91Ó°ÊÓ

The volume swept out equals the area times the distance moved by the centroid if a plane area is rotated about an axis in its plane but which does not intersect the area. The area swept out equals the length times the distance moved by the centroid when a plane curve is rotated about an axis in its plane but does not intersect the curve.

Consider the figure, as given below –

Let the circle have a radiusand let its centroid be at distance R from the z axis. Then due to theorem in problem 12–

$V=A\left(2\mathrm{Ï€}R\right)$

But $A=r^2\mathrm{π}$,so –

$V=2{\mathrm{Ï€}}^2{r}^2\mathrm{R}$

Due to the theorem in problem 13 –

$$\begin{gathered} S=\left(2\mathrm{Ï€}R\right)C \\ =4{\mathrm{Ï€}}^{2}rR \end{gathered}$$

Where$C=2\mathrm{Ï€}r$is the circumference of the circle.

Therefore, the theorem is proved and the volume and surface area is found as -

$V=2{\mathrm{Ï€}}^2{r}^{2}R,S=4{\mathrm{Ï€}}^{2}rR$

Hereis the radius of the cross-section of the torus and R its large radius.