1. The problem statement, all variables and given/known data Evaluate the integral integral from 0 to 1 of (integral from y to sqrt(2-y^2) of (3(x-y))dxdy) by converting to polar coordinates. 2. Relevant equations x = rcos(theta) y = rsin(theta) 3. The attempt at a solution By drawing a picture of the bounds, I concluded that r goes from 0 to sqrt(2) and theta goes from pi/4 to pi/2. I tried to integrate like this: integral from 0 to sqrt(2) of 3(rcos(theta) - rsin(theta) r dr dtheta = integral from 0 to sqrt(2) of 3r^2 (cos(theta) - sin(theta)) dr dtheta = r^3(cos(theta) - sin(theta)) where r is from 0 to sqrt(2) =2sqrt(2) * integral from pi/4 to pi/2 of (cos(theta)-sin(theta)) dtheta =2sqrt(2) * (sin(theta)+cos(theta)) with theta from pi/4 to pi/2 =2sqrt(2) *((1+0)-((sqrt(2)/2) + (sqrt(2)/2)) =2sqrt(2) *(1-sqrt(2)) =2sqrt(2) - 4 The actual answer is 4 - 2sqrt(2) so I missed a sign somewhere, but I cannot find it. Sorry if it's difficult to read. Any help would be greatly appreciated.