How do you use an integral to find the volume of a solid torus?

1 Answer
Wataru
Aug 30, 2014

If the radius of its circular cross section is #r#, and the radius of the circle traced by the center of the cross sections is #R#, then the volume of the torus is #V=2pi^2r^2R#.

Let's say the torus is obtained by rotating the circular region #x^2+(y-R)^2=r^2# about the #x#-axis. Notice that this circular region is the region between the curves: #y=sqrt{r^2-x^2}+R# and #y=-sqrt{r^2-x^2}+R#.

By Washer Method, the volume of the solid of revolution can be expressed as:
#V=pi int_{-r}^r[(sqrt{r^2-x^2}+R)^2-(-sqrt{r^2-x^2}+R)^2]dx#,
which simplifies to:
#V=4piR\int_{-r}^r sqrt{r^2-x^2}dx#
Since the integral above is equivalent to the area of a semicircle with radius r, we have
#V=4piRcdot1/2pi r^2=2pi^2r^2R#

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