Find the volume common to sphere $x^2+y^2+z^2$ and the cylinder $x^2+y^2<ax$

Question

Find the volume common to sphere $x^2+y^2+z^2<1$ and the cylinder $x^2+y^2<ax$

I set up the following integral :

$$I=2\cdot \iiint_{z=0}^{\sqrt{a^2-x^2-y^2}} dz\,dy\,dz = 2\cdot\iint_E\sqrt{a^2-x^2-y^2} \, dy\,dx$$ where $E:x^2+y^2=ax$

Now under polar coordinates $E:r=a\cos(\theta)$

and so $0\leq r \leq a\cos(\theta)$ and $-\pi/2\leq \theta \leq \pi/2$

$$I= 2\cdot\int_{\theta=-\pi/2}^{\pi/2}\,\int_{r=0}^{a\cos(\theta)} \sqrt{a^2-r^2}\,r\,dr\,d\theta=2\cdot (1/2)\cdot(2/3)\int_{\theta=-\pi/2}^{\pi/2}a^3\cdot(1-\sin^3(\theta))d\theta = \frac{2a^3}{3}\cdot\{\int_{\theta=-\pi/2}^{\pi/2} d\theta - \int_{\theta=-\pi/2}^{\pi/2}\sin^3(\theta)d\theta\} = \frac{2\pi a^3}{3}$$

However answer given to me is $$\frac{2a^3}{3}\cdot(\pi - \frac43)$$

Where am I making the mistake? Or is it the case that the answer given to me is incorrect?

Answer 1

$$I= 2\cdot\int_{\theta=-\pi/2}^{\pi/2}\,\int_{r=0}^{a\cos(\theta)} \sqrt{a^2-r^2}\,r\,dr\,d\theta=2\cdot (1/2)\cdot(2/3)\int_{\theta=-\pi/2}^{\pi/2}a^3\cdot(1-\color{red}{|\sin^3(\theta)|})d\theta$$

Specifically,

$$\int_0^{a\cos t} \sqrt{a^2 - r^2}rdr = \frac{1}{2}\int_{a^2\sin^2t}^{a^2} u^{1/2}dr = \frac{a^3}{3}\left(1-(\sin^2t)^{3/2}\right)= \frac{a^3}{3}\left(1-|\sin^3t|\right)$$

The rest follows as you expected.

Answer 2

Assuming $a<1/2$, here's the proper integral:

$$\int\limits_{x=0}^a \int\limits_{y=0}^\sqrt{x(a - x)} \int\limits_{z=0}^\sqrt{1 - x^2 - y^2} 4\ dx\ dy\ dz$$