Revolve $\{(x,y) | x^2 + y^2 \le y\}$ around $y=0$

Question

Find the volume of the solid that is obtained when $\{(x,y) | x^2 + y^2 \le y\}$ is revolved around $x$-axis

I deduced

$$x^2 + y^2 = y$$

$$\iff x^2+(y-\frac12)^2 = (\frac12)^2$$

So I split the circle in two and then...

Answer 1

... this is my set up

$$V=V_1+V_2$$

$$V_1 = \pi \int_{-\frac12}^{\frac12}\left(\frac12 + \sqrt{\frac14-x^2} - 0 \right)^2-\left(\frac12-0 \right)^2 dx$$

$$V_2 = \pi \int_{-\frac12}^{\frac12}\left(\frac12-0 \right)^2 - \left(\frac12 - \sqrt{\frac14-x^2} - 0 \right)^2 dx$$

Hence $$V=\pi^2/8 + \pi/6 + \pi^2/8 - \pi/6 = \pi^2/4$$

Answer 2

The solid is a torus with the radius of the tube $r=1/2$ and the distance from the center of the tube to the center of the torus $R=1/2$. The volume (as can be show using the Pappus centroid theorem), is:

$$ V=2\pi^2Rr^2= \frac{\pi^2}{4} $$