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.
x = rcos(theta)
y = rsin(theta)
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) - 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.