Finding the volume of $(x-4)^2+y^2 \leqslant 4$ Calculate the volume of the ring-formed body you get when the circle formed area $$(x-4)^2+y^2 \leqslant  4$$ rotates around the y-axel.
The answer should be: $32\pi^2$
My approach was:
$$ \pi \int_2^6 \left(\sqrt{(x-4)^2-4}\right)^2 dx $$
but I suspect I've the wrong approach. Partly because I don't get a square $\pi$ in the answer
 A: An easier approach is to see that the body generated is a torus, and its volume is given by
$$V = 2\pi^2Rr^2$$
where $R=4$ is the distance from the center of the ring to the center of the torus, and $r=\sqrt{4}=2$ is the radius of the ring. Hence:
$$V = 2\pi^2\cdot4\cdot2^2 = 32\pi^2$$
A: Unfortunately your approach is wrong. 
This is a normal approach: We need to separate the circle into two curves as
$$x=4\pm\sqrt{4-y^2}=x_{\pm}.$$
Also, rotating around $y$-axis means that we need to have integral about $y$. So,
$$\begin{align}V&=\int_{-2}^{2}\pi (x_+)^2dy-\int_{-2}^{2}\pi (x_-)^2dy\\&=\pi\int_{-2}^{2}(x_++x_-)(x_+-x_-)dy\\&=\pi\int_{-2}^{2}8\times 2\sqrt{4-y^2}dy\\&=16\pi\times \color{red}{\int_{-2}^{2}\sqrt{4-y^2}dy}\\&=16\pi\times \color{red}{\frac{2^2\pi}{2}}\\&=32{\pi}^2.\end{align}$$
Here, $\color{red}{2^2\pi/2}$ represents the half of the area surrounded by $x^2+y^2=4$.
A: You may also use Volumes of Revolution and then get the following integral: 
$$2\times\int_2^62\pi x\sqrt{4-(x-4)^2}dx=32\pi^2$$

