Question

Find the surface area of the torus in Exercise \({\rm{47}}\) in Section \({\rm{7}}{\rm{.2}}\).

Short answer

The surface area of torus is obtained as \({\rm{A = 4}}{{\rm{\pi }}^{\rm{2}}}{\rm{Rr}}\).

Step-by-step solution

Concept Introduction

The area is the amount of space occupied by a two-dimensional flat surface. It is based on square units of measurement. The surface area of a three-dimensional object is the area filled by its outer surface. In addition, it is measured in square units.

Evaluating the distance

The given torus can be obtained by rotating\({{\rm{x}}^{\rm{2}}}{\rm{ + (y - R}}{{\rm{)}}^{\rm{2}}}{\rm{ = }}{{\rm{r}}^{\rm{2}}}\) about the \({\rm{x - }}\)axis.

Consider the area swept out by the circle when rotated through a small angle\({\rm{d\theta }}\).

Take a point on the circle such that the radius to this point makes an angle\(\phi \)with the radius towards the origin.

(This initial radius is along the\({\rm{y - }}\)axis when the circle is in its initial position).

The distance of this point from the\({\rm{x - }}\)axis is –

\({\rm{R - rcos}}\phi \)

And so, in the rotation of the circle through\({\rm{d\theta }}\)this point moves through a distance –

\({\rm{(R - rcos}}\phi {\rm{)d\theta }}\)

Evaluating the area

The length of arc generated by a rotation \({\rm{d}}\phi \) is \({\rm{rd}}\phi \).

To find the area generated by the whole circle by a rotation through\({\rm{d\theta }}\),

Integrate from\(\phi {\rm{ = 0}}\)to\({\rm{2\pi }}\)-

\({\rm{ = }}\int_{\rm{0}}^{{\rm{2\pi }}} {\left( {{\rm{Rr - }}{{\rm{r}}^{\rm{2}}}{\rm{cos}}\phi } \right)} {\rm{d}}\phi {\rm{ = 2\pi Rr}}\).

So, the area is\({\rm{2\pi Rrd\theta }}\).

Hence, integrating from\({\rm{0}}\)to\({\rm{2\pi }}\)the area is –

\(\int_{\rm{0}}^{{\rm{2\pi }}} {{\rm{2\pi Rrd\theta = 4}}} {{\rm{\pi }}^{\rm{2}}}{\rm{Rr}}\)

Therefore, the area is obtained as \({\rm{A = 4}}{{\rm{\pi }}^{\rm{2}}}{\rm{Rr}}\).