00:01
So in this question, we are trying to find the volume of the solid that's bounded below by the xy plane, on the sides by the sphere rho equals 2, and above by the cone phi equals pi over 2.
00:11
So if i want to find the volume of a solid in three -dimensional space, what do i do? i compute the triple integral over the region r in three -dimensional space of the function 1dv.
00:25
So let's see, what is my dv this time? my dv is rho squared times the sine of phi, d rho, d theta, d phi.
00:38
Now this time, my rho's, they are ranging from 0 to 2.
00:43
My theta's, there's no restriction on, so 0 to 2 pi.
00:48
My phi's, well, i'm bounded below by the xy plane, which is phi equals pi over 2, as well as phi equals pi over 3.
00:58
So my phi limits of integration are pi over 3 to pi over 2.
01:05
Now, my integrand is the product of a function of rho, theta, and phi, and i have all constant limits of integration...