INSTRUCTOR: Evaluate the iterated integral, the double integral, of the function x squared plus y squared over the region D in the first quadrant between the functions y equals 2x and y equals x squared. Evaluate the iterated interval by integrating first with respect to y, and then integrating next with respect to x. Now first, because we have two bounds in the y direction and we're going to integrate first with respect to y, we need to find where those two intersect because while we can use the functions for the bounds of y, we need values for the bounds x.

So let's take 2x equals x squared and solve that. So we have 0 equals x squared minus 2x. And clearly, x is equal to 0. And x is equal to 2. So those are going to be our bounds in the x direction.

So this double integral over the region D of the function x squared plus y squared is going to be equal to the integral from 0 to 2 with respect to x of the integral from x squared to 2x with respect to y of the function x squared plus y squared. Now, before I go further, I want to note that between x equals 0 and x equals 2, the function 2x is larger than the function x squared. And that's why I've chosen that order, x squared being my lower bound, 2x being my upper bound.

Now, let's take the integral with respect to y first of our function. And that will mean we have the integral from 0 to 2 with respect to x of x squared y plus y cubed over 3 as our function, evaluated from x square to 2x. Evaluating this at those two endpoints, this is equal to the integral from 0 to 2 with respect to x of the function 14/3 times x cubed minus x to the fourth minus 1/3 x to the sixth. Now, finding the antiderivative, that function, and evaluating those endpoints, means that we are going to evaluate the function 7/6 times x to the fourth, minus 1/5 x to the fifth, minus 1/21 multiply by x to the seventh, evaluating that from 0 to 2 will be 216/35.