INSTRUCTOR: Use the triangular region R with vertices 0 comma 0, 2 comma 2, and 2 comma 0, with the density function rho of x comma y equal to the square root of xy, end square root. Find the moments of inertia. Now, we have some definitions for the inertia with respect to x and y. But first, let's think about what these three points are, this triangular region.

So I'm going to draw these three points, 0, 0, the point 2, 2, and the point 2, 0. Now, thinking about the area bounded between those 3's in a triangle, we see that our x boundary is from 0 to 2, and the y boundary is from 0 to a line with a slope of 1. So we are going through the origin. So this is going to be a boundary in the y direction from 0 to x, to y equals x.

All right, now let's set these up, so I sub x, with an inertia with respect to the x-axis. This will be the integral from 0 to 2 with respect to x, based on our boundaries of the integral from 0 to x with respect to y of the function y squared, square root of xy, end square root. Now, remembering that the square root is a 1/2 half power with these, this integral ends up becoming integral from 0 to 2 with respect to x of the function 2/7 times x to the fourth, which, evaluating that, will be equal to 64/35.

Next, let's find I sub y, that other moment of inertia. This would be the integral from 0 to 2 with respect to x of the integral from 0 to x with respect to y of x squared, square root of xy, end square root. Now again, thinking of the square root as a 1/2 power, this becomes the integral from 0 to 2 with respect to x of 2/3 times x to the fourth. And evaluating that further, that is 64/15.

Now there is one final moment of inertia. And that's with respect to the origin. By definition, that is equal to I sub x plus I sub y. So taking these two values we found, 64/35 and 64/15 and summing those, I sub 0 is 128/21.