INSTRUCTOR: Evaluate with the integral of the function 4 minus x squared minus y squared over the region R, where R is the circle of radius 2 on the xy plane. Now first, let's notice what the region R is in polar coordinates, a circle of radius 2 on the xy plane. R is equal to the set of all points r comma theta, such that 0 is less than or equal to r, is less than or equal to 2, because we have a circle of radius 2, where also, 0 is less than or equal to theta is less than or equal to 2 pi because we are talking about a circle, after all, so a circle, 0 to 2 pi, in radians, and a radius of 2.

So we have our restrictions on R. Now also, our function, 4 minus x squared minus y squared, can be rewritten in terms of r because we have the relationship that x squared plus y squared equals r squared. And also, da in our integral is equal to r dr d theta. So putting all of this together, this integral is from 0 to 2 pi with respect to theta of the integral from 0 to 2 with respect to r of the function 4 minus r squared in parentheses multiplied by r, so parentheses 4 minus r squared close parentheses times r.

Now, distributing the r, our function becomes 4r minus r cubed. That is, we have the integral from 0 to 2 pi with respect to theta of the integral from 0 to 2 with respect to r of the function for 4r minus r cubed. Finding the antiderivative and evaluating on those bounds on r, we get, the function is going to be-- Well, the integral is going to be from 0 to 2 pi with respect to theta.

And our function will be 2r squared minus 1/4 times r to the fourth. Evaluating that from 0 to 2, that becomes the integral from 0 to 2 pi of 4 with respect to theta. And find the antiderivative with respect to theta and evaluating, this becomes 4 theta evaluated from 0 to 2 pi, which is 8 pi.