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.