# Thread: area between polar curve

1. ## area between polar curve

Find the area of the region that lies inside the curve r=1 + cosΘ and outside of the curve r= 3cosΘ

I set them equal to eachother and solved for Θ = pi/3, 5pi/3
I then set up the integral from pi/3 to pi of (0.5)(1+cosΘ)^2 and got

can anyone help me solve this problem

thanks

2. ## Solved

Ok, so what you need to do is to take the integral of the $\displaystyle r=1+cos\Theta$ minus $\displaystyle r=3cos\Theta$

So, it'll look like this:
$\displaystyle 2\int_{\pi/3}^{\pi} (1+cos\Theta - 3cos\Theta) d\Theta$
$\displaystyle 2\int_{\pi/3}^{\pi} (1-2cos\Theta) d\Theta$
$\displaystyle 2 \times [(\int_{\pi/3}^{\pi} 1 d\Theta) - 2 (\int_{\pi/3}^{\pi} cos\Theta d\Theta)]$
$\displaystyle 2 \times [\Theta |_{\pi/3}^{\pi} - 2sin\Theta |_{\pi/3}^{\pi}]$
$\displaystyle 2[\pi - \pi/3 - 2(0-\sqrt{3}/2)]$
$\displaystyle 2[(2\pi)/3 + \sqrt{3}]$
$\displaystyle (4\pi + 6\sqrt{3})/3$

there ya are, have fun

3. I don't think that is the correct answer. I believe you need to use the formula
integral of .5[f(x)^2 - g(x)^2]. Can anyone else guide me to the correct answer.

4. ## solved for real

You didn't like my work before? lol. Ok, so to do it correctly this time. The trick is that the function $\displaystyle r=1+cos\theta$ will need to be integrated from $\displaystyle \pi to \pi/3$ whereas the other function $\displaystyle r=3cos\theta$ needs to be integrated from $\displaystyle \pi/2 to \pi/3$

This actually comes out very nicely:
$\displaystyle 2 \times ([1/2 \times \int_{\pi/3}^{\pi} (1 + cos\theta)^2 d\theta] - [1/2 \times \int_{\pi/3}^{\pi/2} (3cos\theta)^2 d\theta])$
$\displaystyle \int_{\pi/3}^{\pi} (1 + 2cos\theta + cos^2\theta) d\theta] - \int_{\pi/3}^{\pi/2} (9cos^2\theta) d\theta$
$\displaystyle (3\theta/2 + 2sin\theta + sin(2\theta)/4) |_{\pi/3}^{\pi} - (9\theta/2 + 9sin(2\theta)/4) |_{\pi/3}^{\pi/2}$
$\displaystyle 3\pi/2 - \pi/2 - 9\sqrt{3}/8 - (9\pi/4 -6\pi/4 - 9\sqrt{3}/8)$
$\displaystyle \pi/4$

there ya go. That's the correct formula and the correct answer.