2 volumes of viviani's vault

Question

I need some help calculating the following volume with triple integral. Te volume is half viviani's vault, given by $$ 0\leq z \leq \sqrt{16-x^2-y^2}, \ x^2+y^2-4x=0 $$ I tried with polar coordinates (I take $\theta\in]-\pi,\pi[$) and got the following integral: $$ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_0^{4\cos(\theta)}\int_0^\sqrt{16-\rho^2}\rho dzd\rho d\theta $$ I think that the limits are ok, but calculating: $$ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_0^{4\cos(\theta)}\int_0^\sqrt{16-\rho^2}\rho dzd\rho d\theta= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_0^{4\cos(\theta)}\sqrt{16-\rho^2}\rho d\rho d\theta= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{-1}{3}(16-\rho^2)^{3/2}|_0^{4\cos(\theta)}d\theta= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{64-64\sin(\theta)^3}{3}d\theta=\frac{64\pi}{3} $$ However, that is not the volume, it is supposed to be $\frac{64\pi}{3}-\frac{256}{9}$

Answer 1

There is a mistake in your computation. Remember that $\sqrt{x^2}=|x|$ for all $x$. In this case, when $\theta\in\left[-\frac\pi2,0\right)$ we have $\sin(\theta)<0$, so that $\left(\sin^2(\theta)\right)^{3/2} = -\sin^3(\theta)$.

$$\begin{align*} V &= -\frac{64}3 \int_{-\tfrac\pi2}^{\tfrac\pi2} \left(\left(\sin^2(\theta)\right)^{3/2} - 1\right) \, d\theta \\ &= -\frac{64}3 \left(\int_{-\tfrac\pi2}^0 \left(-\sin^3(\theta) - 1\right) \, d\theta + \int_0^{\tfrac\pi2} \left(\sin^3(\theta) - 1\right) \, d\theta\right) \\ &= \frac{64}3 \left(\pi - 2 \int_0^{\tfrac\pi2}\sin^3(\theta) \, d\theta\right) \end{align*}$$