# Convert to Rectangular coordinates and Eval

• Oct 30th 2012, 11:20 AM
cbull373
Convert to Rectangular coordinates and Eval
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.
• Oct 30th 2012, 08:01 PM
Prove It
Re: Convert to Rectangular coordinates and Eval
Quote:

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.

Remember that to convert from Cartesians to Polars we make the transformation \displaystyle \begin{align*} dx \, dy \to r \, dr \, d\theta \end{align*}, so you only need to convert \displaystyle \begin{align*} \sin{2\theta} \end{align*} to Cartesians, and recall

\displaystyle \begin{align*} \sin{2\theta} &= 2\sin{\theta}\cos{\theta} \\ &= 2\left( \frac{y}{r} \right) \left( \frac{x}{r} \right) \\ &= \frac{2\,x\,y}{r^2} \\ &= \frac{2\,x\,y}{x^2 + y^2} \end{align*}