Areas under and between polar curves

**mistykz** · April 25th 2008, 12:46 PM

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

**galactus** · April 25th 2008, 04:46 PM

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.

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

**Soroban** · April 25th 2008, 05:40 PM

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?

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