INSTRUCTOR: Evaluate the improper integral, the double integral of the function y divided by the square root of 1 minus x squared minus y squared, end square root, where D is equal to the set of all ordered pairs, x comma y, such that x is greater than or equal to 0, y is greater than or equal to 0, and x squared plus y squared is less than or equal to 1. Note first that this region is actually a quarter of a circle. That's worth noting. So it's a quarter of a circle.

So that means that the integral boundaries that we're going to set up are going to be over the interval bracket 0 comma 1 end bracket, by the interval bracket 0 comma square root of 1 minus x squared, end square root, end bracket. That's how I want to interpret that. So when I go to write my integral, and notice my y values are in terms of x intentionally.

OK, so this would be the integral, from 0 to 1 with respect to x, of the integral from 0 to the square root of 1 minus x squared, end square root, with respect to y, of our function y over the square root of 1 minus x squared minus y squared, end square root. Before we go any further, I would like to point out that there actually is a discontinuity in this interval. And it's really at the point 1 comma 0 the way that I have the setup. 1 comma 0 is the issue.

If you set this up in a different way, you'd have a different discontinuity. But there is a discontinuity. However, it's on the boundary. It's on the boundary of that quarter circle.

So again, thinking about it as a quarter circle in the first quadrant, it would go from 0, 0 to the right at 1 comma 0. And the upper limit is 0 comma 1. And it's that circle.

However, because of Fubini's theorem that we've seen in this section, we know that that's actually not an issue. We can go ahead and evaluate this integral as an iterated integral. All right, now, to evaluate the inner integral here, let's make a u substitution, so u being equal to 1 minus x squared minus y squared. Which means that du, and we're actually taking this with respect to y because our inner integral is with respect to y, that is negative 2y dy, which means that negative 1/2 du equals y dy, which is part of what we have here.

So this inner integral, or actually the whole integral, becomes integral from 0 to 1 with respect to x of the integral from 1 minus x squared to 0 with respect to u of the function negative 1/2 times u to the negative 1/2. I've changed the boundaries based on my definition of what u is in relation to y. So this is going to become the integral from 0 to 1 with respect to x of the square root of 1 minus x squared, once we make all of our substitutions, and that's all inside that square root of 1 minus x squared, end square root. Now, if we appeal to an integral table or possibly use trig substitution, this would then be equal to the function x over 2, end fraction, square root of 1 minus x squared, end square root, plus 1/2 times arcsin of x, end parentheses. That function evaluated from 0 to 1, which is pi over 4.