# find area of polar graph

• Feb 1st 2010, 03:53 PM
isuckatcalc
find area of polar graph
the area shared by the circle r=2 and the cardiod r=2(1-cos(x))

so it looks like the limits of integration are from -pi/2 to pi/2.

squaring both gets 4 and 4-8cos(x)+4cos(x)^2

i'm not sure which is supposed to subtracted from which... but i'll go 4-(4-8cos(x)+4cos(X)^2.

which gives 8cos(x) - 4cos(x)^2

half of that is 4cos(x) -2cos(x)^2

then i can get 4cos(x) - 1 - cos(2x)/2

integrating gives

4sin(x) -x - sin(2x)/4

and plugging in pi/2 and -pi/2 i get 8-pi. but that's not the correct answer that i've been given. where am i going wrong?
• Feb 1st 2010, 04:32 PM
skeeter
first off, your limits of integration are wrong. you really need to start looking at the graph of the situation.

let $\displaystyle r_1(\theta) = 2$

$\displaystyle r_2(\theta) = 2(1-\cos{\theta})$

using symmetry ...

$\displaystyle A = 2 \int_0^{\frac{\pi}{2}} \frac{[r_2(\theta)]^2}{2} \, d\theta + 2\int_{\frac{\pi}{2}}^{\pi} \frac{[r_1(\theta)]^2}{2} \, d\theta$