Question

Evaluate this integral using spherical coordinates

Answer 1

Please help! Medals here!

Answer 2

Pretty much everything you need here:
\[\begin{cases}
x=\rho\cos\theta\sin\phi\\
y=\rho\sin\theta\sin\phi\\
z=\rho\cos\phi\\
x^2+y^2+z^2=\rho^2
\end{cases}\]
To determine the proper limits, think about the bounds for each variable. \(z\) ranges from the bottom half of a sphere of radius \(a\) to the top half. In the x-y plane, \(x\) ranges from the left half of a circle with radius \(a\) to the right half, while \(y\) ranges from \(-a\) to \(a\) (i.e. spans the diameter of the circle).

Answer 3

Yes, but I stuck halfway in my calculation

Answer 4

Alright, what do you have so far?

Answer 5

\[\int\limits_{0}^{\pi}\int\limits_{0}^{2\pi}\int\limits_{-a}^{a}\rho ^{4}\sin ^{3}\phi \cos \phi \cos ^{2}\theta+\rho ^{4}\sin ^{3}\phi \cos \phi \sin ^{2}\theta+ \rho ^{5} \sin \phi \cos ^{3}\phi d \rho d \theta d \phi\]

Answer 6

I cannot handle this crazy stuff

Answer 7

Your \(\rho\)s should all be raised to the 5th

Answer 8

I don't see that

Answer 9

oh, I see now!

Answer 10

It comes from the Jacobian for the spherical conversion:
\[dx~dy~dz=\rho^2\sin\phi\]
You might also consider taking advantage of symmetry.
\[\int_{-a}^a=2\int_0^a\]Anyway, the integration with respect to \(\rho\) shouldn't be too difficult. Power rule gives you
\[2\int_0^{2\pi}\int_0^\pi\left[\frac{\rho^6}{6}\right]_0^a\left(\cos^2\theta\sin^3\phi\cos\phi+\sin^2\theta\sin^3\phi\cos\phi+\cos^3\phi\sin\phi\right)~d\phi~d\theta\]
\[\frac{a^6}{3}\int_0^{2\pi}\int_0^\pi\left(\color{red}{\cos^2\theta}\sin^3\phi\cos\phi+\color{red}{\sin^2\theta}\sin^3\phi\cos\phi+\cos^3\phi\sin\phi\right)~d\phi~d\theta\]
What identity can you use to consolidate the first two terms?

Answer 11

so sin^3phi cos phi + cos^3phi sin phi from 0 to pi

Answer 12

my calculator gives 0, but I don't know why

Answer 13

Shouldn't be zero, you're finding the volume of a sphere.
\[\frac{a^6}{3}\int_0^{2\pi}\int_0^\pi\left(\sin^3\phi\cos\phi+\cos^3\phi\sin\phi\right)~d\phi~d\theta\]
Notice that there are no \(\theta\)s remaining, so you can split the integrals:
\[\frac{a^6}{3}\int_0^{2\pi}d\theta\int_0^\pi\left(\sin^3\phi\cos\phi+\cos^3\phi\sin\phi\right)~d\phi\]
and the \(\theta\) integral simplifies to \(2\pi\):
\[\frac{2\pi a^6}{3}\int_0^\pi\left(\sin^3\phi\cos\phi+\cos^3\phi\sin\phi\right)~d\phi\]
For these two terms, consider two substitutions: for the first, \(t=\sin\phi\), and for the second, \(u=\cos\phi\). You then have
\[\frac{2\pi a^6}{3}\left(\underbrace{\int_0^0 t^3~dt}_{\text{this integral is zero}}-\underbrace{\int_{1}^{-1}u^3~du}_{\text{this one is not}}\right)\]

Answer 14

http://www.wolframalpha.com/input/?i=integrate+sin%5E3%28phi%29*cos%28phi%29%2Bcos%5E3%28phi%29*sin%28phi%29+from+0+to+pi+

Answer 15

I check my work on wolfram alpha, and it agrees with my calculator

Answer 16

Anything going wrong?

Answer 17

Only on my end! I must have forgotten that we're NOT taking the volume of a sphere, we have that nasty function that's being integrated. Sorry for the confusion!

Ignore the note I made about the second integral being non-zero, that's just not true. The answer is indeed 0.

Answer 18

I was just checking here:
http://www.wolframalpha.com/input/?i=Integrate%5BIntegrate%5BIntegrate%5Bx%5E2z%2By%5E2z%2Bz%5E3%2C%7Bz%2C-Sqrt%5Ba%5E2-x%5E2-y%5E2%5D%2CSqrt%5Ba%5E2-x%5E2-y%5E2%5D%7D%5D%2C%7Bx%2C-Sqrt%5Ba%5E2-y%5E2%5D%2CSqrt%5Ba%5E2-y%5E2%5D%7D%5D%2C%7By%2C-a%2Ca%7D%5D

Answer 19

Great thx!

Answer 20

You're welcome!