Triple integrals using spherical co-ordinates

Question

I need to calculate the following:

$ \int_{-2}^{2} \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} \int_{0}^{\sqrt{4-x^2-y^2}} z^2\sqrt{x^2+y^2+z^2} dzdydx $

I sketched this and I can see this is some kind a of a dome, which has ${0\leq z \leq 2}$.

I know how to work with polar coordinates, this seems like a problem fitting for spehrical coordinates, i.e. :

$$ x \to p\sin\phi \cos\theta $$

$$y \to p\sin\phi \sin\theta$$

$$z \to p\cos\phi $$

knowing that $ p = \sqrt{x^2+y^2+z^2} $ I can deduce that ${ 0 \leq p \leq 2}$, but how do I continue to find the angles?

Answer 1

Here, from the integrals, you can easily deduce that:

The function to integrate is $f(x,y,z)=z^2\sqrt{x^2+y^2+z^2}$ and the boundaries to integrate over are: $0 \leq z \leq \sqrt{4-x^2-y^2}$, $-\sqrt{4-x^2} \leq y \leq \sqrt{4-x^2}$ and $-2 \leq x \leq 2$.

Basically, you are integrating over a hemisphere of radius $2$, which lies on the positive side of the $z$ axis.

So, let

$$ x \to p\sin\phi \cos\theta $$

$$y \to p\sin\phi \sin\theta$$

$$z \to p\cos\phi $$

with the limits $ 0 \leq p \leq 2$, $ 0 \leq \phi \leq \pi/2$ and $ 0 \leq \theta \leq 2\pi$.

So the integral will become,

$ \displaystyle \int_{0}^{2} \int_{0}^{\pi/2} \int_{0}^{2\pi} (p^2\cos^2 \phi)(p) p^2\sin \phi \ d\theta d\phi dp$

$= (2\pi) \ \dfrac{p^6}{6}\bigg|_{0}^{2} \ \dfrac{-\cos^3 \phi}{3}\bigg|_{0}^{\pi/2} $

$=\dfrac{64}{9} \pi$

Answer 2

A sphere is given by

$$r^2=x^2+y^2+z^2$$

You have spotted that the radius is 2, and that z ranges from 0 to 2. Now if you look at the bounds of the integrations wrt y and x, you see a disk - again radius 2, but ranging from -2 to 2. This is a hemisphere in the positive z-direction.

Now in spherical coordinates the angles are (in the convention that you used - I f.e. work with the reverse convention, more commonly used in physics)

$$0 \leq \phi \leq \pi$$ $$0 \leq \theta \leq 2\pi$$

These are for a full sphere - to get a hemisphere, just take

$$0 \leq \phi \leq \frac{\pi}{2}$$

And together with the radial one you determined, you have your set of boundaries. The last step to change from cartesian coordinates to spherical coordinates is to evaluate the Jacobian of the transformation which is

$$\mathrm{d}x\mathrm{d}y\mathrm{d}z = \rho^2 \mathrm{sin} \phi \;\mathrm{d}r \mathrm{d}\phi \mathrm{d}\theta $$