Volume of a Torus
Let a circle of radius
\[a\]
be centred at \[(0,b)\]
with \[b \gt a\]
. The circle is rotated about the \[x\]
axis.The upper half of the circle has equation
\[y_2=b+ \sqrt{a^2-x^2}\]
and the lower half has equation \[y_1= b - \sqrt{a^2-x^2}\]
.A volume element is given by
\[\begin{equation} \begin{aligned} dV &= (y^2_2-y^2_1) dx \\ &= (y_2+y_1)(y_2-y_1)dx \\ &= ((b+ \sqrt{a^2-x^2})+ (b - \sqrt{a^2-x^2}))((b+ \sqrt{a^2-x^2})- (b - \sqrt{a^2-x^2}))dx \\ &= (2b)( 2 \sqrt{a^2-x^2})dx \\ &= 4b \sqrt{a^2-x^2}dx \end{aligned} \end{equation}\]
The volume of the torus is
\[\begin{equation} \begin{aligned} V &= \pi \int^a_{-a} 4b \sqrt{a^2-x^2} dx \\ &= 8b \pi \int^a_0 \sqrt{a^2-x^2}dx \\ &= 8b \pi [ \frac{x \sqrt{a^2-x^2}}{2} + a sin^{-1} ( \frac{x}{a}) ]^a_0 \\ &= 8b \pi ( \frac{a^2}{2} \frac{\pi}{2}) \\ &=2 \pi^2 a^2b \end{aligned} \end{equation}\]
