00:01
So for this question we are evaluating a triple integral over a region e, and in particular because the region is defined using a paraboloid, it's very circular in nature, it would be in our benefit to use cylindrical coordinates to figure this out.
00:19
So what that means is we can rewrite the equation for the paraboloid as r squared in cylindrical coordinates, and in particular the way that cylindrical coordinates work, just so we remember, is that your x and y are the polar coordinates, r cosine theta and r sine, and then your z coordinate remains cartesian.
00:51
So let's go ahead and write out the triple integral.
01:01
Let's put the bounds on these.
01:08
The radius bounds, since we're bounded below by the paraboloid, will go from, excuse me, the z bounds, will go from r squared up until 16, and don't forget when you convert from cartesian to cylindrical, you have to include this additional factor of r inside the integral.
01:40
Okay, so we have our z bounds, and all we're left with are the radius and theta bounds.
01:46
So we're going to project the intersection of the two surfaces into the xy plane, and that's done by setting the two equal to one another.
01:56
You'll get that the x squared plus y squared is equal to 16, or better yet, r squared equals 16, which tells you r equals 4, or in other words, that it's a circle with radius 4, so the radius bounds will then go from 0 to 4, and because there are no restrictions on going full circle, that tells us that our theta bounds will be 0 to 2 pi.
02:26
Alright, so this is the triple integral that we're going to evaluate...