Multiple Integration - Spherical Coordinates

Sorry for the late post (haven't gotten around to updating the blog over spring break)!

But yes, it's another coordinate system. Just as with cylindrical coordinates mentioned in the last post, spherical is somewhat of a cousin of our old polar coordinates.

(Refer to the blog archive for a refresher on polar coordinates.)

There are three components to coordinatizing points in spherical:

Both "rho" and "theta" are exactly the same as polar and cylindrical. The only difference between spherical and cylindrical, however, is the final component. This difference makes sense, though. Take a moment and imagine a cylinder. It is, in essence, a circle extruded infinitely in the z-direction. Therefore, it only makes sense for it to have a z-component to its point-coordinates. For a sphere, however, you have "phi", the extent of the surface from the positive z-axis.

Imagine a circle on the yz-plane extruded down to the xy-plane as depicted in the image below. If the top of the circle is only brought down to the xy-plane, it has only completed 1/4 of its way around the coordinate plane, so "phi" is pi/4. Keep in mind, however, that since "phi" is only measured from the positive z-axis, the highest possible value for "phi" is pi. This prevents doubling-over when measuring "phi".

To integrate, simply keep the following conversions from rectangular to spherical in mind and integrate accordingly:

Also, just as with cylindrical coordinates, there is a new "dV" in spherical, which may seem a little daunting at first sight, but isn't that bad once the trigonometric functions eliminate the angles involved with "theta" and "phi":

A short sample:
Consider the following sphere of radius 4 (it looks like a cracked egg, but bear with me):

Let's write the triple integral that will result in it's enclosed volume.

• It extends from the positive z-axis to the negative z-axis, so "phi" is pi.
• It extends all the way around the xy-plane, so "theta" is two-pi.
• And it's radius is 4, so "rho" is 4.

And our integral should look something like this:

It may seem a little confusing at first, but with a little practice, it'll get easier.
Happy spherical integrating!