Double integration when switching to polar coordinates

[Colts]

Homework Statement

Take the double integration of
http://webwork.usi.edu/webwork2_files/tmp/equations/08/1294e87299342c0ccfe2f8a97055da1.png
when f(x)=sqrt(4x-x^2)

Homework Equations

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

The Attempt at a Solution

I know I plug in the r*cos(theta) and r*sin(theta) for the x and y in the equation I am integrating, but other than that i can't remember how to switch the limits about which I integrate. After I substitute and simplify I get double integral of 1 dr dtheta. How do I change the limits of the integration to polar?

 

[SammyS]

Homework Statement

Take the double integration of
http://webwork.usi.edu/webwork2_files/tmp/equations/08/1294e87299342c0ccfe2f8a97055da1.png
when f(x)=sqrt(4x-x^2)

Homework Equations

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

The Attempt at a Solution

I know I plug in the r*cos(theta) and r*sin(theta) for the x and y in the equation I am integrating, but other than that i can't remember how to switch the limits about which I integrate. After I substitute and simplify I get double integral of 1 dr dtheta. How do I change the limits of the integration to polar?

Sketch the region to be integrated over.

Write the boundary as r equal to some function of θ, r = g(θ) .

The element of area in polar coordinates is r∙dr∙dθ .