Evaluate the triple integral H of (x^2 + y^2 + z^2)^2 d, where H is the solid hemisphere x^2 + y^2 + z^2 less than or equal to 9 with z greater than or equal to 0

Question

amistre64 · Answer

what are your thoughts on it?

anonymous · Answer

well I understand how to take a triple integral, but I always get confused on the limits.  Would I put it in terms of the spherical coordinate system? such as 

$$\int\limits_{0}^{\pi} \int\limits_{0}^{2\pi} \int\limits_{0}^{3} (e^2 \sin \phi)^2 de d \Theta d \phi$$

anonymous · Answer

I would say that's almost right. Check your upper bound on $\phi$. What you have suggests that we also consider the half of the sphere below the $x$-$y$ plane.

anonymous · Answer

$$z \ge0$$, so it wouldn't be below the xy-plane right?

anonymous · Answer

That's right. However, $z=e\cos\phi$. But if $\dfrac{\pi}{2}<\phi<\pi$, you have $-1<\cos\phi<0$, which would mean $z$ is negative. This means you have to use $0\le \phi\le\dfrac{\pi}{2}$ as your interval.

anonymous · Answer

ohh okay that makes a lot of sense, so if I change phi the rest of the set up is correct?

anonymous · Answer

Yes

anonymous · Answer

is the answer $$\frac{ 243\pi^2 }{ 10 }$$

anonymous · Answer

Yep I'm getting the same solution.