INSTRUCTOR: Evaluate the integral over the region D with respect to r, and then with respect to theta of the function r squared, sine squared open parentheses 2 theta close parentheses, multiplied by r where the region D is equal to the set of all ordered pairs r comma theta, such that negative pi over 4 is less than or equal to theta, is less than or equal to pi over 4. And 0 is less than or equal to r, which is less than or equal to 2 square root of cosine parentheses 2 theta, close parentheses and end squared root. We can set up this integral as the integral from negative pi over 4, 2 pi over 4 with respect to theta of the integral from 0 to 2 square root of cosine, open parentheses 2 theta close parentheses and then end squared root of the function r cubed sine squared open parentheses 2 theta close parentheses. And of course, that inner integral is with respect to r.

Now, taking the antiderivative of the innermost function, this becomes the integral from negative pi over 4 to pi over 4 of the function 1/4 r to the fourth power times sine squared in parentheses 2 theta close parentheses, evaluated from 0 to 2 square root of cosine, open parentheses 2 theta close parentheses, and then end the square root with respect to theta. Now, when we evaluated these endpoints, this becomes the integral, from negative pi over 4, 2 pi over 4 with respect to theta of the function 4 cosine squared, open parentheses 2 theta close parentheses, multiplied by sine squared, open parentheses 2 theta close parentheses.

Now, referring to some of our Calculus 2 integration techniques, trig substitution and using some of those identities, this will be equal to the function, negative 1/8 co-sign, open parentheses 4 theta close parentheses, sine open parentheses 4 theta close parentheses, plus 1/2 theta evaluated from negative pi over 4 to pi over 4. And evaluating this, we find that this integral is equal to pi over 4.