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gayaneshka [121]
3 years ago
14

A torus is formed by rotating a circle of radius r about a line in the plane of the circle that is a distance R (> r) from th

e center of the circle. Find the volume of the torus.
Mathematics
1 answer:
jeyben [28]3 years ago
6 0

Consider a circle with radius r centered at some point (R+r,0) on the x-axis. This circle has equation

(x-(R+r))^2+y^2=r^2

Revolve the region bounded by this circle across the y-axis to get a torus. Using the shell method, the volume of the resulting torus is

\displaystyle2\pi\int_R^{R+2r}2xy\,\mathrm dx

where 2y=\sqrt{r^2-(x-(R+r))^2}-(-\sqrt{r^2-(x-(R+r))^2})=2\sqrt{r^2-(x-(R+r))^2}.

So the volume is

\displaystyle4\pi\int_R^{R+2r}x\sqrt{r^2-(x-(R+r))^2}\,\mathrm dx

Substitute

x-(R+r)=r\sin t\implies\mathrm dx=r\cos t\,\mathrm dt

and the integral becomes

\displaystyle4\pi r^2\int_{-\pi/2}^{\pi/2}(R+r+r\sin t)\cos^2t\,\mathrm dt

Notice that \sin t\cos^2t is an odd function, so the integral over \left[-\frac\pi2,\frac\pi2\right] is 0. This leaves us with

\displaystyle4\pi r^2(R+r)\int_{-\pi/2}^{\pi/2}\cos^2t\,\mathrm dt

Write

\cos^2t=\dfrac{1+\cos(2t)}2

so the volume is

\displaystyle2\pi r^2(R+r)\int_{-\pi/2}^{\pi/2}(1+\cos(2t))\,\mathrm dt=\boxed{2\pi^2r^2(R+r)}

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Hi can some please help?
Oduvanchick [21]
So I'm assuming that you're taking Calculus.
The first thing you want to do is take the integral of f(x)...
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