# Math Help - Areas under and between polar curves

1. ## Areas under and between polar curves

Find the area of the region outside r=9+9sin(theta), but inside r=27sin(theta)
I did the integral of .5((27sin(theta))^2-(9+9sin(theta))^2), from pi/6 to 5pi/6, but got the wrong answer. Any hints?

Find the area inside the curve r^2 = 72cos(2theta)
Do I just take the square root of the entire equation? If so, how do I integrate the square root of cosine?

Find the area inside one loop of the curve r^2=4sin(2theta)
Same question as the previous one

2. Originally Posted by mistykz
Find the area of the region outside r=9+9sin(theta), but inside r=27sin(theta)
I did the integral of .5((27sin(theta))^2-(9+9sin(theta))^2), from pi/6 to 5pi/6, but got the wrong answer. Any hints?
Did you get $81{\pi}$?.

Find the area inside one loop of the curve r^2=4sin(2theta)
Same question as the previous one
When you set up the polar form of integration, the square eliminates the radical.

$\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\left(\sqrt{4si n(2{\theta})}\right)^{2}d{\theta}=2\int_{0}^{\frac {\pi}{2}}sin(2{\theta})d{\theta}$

3. Hello, mistykz!

You seem to be forgetting the area formula: . $A \;=\;\frac{1}{2}\int^{\beta}_{\alpha} r^2\,d\theta$

Find the area inside the curve: . $r^2 \:= \:72\cos(2\theta)$

Do I just take the square root of the entire equation?
If so, how do I integrate the square root of cosine?
Look at the formula again!
It says: . $A \;=\;\frac{1}{2}\int^{\beta}_{\alpha}{\color{red}r ^2}\,d\theta$

So we have: . $A \;=\;\frac{1}{2}\int^{\beta}_{\alpha}72\cos2\theta \,d\theta \;=\;36\int^{\frac{\pi}{4}}_{\text{-}\frac{\pi}{4}} \cos2\theta\,d\theta$

Got it?