Originally Posted by

**cbull373** My problem is this:

Convert to rectangular coordinates and evaluate: The integral [0 to pi/2] of the integral [0 to 2sin(theta)] of (r*sin(2theta)) dR dTheta

The area being integrated is a circle of radius 1, shifted up the y axis by 1, only integrated in the first quadrant from 0 to pi/2.

Work:

For converting to rectangular coordinates, I have solved that the new limits of integration should be from [0 to 2] and [-sqroot(1-(y-1)^2) to +sqroot(1-(y-1)^2)] dydx

I am having a lot of trouble finding the new f(r,theta) equation though. I calculated it as follows

z=r*sin(2theta)

z=2sin(theta)cos(theta)

z=2(y/r)(x/r)

z=2yx/(r^2)*

z=2yx/(x^2+y^2)

Is that correct? this seems impossible to integrate though! Anyone's help or ideas would be extremely appreciated. Worked hard on this for hours.