use the so called 
.
multipole
expansion
Let and be the angle between the two vectors and .
Let and , then
where are the .
→p = (a, b, c) θ →x →p
r < = min ( |
→x | , | →p | ) r > = max ( |
→x | , | →p | )
1
| →x − →p |
=
∞
∑
n= 0
rn<
rn+ 1>
Pn(cosθ)
Pn(cosθ) Legendre's polynomials
In our cases, is outside the sphere, the triple integral can be rewritten as→p
∞
∑
n= 0
2π
| →p | n+ 1
(∫R0 rn+ 2dr)(∫π0Pn(cosθ)sinθdθ)
Using
By using our site, you acknowledge that you have read and understand our , , and our
.
Cookie Policy Privacy Policy
Terms of Service
11/7/2019 calculus - Triple Integral $ \iiint\limits_S\frac1{\sqrt{\left(x-a\right)^{2}+\left(y-b\right)^{2}+\left(z-c\right)^{2}}}\;d…
https://math.stackexchange.com/questions/474441/triple-integral-iiint-limits-s-frac1-sqrt-leftx-a-right2-lefty-b-rig 3/3
∫π0Pn(cosθ)sinθdθ = {2, n = 00, n > 0
one can evaluate the triple integral as:
4π
| →p |
(∫R0 r2dr) = 4πR3
3√a2 + b2 + c2
Method 3
If one really want to evaluate the integral in the most elementary manner, then one should use rotational
symmetry to simplify the problem. We will assume and in terms of cylindrical polar
coordinate, the triple integral becomes:
(a, b, c) = (0, 0, − p)
2π∫R−R(∫√R2 − z20 rdr√r2 + (z + p)2 )dz
= 2π∫
R
−R[√r2 + (z + p)2]√R
2 − z2
0 dz
= 2π∫R−R(√R2 + 2zp + p2 − (z + p))dz
= 2π[ 13p√R2 + 2zp + p23 − ( z22 + pz)]R−R
= 2π( 13p ((p + R)3 − (p − R)3) − 2pR)
=
4πR3
3√a2 + b2 + c2
edited Apr 22 '15 at 9:47
user91500
3,676 9 46 105
answered Aug 23 '13 at 17:54
achille hui
98.5k 5 134 268
1
 – 
Thanks for the answer. But this question is a problem in the Apostol calculus vol II (11.34.15) and it's before the surface
integral. user91500 Aug 23 '13 at 18:10
3 – @Artin, I updated answer to include the method which should be covered by Apostol.3rd achille hui Aug 23 '13 at
19:02
2 – What does means ?r > Hexacoordinate-C Oct 5 '16 at 16:17
 – OP asks method 3 obviously. Takahiro Waki Jun 16 '17 at 17:40
By using our site, you acknowledge that you have read and understand our , , and our
.
Cookie Policy Privacy Policy
Terms of Service