Question

Are triple integrals in spherical coordinates easier than those in rectangular coordinates?

Answer 1

It truly depends on the integrand.

Answer 2

^

Answer 3

Yah it really depends :) As these guys have said.

I think the most obvious illustration of this would be when you want to find the volume of a sphere.

The equation of a sphere:
\[\large x^2+y^2+z^2=R^2 \qquad \rightarrow \qquad z=\sqrt{R^2-x^2-y^2}\]

Here's what it looks like in the xy-plane:
\[z=0 \qquad \rightarrow \qquad y=\sqrt{R^2-x^2}\]

So our integral would look something like this:\[\huge \int\limits_{x=-R}^{R}\quad\int\limits_{y=-\sqrt{R^2-x^2}}^{\sqrt{R^2-x^2}}\qquad\int\limits_{z=-\sqrt{R^2-x^2-y^2}}^{\sqrt{R^2-x^2-y^2}}\;dz\;dy\;dx\]

You can use symmetry to simplify it down a bit.
But it's still going to involve using Trig subs and is a bit of work.

Answer 4

If we look at it in spherical coordinates:

We're integrating with respect to \(\large r\), \(\large \theta\) and \(\large \phi\) instead of \(\large x\), \(\large y\) and \(\large z\) now.

Radially we'll want to integrate from the center of the circle out to the surface.
\(\large 0\le r\le R\)

Phi is measured from the z-axis rotating downward. We'll want to integrate from an angle of 0 out to pi, a half circle.
\(\large 0\le \phi\le \pi\)

And then we also integrate around in the xy direction, a full rotation.
\(\large 0\le\theta\le2\pi\)

Then you need to remember how the differentials change when you're switching to spherical.

\(\large dV=dx\;dy\;dz\)
\(\large dV=r^2\sin\phi\;dr\;d\phi\;d\theta\)

So our integral would look something like this:\[\large \int\limits_{r=0}^R \quad\int\limits_{\theta=0}^{2\pi}\quad\int\limits_{\phi=0}^{\pi}\;r^2\sin\phi\;d\phi\;d\theta\;dr\]

And as you can see, all of our boundaries are constant, so we can rewrite it like this if we want:

\[\large \int\limits\limits_{r=0}^R r^2\;dr\cdot\int\limits\limits_{\theta=0}^{2\pi}d\theta\cdot\int\limits\limits_{\phi=0}^{\pi}\sin\phi\;d\phi\]

And from here it's a piece of cake to work out.
With spherical, I guess the tricky part is setting it up correctly. :)

Answer 5

Couldn't have said it better Zep!