Question

Using the method of cylindrical shells, set up an integral for the volume of the torus formed when the circle of radius 2 units covered at (3,0) is revolved about the line x=6

Answer 1

The equation of the circle is-

$(x-3)^2+y^2=4\Rightarrow y=f(x)=\sqrt{4-(x-3)^2}$

The Volume of the torus formed about the line x=6 is-

$V=2\pi\int_{0}^2xf(x)dx$

$=2\pi\int_{0}^2x\sqrt{4-(x-3)^2}dx$

$=2\pi\int_0^2(x-3+3)\sqrt{4-(x-3)^2}dx$

$=2\pi[\int_0^2(x-3)\sqrt{4-(x-3)^2}dx+\int_0^23\sqrt{4-(x-3)^2}dx]$

$=2\pi[-\dfrac{(4-(x-3)^2)^{\frac{3}{2}}}{3}+\dfrac{3(x-3)}{2}\sqrt{4-(x-3)^2}+6sin^{-1}(\dfrac{x-3}{2})]_0^2$

$=2\pi[-\dfrac{-1-\dfrac{3}{2}\sqrt{3}}{1}+6sin^{-1}(-1)-(-\dfrac{5}{3}+\dfrac{-9}{2}\sqrt{4}+6sin^{-1}(1)$

$=2\pi (1+\dfrac{3\sqrt{3}}{2}+\dfrac{5}{3}+9)=2\pi(\dfrac{35}{3}+\dfrac{3\sqrt{3}}{2})$