How to find the area inside a polar curve

Formula for finding the area inside a polar curve

The area inside a polar curve is given by

???A=\int^{\beta}_{\alpha}\frac{1}{2}r^2\ d\theta???

where ???[\alpha,\beta]??? is the interval

where ???r??? is the equation of the polar curve

Example

Find the area inside the polar curve.

???r=4\sin{\theta}???

We need to find the interval, which we’ll do by setting ???r=0??? and solving for any values of ???\theta???.

???r=4\sin{\theta}???

???0=\sin{\theta}???

???\theta=\pi??? and ???\theta=2\pi???

???\alpha=\pi??? and ???\beta=2\pi???

Now we can plug the interval we found and the given polar equation into the formula for the area inside a polar curve.

???A=\int^{2\pi}_{\pi}\frac{1}{2}\left(4\sin{\theta}\right)^2\ d\theta???

???A=\int^{2\pi}_{\pi}\frac12\cdot16\sin^2{\theta}\ d\theta???

???A=8\int^{2\pi}_{\pi}\sin^2{\theta}\ d\theta???

Since ???\sin^2{\theta}=\frac12\left[1-\cos{(2\theta)}\right]???, we get

???A=8\int^{2\pi}_{\pi}\frac12\left[1-\cos{(2\theta)}\right]\ d\theta???

???A=4\int^{2\pi}_{\pi}1-\cos{(2\theta)}\ d\theta???

???A=4\left(\theta-\frac{\sin{(2\theta)}}{2}\right)\bigg|^{2\pi}_{\pi}???

???A=4\left[2\pi-\frac{\sin{\left(2(2\pi)\right)}}{2}-\left(\pi-\frac{\sin{(2\pi)}}{2}\right)\right]???

???A=4\left[2\pi-\frac{\sin{(4\pi)}}{2}-\pi+\frac{\sin{(2\pi)}}{2}\right]???

???A=4\left(2\pi-\frac{0}{2}-\pi+\frac{0}{2}\right)???

???A=4\left(2\pi-\pi\right)???

???A=4\pi???