00:01
We're given an integral and we're asked to use spherical coordinates to evaluate it.
00:06
The integral is the triple integral over the region e of y squared, dv, where e is the solid hemisphere, x squared plus y squared plus c squared is less than or equal to 9, and y is greater than or equal to 0.
00:36
So it's the, if you're looking straight on the positive x -axis, the right side of the sphere.
00:50
First we'll want to rewrite our region in terms of spherical coordinates.
00:55
So we have that row squared is less than equal to 9, or that row is less than or equal to 3.
01:02
This is our solid sphere of radius 3.
01:06
Now, for y to be greater than are equal to 0, this means that plugging in spherical coordinates we have row times the sine of phi times the sign of theta is greater than or equal to zero or more simply because y is greater than or equal to zero we know that we're going to restrict theta to be greater than equal to zero and less than or equal to pi as for phi there's no restriction it goes from zero to pi and row goes from zero to three and so for our integral in spherical coordinates is the integral from 0 to pi, integral from 0 to pi, integral from 0 to 3 of our function y squared in spherical coordinates...