Chapter 7.3, Problem 62E

To determine

To show: The area of the surface generated by revolving the curve $x=g\left(y\right)>0$ on interval $\left[c,d\right]$ about $y$ -axis is $S={\displaystyle \underset{c}{\overset{d}{\int }}2\pi g\left(y\right)\sqrt{1+{\left(g^{\prime }\left(y\right)\right)}^2}dy}$ .

Explanation of Solution

Given information:

The areas of cross-sections perpendicular to the y-axis are given for particular values of $y$ in the table.

Calculation:

Draw the curve $x=g\left(y\right)$ and surface that is generated when this curve is rotated about the $y$ -axis. Also, draw a typical slice of slant height $\Delta s$ as shown below.

  

The surface area of the typical slice is $2\pi \cdot g\left(y\right)\cdot \Delta s$ . Remember that the slant height can be given as $\Delta s=\sqrt{1+{\left(g^{\prime }\left(y_k\right)\right)}^2}\Delta y$

Thus, the surface area can be approximated using the Riemann sum ${\displaystyle \sum _{k=1}^{n}2\pi g\left(y_k\right)\sqrt{1+{\left(g^{\prime }\left(y_k\right)\right)}^2}\Delta y}$ .

Now, by taking the limit of the Riemann sum, the surface area can be given as the integral as shown below.

  $\begin{array}{c}S=\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^{n}2\pi g\left(y_k\right)\sqrt{1+{\left(g^{\prime }\left(y_k\right)\right)}^2}\Delta y}\\ ={\displaystyle \underset{c}{\overset{d}{\int }}2\pi g\left(y\right)\sqrt{1+{\left(g^{\prime }\left(y\right)\right)}^2}dy}\end{array}$

Therefore, the surface are of the solid of revolution is $S={\displaystyle \underset{c}{\overset{d}{\int }}2\pi g\left(y\right)\sqrt{1+{\left(g^{\prime }\left(y\right)\right)}^2}dy}$ .