# Setting up a triple integral in spherical coordinates to find volume

• July 5th 2010, 03:16 AM
sherryl
Setting up a triple integral in spherical coordinates to find volume
I want to use spherical coordinates for the volume of a solid contained within the sphere x²+y²+z²=16 and outside the cone z=4-√(x²+y²).

Could somebody show me just how to set up the triple integral. The integration I'll plow through myself.
• July 5th 2010, 05:05 AM
HallsofIvy
But setting up the integral is the only interesting part of this problem!

Since this involves an upside down cone, it's probably best to use cylindrical coordinates (if it had been a "regular" cone, spherical coordinates). In cylindrical coordinates the sphere is given by, of course, $r^2+ z^2= 16$ and the cone is given by z= 4- r. The differential of volume is $r drd\theta dz$. The cone and the sphere intersect when $r^2+ (4-r)^2= 2r^2- 8r+ 16= 16$ or $r^2- 4r= r(r- 4)= 0$. That is, they intersect at r= 0 (the top of the sphere) and at r= 4, the circumference.

To cover the entire sphere, r must go from 0 to 2 and $\theta$ must go from 0 to $2\pi$. For each r, z must go from the cone z= 4- r up to the upper half of the sphere, z= \sqrt{16- r^2}.
• July 5th 2010, 10:04 AM
sherryl
in spherical though...
Thanks HallsofIvy for your comprehensive explanation.

I'm doing these practice problems to get more comfortable with spherical though... Hence the emphasis on setting up in spherical even though cylindrical is the way to go.

In Spherical, does this look good?

$\int\nolimits_{0}^{2\pi } {d\theta }\int\nolimits_{0}^{\tfrac{\pi }{2}} {\sin \phi d\phi }\int\nolimits_{\frac{4}{\sin \phi +\cos \phi }}^{4} {\rho ^{2}d\rho }$

Mainly, I'm wondering about the lower limit of $\rho$. The upper bound is 4 which is the edge of the sphere. For the lower limit, is it correct to refer to the edge of the upside down cone with: ${\frac{4}{\sin \phi +\cos \phi }}$
• July 6th 2010, 01:00 PM
sherryl
Quote:

Originally Posted by sherryl
In Spherical, does this look good?

$\int\nolimits_{0}^{2\pi } {d\theta }\int\nolimits_{0}^{\tfrac{\pi }{2}} {\sin \phi d\phi }\int\nolimits_{\frac{4}{\sin \phi +\cos \phi }}^{4} {\rho ^{2}d\rho }$

The idea behind ${\frac{4}{\sin \phi +\cos \phi }}$ is that:

$\rho \cos \varphi =4-\rho \sin \varphi$

which itself comes from the following two premises:

$z=\rho \cos \varphi \\$
$r=\rho \sin \varphi$

Does this seem right?