INSTRUCTOR: Evaluate the integral, the double integral over r squared, with respect to x, and then with respect to y of the function e to the exponent of negative 4, open parenthesis x squared plus y squared close parentheses, end exponent. First, we need to understand that r squared is equal to the set of all points r comma theta such that 0 is less than or equal to theta is less than or equal to 2 pi and 0 is less than or equal to r, which is less than infinity in every direction, every distance from 0 to infinity in all directions, theta being from 0 to 2 pi. So this means that our integral can be written as integral from 0 to 2 pi with respect to theta of the integral from 0 to infinity with respect to r of e to the negative 4 r squared. That is negative 4 r squared is our exponent. And out of our exponent, we have multiplied by r.

Now, we want to treat this as a limit as b going to infinity. So let me just rewrite that again. The integral from 0 to 2 pi with respect to theta of the limit as b goes to infinity of the integral 0 to b with respect to r, finally, of our function, e to the exponent of negative 4 r squared, end exponent, times r. Now, because this inner function is only in terms of r, not in terms of theta, we can actually rewrite this as being a product of two integrals. So the integral from 0 to 2 pi, d theta, with a function there of 1, multiplied by the limit as b goes to infinity of the integral from 0 to b, e to the exponent of negative 4 r squared, end exponent, r dr.

Now, applying use substitution, we can see that our second integral, our second integral is equal to negative 1/8 e to the exponent of negative 4r squared, end exponent. And then evaluating that from 0 to b, its function becomes negative 1/8 e to the exponent of negative 4 r squared. Or sorry, it becomes negative 1/8 e to the exponent of negative 4b squared plus 1/8.

Now, following the limit of that, if we take the limit as b goes to infinity, our second integral is going to be equal to 1/8. Now, going back to our first integral, integral from 0 to 2 pi d theta, that's equal to 2 pi. So since we were finding the product of these two, our integral is equal to 2 pi multiplied by 1/8 or pi over 4.