# I wanna calculate the volume of a torus that is created when the circlearea (x-4)^2+y^2 leq 4 revolves around the y-axis.

Volume of torus
I wanna calculate the volume of a torus that is created when the circlearea $\left(x-4{\right)}^{2}+{y}^{2}\le 4$ revolves around the y-axis. I don't understand what the difference would be if I calculated the volume of the circle $\left(x-4{\right)}^{2}+{y}^{2}=4$ revolving around the y-axis. Doesn't the circlearea just create a solid torus instead of a hollow? Then comes the practical problem of finding a formula for the volume. My idea is that if the distance from the "right" and "left" side of the circle to origo is R and r respectively, then the area of the thin circles that constitute the torus is $\pi \left({R}^{2}-{r}^{2}\right)$. Then the height is dy and so we integrate from, if x denotes the radius of the circle, -r to r. Would this work?
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Step 1
$V=\pi {\int }_{-2}^{2}{\left(4+\sqrt{\left(4-{y}^{2}\right)}\right)}^{2}-{\left(4-\sqrt{\left(4-{y}^{2}\right)}\right)}^{2}dy$
Step 2
Outer radius $=4+\sqrt{4-{y}^{2}}$
Inner radius $=4-\sqrt{4-{y}^{2}}$
$V={\int }_{-2}^{2}\left(16\pi \sqrt{4-{y}^{2}}\right)dy$
${\int }_{-2}^{2}\sqrt{4-{y}^{2}}dy=2\pi$
$V=16\pi \ast 2\pi =32{\pi }^{2}$