Question

Let W be the region between concentric spheres centered at the origin of radii a and b, a < b. Consider the triple integral
dV/(x^2 + y^2 + z^2)^(3/2) Convert this integral to spherical coordinates.
:
(a) Convert this integral to spherical coordinates.

Answer 1

I was able to get the integral, but I am confused about how to compute it

Answer 2

\[x=\rho\sin\phi\cos\theta\]\[y=\rho\sin\phi\sin\theta\]\[z=\rho\cos\phi\]bounds\[a\le\rho\le b\]\[-\frac\pi2\le\phi\le\frac\pi2\]\[0\le\theta\le2\pi\]

Answer 3

\[x^2+y^2+z^2=\rho^2\sin^2\phi\cos^2\theta+\rho^2\sin^2\phi\sin^2\theta+\rho^2\cos\phi\]\[=\rho^2\sin^2\phi\cancel{(\cos^2\theta+\sin^2\theta)}^1+\rho^2\cos\phi\]\[=\rho^2\cancel{(\sin^2\phi+\cos^2\phi)}^1=\rho^2\]

Answer 4

part b of the question ask to compute the integral.
i got the bound \[0\le \Phi \le \pi \]

Answer 5

that should work out to the same answer

Answer 6

for the integral i got \[\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi}\int\limits_{a}^{b}\rho ^{2/3}*\sin \Phi \delta \rho \delta \phi \delta \theta\]

Answer 7

so\[(x^2+y^2+z^2)^{3/2}=\rho^3\]so\[\int\int\int{dV\over\rho^3}=\int\int\int{\rho^2\sin\phi d\rho d\phi d\theta\over\rho^3}=\int_0^{2\pi}\int_{0}^{\pi}\int_a^b{\sin\phi d\rho d\phi d\theta\over\rho}\]

Answer 8

I think you made an algebra error

Answer 9

does the answer come out to 0?

Answer 10

I dunno, I haven't tried it yet...

Answer 11

...shouldn't be though, it's a volume

Answer 12

thats what i thought

Answer 13

\[\int_0^{2\pi}\int_0^\pi\ln(\frac ba)\sin\phi d\phi d\theta=\ln(\frac ba)\int_0^{2\pi}\left.-\cos\phi\right]_0^\pi d\theta\]\[=2\ln(\frac ab)\int_0^{2\pi}d\theta=4\pi\ln(\frac ab)\]at least that's what I get the first time around

Answer 14

thank you that was very helpful!

Answer 15

welcome, I can only hope it's correct :)

Answer 16

thats what i got after fixing my algebra, just ln(b/a) in the last part