1. Area of polar curve

Find the area of the region that is inside $\displaystyle r^2=2sin2\theta$ and inside r=1

Attempt to solution:

$\displaystyle r^2=r^2$

$\displaystyle 2sin2\theta=1$

$\displaystyle \theta= \frac{\pi}{12}, \frac{5\pi}{12}$

$\displaystyle \int_\frac{\pi}{12}^\frac{5\pi}{12} \frac{1}{2}((2sin2\theta)^2-(1)^2)d\theta)$

I don't understand how this is wrong. My answer is totally different from the one in the book

2. Originally Posted by nyasha
Find the area of the region that is inside $\displaystyle r^2=2sin2\theta$ and inside r=1

Attempt to solution:

$\displaystyle r^2=r^2$

$\displaystyle 2sin2\theta=1$

$\displaystyle \theta= \frac{\pi}{12}, \frac{5\pi}{12}$

$\displaystyle \int_\frac{\pi}{12}^\frac{5\pi}{12} \frac{1}{2}((2sin2\theta)^2-(1)^2)d\theta)$

I don't understand how this is wrong. My answer is totally different from the one in the book
note the symmetrical areas in quads I and III ...

$\displaystyle A = 4\int_0^{\frac{\pi}{12}} \sin(2\theta) \, d\theta + 4\int_{\frac{\pi}{12}}^{\frac{\pi}{4}} \frac{1}{2} \, d\theta$