Volume of Solids of Revolution using Polar Coordinates 2

Note: The volume $V = 2\pi\int \int_{R} y dydx =$ $2\pi\int_{\theta_{1}}^{\theta_{2}} \int_{0}^{r(\theta)} r^2\sin(\theta) dr d\theta =$ $\dfrac{2\pi}{3}\int_{\theta_{1}}^{\theta_{2}} (r(\theta))^3 \sin(\theta) d\theta$ .

$r = \sin(\theta)$ on the branch $(0 \leq \theta \leq \dfrac{\pi}{3})$ and $r = \sin(2\theta)$ on the branch $(\dfrac{\pi}{3} \leq \theta \leq \dfrac{\pi}{2})$ .

$V = \dfrac{4\pi}{3}(\int_{0}^{\dfrac{\pi}{3}} \sin^{4}(\theta) d\theta + \int_{\dfrac{\pi}{3}}^{\dfrac{\pi}{2}} \sin^{3}(2\theta) \sin(\theta) d\theta)$ .

Let $I_{1} = \int_{0}^{\dfrac{\pi}{3}} \sin^{4}(\theta) d\theta$ :

$I_{1} = \dfrac{1}{4}\int_{0}^{\dfrac{\pi}{3}} \dfrac{3}{2} - 2\cos(2\theta) + \dfrac{1}{2}\cos(4\theta) \:\ d\theta = \dfrac{1}{4}(\dfrac{3}{2}\theta -\sin(2\theta) + \dfrac{1}{8}\sin(4\theta))|_{0}^{\dfrac{\pi}{3}} = \dfrac{\pi}{8} - \dfrac{9\sqrt{3}}{64}$ .

Let $I_{2} = \int_{\dfrac{\pi}{3}}^{\dfrac{\pi}{2}} \sin^{3}(2\theta) \sin(\theta) d\theta$ :

$I_{2} = 8\int_{\dfrac{\pi}{3}}^{\dfrac{\pi}{2}} (\sin^4(\theta) - \sin^6(\theta))\cos(\theta) d\theta =$ $8(\dfrac{\sin^5(\theta)}{5} - \dfrac{\sin^7(\theta)}{7})|_{\dfrac{\pi}{3}}^{\dfrac{\pi}{2}} =$ $\dfrac{256 - 117\sqrt{3}}{560}$

$\implies V = \dfrac{4\pi}{3}(I_{1} + I_{2}) = \pi(\dfrac{280\pi - 783\sqrt{3} + 1024}{1680}) \approx$ $\boxed{1.0237294}$ .