Volume of solids of revolution using polar coordinates

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 $(0 \leq \theta \leq \dfrac{\pi}{6})$ and $r = 1 - \sin(\theta)$ on the branch $(\dfrac{\pi}{6} \leq \theta \leq \dfrac{\pi}{2})$ .

$V = \dfrac{4\pi}{3}(\int_{0}^{\dfrac{\pi}{6}} \sin^{4}(\theta) d\theta + \int_{\dfrac{\pi}{6}}^{\dfrac{\pi}{2}} (1 - \sin(\theta))^3 \sin(\theta) d\theta)$ .

Let $I_{1} = \int_{0}^{\dfrac{\pi}{6}} \sin^{4}(\theta) d\theta$ :

$I_{1} = \dfrac{1}{4}\int_{0}^{\dfrac{\pi}{6}} \dfrac{3}{2} - 2\cos(2\theta) + \dfrac{1}{2}\cos(4\theta) \:\ d\theta = \dfrac{1}{4}(\dfrac{3}{2}\theta -\sin(2\theta) + \dfrac{1}{8}\sin(4\theta))_{0}^{\dfrac{\pi}{6}} = \dfrac{\pi}{16} - \dfrac{7\sqrt{3}}{64}$

Let $I_{2} = \int_{\dfrac{\pi}{6}}^{\dfrac{\pi}{2}} (1 - \sin(\theta))^3 \sin(\theta) d\theta$ .

$I_{2} = \int_{\dfrac{\pi}{6}}^{\dfrac{\pi}{2}} 4\sin(\theta) + 2\cos(2\theta) + 3(\cos(\theta)^2(-\sin(\theta)) - \dfrac{\cos(4\theta)}{8} - \dfrac{15}{8}) \:\ d\theta =$ $(-4\cos(\theta) + \sin(2\theta) + (\cos(\theta))^3 - \dfrac{1}{32}\sin(4\theta) - \dfrac{15}{8}\theta)|_{\dfrac{\pi}{6}}^{\dfrac{\pi}{2}} = \dfrac{73\sqrt{3}}{64} - \dfrac{5\pi}{8}$

$\implies V = \dfrac{4\pi}{3}(I_{1} + I_{2}) = \pi(\dfrac{11\sqrt{3}}{8} - \dfrac{3\pi}{4}) = \pi(\dfrac{11\sqrt{3}}{2^3} - \dfrac{3\pi}{2^2}) =$ $\pi(\dfrac{\alpha\sqrt{\beta}}{\lambda^{\beta}} - \dfrac{\beta\pi}{\lambda^{\lambda}})$ $\implies \alpha + \beta + \lambda = \boxed{16}$ .