More polar help needed

**leinadwerdna** · April 28th 2009, 02:32 PM

Find the area between the curves r=sin(theta) and r=2sin(theta).

Find the area of the region enclosed by r=3cos3(theta)

**skeeter** · April 28th 2009, 03:08 PM

Originally Posted by leinadwerdna

Find the area between the curves r=sin(theta) and r=2sin(theta).

make a sketch?

geometric method ...

A = area of big circle - area of little circle = $\frac{3\pi}{4}$

calculus method ...

$2 \int_0^{\frac{\pi}{2}} \frac{1}{2}(2\sin{\theta})^2 - \frac{1}{2}(\sin{\theta})^2 \, d\theta$

$\int_0^{\frac{\pi}{2}} 3\sin^2{\theta} \, d\theta$

Find the area of the region enclosed by r=3cos3(theta)

$6\int_0^{\frac{\pi}{6}} \frac{1}{2}[3\cos(3\theta)]^2 \, d\theta$

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