Calculate the flux of capital F of x comma y equal to angle bracket x cubed comma y cubed close angle bracket across a unit circle oriented counterclockwise. So this is going to become the double integral over the region D with respect to A lot of P sub x plus Q sub y. Going back to Green's theorem.

Now in this case P is x cubed and Q is y cubed. So taking those derivatives respectively, this becomes the integral, double integral over the region D with respect to A of 3x squared y squared. 3x squared plus 3y squared. And in fact, I'm going to go ahead and change those boundaries to make that double integral.

So we'll have the integral from negative 1 to 1 with respect to x of the integral from negative square root of 1 minus x squared end square root to square root of 1 minus x squared end square root with respect to y of the function 3x squared plus 3y squared. Might be a good idea to go ahead and change to polar coordinates. Since we have an x squared plus y squared we are dealing with a circle after all.

So let's see using some of our identities, this will be equal to the integral from 0 to 2 pi with respect to theta of the integral from 0 to 1 with respect to r of the function 3r squared times r. And we have two independent functions here. So this will be equal to parentheses integral from 0 to 2 pi with respect to theta of the function 1 close parentheses. Open parentheses integral from 0 to 1 with respect to r of 3r to the fourth power. Well, 3r to the third power. Close parentheses.

First integral is equal to 2 pi. And the second integral be equal to 3/4, which makes the flux of capital F of x comma y equal to 3 pi over 2.