Volume of solids of Revolution and Polar Functions.

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

In the first quadrant $\sin(3\theta) = \sin(\theta) \implies 3\sin(\theta) - 4\sin^3(\theta) = \sin(\theta) \implies 2\sin(\theta)(1 - 2\sin^2(\theta)) = 0 \implies \theta = 0, \dfrac{\pi}{4}$ and $\sin(3\theta) = 0 \implies \theta = \dfrac{\pi}{3} \implies$ .

$r = \sin(\theta)$ on the branch $(0 \leq \theta \leq \dfrac{\pi}{4})$ and $r = \sin(3\theta)$ on the branch $(\dfrac{\pi}{4} \leq \theta \leq \dfrac{\pi}{3})$ .

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

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

$I_{1} = \dfrac{1}{4}\int_{0}^{\dfrac{\pi}{4}} \dfrac{3}{2} - 2\cos(2\theta) + \dfrac{1}{2}\cos(4\theta) \:\ d\theta = \dfrac{1}{4}(\dfrac{3}{2}\theta -\sin(2\theta) + \dfrac{1}{8}\sin(4\theta))|_{0}^{\dfrac{\pi}{4}} = \dfrac{3\pi}{32} - \dfrac{1}{4}$ .

Let $I_{2} = \int_{\dfrac{\pi}{4}}^{\dfrac{\pi}{3}} \sin^{3}(3\theta) \sin(\theta) d\theta$ :

$I_{2} = \dfrac{1}{8}\int_{\dfrac{\pi}{4}}^{\dfrac{\pi}{3}} \int (3\cos(2\theta) - 3\cos(4\theta) - \cos(8\theta) + \cos(10\theta)) d\theta$ $= \dfrac{3}{16}\sin(2\theta) - \dfrac{3}{32}\sin(4\theta) - \dfrac{1}{64}\sin(8\theta) + \dfrac{1}{80}\sin(10\theta)|_{\dfrac{\pi}{4}}^{\dfrac{\pi}{3}} =$ $\dfrac{81\sqrt{3}}{640} - \dfrac{1}{5}$

$\implies V = \dfrac{4\pi}{3}(I_{1} + I_{2}) = \dfrac{\pi}{480}(81\sqrt{3} - 288 + 60\pi) \approx$ $\boxed{0.26698}$ .