00:01
All right, so we want to use spherical coordinates to compute the volume of this solid given in cartesian coordinates.
00:08
Step one then, of course, will be to convert this expression into spherical coordinates.
00:15
First the cone, have this z equals two -thirds times the square root of x squared plus y squared.
00:21
Well, z equals rho times cosine phi, that is two -thirds times the square root of x squared will be rho squared sine squared phi cos squared theta plus y squared is rho squared sine squared phi sine squared theta.
00:47
We can pull out both a rho squared and a sine squared phi from the square root.
00:54
That leaves us with the square root of cos squared theta plus sine squared theta.
01:02
By the fundamental theorem of trigonometry, this is one, and our cone expression becomes rho cos phi equals two -thirds, oh i forgot to copy that there, times rho sine phi.
01:18
Cancel out the rows, throw some symbols around a little bit, and we get three halves equals tan phi, thus phi equals the arc tangent of three halves.
01:38
And this is the equation for our cone.
01:43
For the sphere, we have x squared plus y squared plus z squared equals three halves z.
01:58
Well, x squared plus y squared plus z squared is rho squared, so we can simply write here rho squared equals three halves times rho cosine phi.
02:15
Cancel the rows a little bit, and we get rho equals three halves cosine phi.
02:24
And that is our sphere.
02:27
Now to find the volume, we just need to set up a triple integral.
02:32
I'll do first the theta ranging from zero to two pi, because that is the typical range for theta and we have no constraints on it.
02:43
Then i will do rho zero to, and i don't want to write out arc tan three halves all this time, so i'm just going to make a substitution, call that alpha, and rho ranges from zero to three halves cosine phi.
03:05
This is just a volume integral, so we have no integrand.
03:09
We get d rho d phi d theta...