INSTRUCTOR: Find the volume of the solid bounded above by the graph of f of x, y equal to xy sine open parenthesis x squared times y close parentheses and below by the xy-plane on the rectangular region R equal to the interval from 0 to 1 by the interval from 0 to pi. And set this integral up as v, volume, is equal to the integral from 0 to 1 with respect to x of the integral from 0 to pi with respect to y of the function xy sine parentheses x squared y close parentheses.

To take the antiderivative with respect to y, this requires integration by parts. And I will leave that to you, so that you can double check your steps. Applying integration by parts one time, one iteration of that, this volume is equal to the integral from 0 to 1 with respect to x of a fraction with the numerator of negative y cosine open parenthesis x squared y close parentheses and a denominator of x end fraction plus a numerator of sine open parenthesis x squared y denominator of x cubed end fraction. That function evaluated from 0 to pi. And then finally our integral is with respect to x.

Evaluating from 0 to pi causes our integral to become integral from 0 to 1 of-- or with respect to x. My function is, first, a fraction with a numerator of negative pi cosine open parentheses pi x squared close parentheses. And we have a denominator of x end fraction. Plus numerator of sine open parenthesis x or-- sorry-- pi x squared close parentheses and the denominator of x cubed.

Now, I need to change one thing about the limits on my integration. Because, at this point, this new function, this new integrand is undefined at 0. And so I'm going to change this into a limit. OK, so we need to change my lower endpoint to be a b.

And outside, applied to this whole integral, I'm going to have a limit as b approaches 0. And, specifically, it's from above. But now the antiderivative of the new inner function requires integration by parts again. Each of those two terms, I believe, will require one step of integration by parts.

So this is my volume is equal to the limit as b approaches 0 from above of the function numerator negative sine open parentheses pi x squared close parentheses, the denominator of 2x squared end fraction, and evaluated from b to 1. So this is going to be equal to the limit as b approaches 0 from above of a fraction with a numerator of sine open parentheses pi b squared close parentheses and a denominator of 2b squared.

Now, this limit, all on its own, does not exist. But if we apply L'Hopital's rule, this will be equal to the limit as b approaches 0 from above of a fraction of the numerator of 2 pi b cosine open parentheses pi b squared close parentheses with a denominator of 4b. Now, this reduces to equal the limit as b approaches 0 from above of pi over 2 multiplied by cosine open parentheses pi b squared. And if we take that limit, this will be equal to pi over 2. And so our volume of that region bounded above by f of x, y and below by the xy-plane is equal to pi over 2.