Finding the area in the polar coordinate system

[Guest]

r=1+cos(t) r=3cos(t) Find the area insider=1+cos(t) and outside r=3cos(t)

I set up the integral for each. integrate from 0 to pi (1+cos(t))^2 dt using symmetry.

I did the same for the other... integrate from 0 to pi/2 (3cos(t))^2

What I got was t+2sin(t)+1/2t+1/4sin(2t) and 9/2t+9/4sin(2t)

Initially I intended to subtract the latter from the first but the more I looked at it, the more I questioned myself. I know I integrated correctly and had the right integral for each function. I just wasn't sure how to get the area requested.

Thanks

 

[galactus]

\(\displaystyle \L\\1+cos(t)=3cos(t)\)

Solving, we find \(\displaystyle \L\\cos(t)=\frac{1}{2}\).

Therefore, your limits of integration are \(\displaystyle \L\\\frac{-\pi}{3}\) and \(\displaystyle \frac{\pi}{3}\)

\(\displaystyle \L\\2\left[\frac{1}{2}\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\left((1+cos(t))^{2}-(3cos(t))^{2}\right)dt+\frac{1}{2}\int_{\frac{\pi}{2}}^{\pi}(1+cos(t))^{2}dt\right]\)

=\(\displaystyle \L\\\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\left[-4cos(2t)+2cos(t)-3\right]dt+\int_{\frac{\pi}{2}}^{\pi}\left[\frac{1}{2}cos(2t)+2cos(t)+\frac{3}{2}\right]dt\)

This set up gives the same result as skeets, just a different approach.

 

[skeeter]

\(\displaystyle \L A = \int_{\frac{\pi}{3}}^{\pi} (1 + \cos{\theta})^2 d\theta - \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} (3\cos{\theta})^2 d\theta\)

 

[soroban]

Hello, ezrajoelmicah!

skeeter has the correct set-up . . .

\(\displaystyle r\:=\:1\.+\,\cos\theta.\;r\:=\:3\cos\theta\)
Find the area inside the cardioid and outside the circle.

The shaded region is the desired area.

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