Volume of Solids using Polar functions

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

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

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

$I_{1} = \int_{0}^{\dfrac{\pi}{2}} (1 - \cos(\theta))^3 \sin(\theta) d\theta = \int_{0}^{\dfrac{\pi}{2}} (1 - 3\cos(\theta) + 3\cos^2(\theta) - cos^3(\theta)) \sin(theta) d\theta =$

$-\cos(\theta) + \dfrac{3}{2}\cos^2(\theta) - \cos^3(\theta) + \dfrac{1}{4}\cos^4(\theta)|_{0}^{\dfrac{\pi}{2}} = \dfrac{1}{4}$

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

$\implies V = \dfrac{4\pi}{3}(I_{1} + I_{2}) = \dfrac{4\pi}{3}(\dfrac{1}{2}) = \dfrac{1}{2}V_{s} \implies$ $\dfrac{V}{V_{s}} = \dfrac{1}{2} = \dfrac{a}{b} \implies a + b = \boxed{3}$ .