Volume of Solids of Revolution and Polar Functions Revisited.

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

$r = 1 + \cos(\theta)$ on the branch $(\dfrac{\pi}{4} \leq \theta \leq \dfrac{5\pi}{4})$ and $r = 1 + \sin(\theta)$ on the branch $(\dfrac{5\pi}{4} \leq \theta \leq \dfrac{9\pi}{4})$

Let $I_{1} = \dfrac{2\pi}{3}\int_{\dfrac{\pi}{4}}^{\dfrac{5\pi}{4}} (1 + \cos(\theta))^3 \sin(\theta) d\theta =$ $\dfrac{2\pi}{3}\int_{\dfrac{\pi}{4}}^{\dfrac{5\pi}{4}} (1 + 3\cos(\theta) + 3\cos^2(\theta) + \cos^3(\theta)) \sin(\theta) d\theta = \dfrac{2\pi}{3}(\cos(\theta) - \dfrac{3}{2}\cos^2(\theta) - \cos^3(\theta) - \dfrac{1}{4}\cos^4(\theta))|_{\dfrac{\pi}{4}}^{\dfrac{5\pi}{4}} =$

$\dfrac{2\pi}{3}(\dfrac{1}{\sqrt{2}} - \dfrac{3}{4} + \dfrac{1}{2\sqrt{2}} - \dfrac{1}{16} - (-\dfrac{1}{\sqrt{2}} - \dfrac{3}{4} - \dfrac{1}{2\sqrt{2}} - \dfrac{1}{16}) = \dfrac{2\pi}{3}(\sqrt{2} + \dfrac{1}{\sqrt{2}}) = \dfrac{2\pi}{3}(\dfrac{3}{\sqrt{2}}) = \boxed{\sqrt{2}\pi}$ .

Let $I_{2} = \dfrac{2\pi}{3}\int_{\dfrac{5\pi}{4}}^{\dfrac{9\pi}{4}} (1 + \sin(\theta))^3 \sin(\theta) d\theta =$

$\dfrac{2\pi}{3}\int_{\dfrac{5\pi}{4}}^{\dfrac{9\pi}{4}} (4\sin(\theta) + 2\cos(2\theta) - 3(\cos(\theta))^2 \sin(\theta) + \dfrac{1}{8}\cos(4\theta) + \dfrac{15}{8}) d\theta =$

$\dfrac{2\pi}{3}(-4\cos(\theta) + \sin(2\theta) + \cos^3(\theta) + \dfrac{1}{32}\sin(4\theta) + \dfrac{15}{8}\theta)|_{\dfrac{5\pi}{4}}^{\dfrac{9\pi}{4}} = \boxed{\dfrac{2\pi}{3}(\dfrac{15\pi}{8} - \dfrac{7}{\sqrt{2}})}$

$\implies V = I_{1} + I_{2} = \dfrac{2\pi}{3}(\dfrac{15\pi}{8} - 2\sqrt{2}) \approx \boxed{6.41316158}$