Find the area between the curves r=sin(theta) and r=2sin(theta).
Find the area of the region enclosed by r=3cos3(theta)
make a sketch?
geometric method ...
A = area of big circle - area of little circle = $\displaystyle \frac{3\pi}{4}$
calculus method ...
$\displaystyle 2 \int_0^{\frac{\pi}{2}} \frac{1}{2}(2\sin{\theta})^2 - \frac{1}{2}(\sin{\theta})^2 \, d\theta$
$\displaystyle \int_0^{\frac{\pi}{2}} 3\sin^2{\theta} \, d\theta$
$\displaystyle 6\int_0^{\frac{\pi}{6}} \frac{1}{2}[3\cos(3\theta)]^2 \, d\theta$Find the area of the region enclosed by r=3cos3(theta)