2 The area of a surface of revolution

In Section 14.2 we found an expression for the volume of a solid of revolution. Here we consider the more complicated problem of formulating an expression for the surface area of a solid of revolution.

Figure 12

Figure 12 shows the portion of the curve $y\left(x\right)$ between $x=a$ and $x=b$ which is rotated around the $x$ axis through $360^{\circ }$ . A small disc, of thickness $\delta x$ , of the solid of revolution has been selected. Its radius is $y$ and so its circumference has length $2\pi y$ . (As usual we assume $\delta x$ is ‘small’ so that the curved part of $y\left(x\right)$ representing the hypotenuse of the highlighted ‘triangle’ can be regarded as straight ). This surface ‘ribbon’, shown shaded, has a length $2\pi y$ and a width $\sqrt{{\left(\delta x\right)}^2+{\left(\delta y\right)}^2}$ and so its area is, to a good approximation, $2\pi y\sqrt{{\left(\delta x\right)}^2+{\left(\delta y\right)}^2}$ . We now let $\delta x\to 0$ to obtain the result in Key Point 8:

Task!

Find the area of the surface generated when the part of the curve $y=x^3$ between $x=0$ and $x=4$ is rotated around the $x$ axis.

Using Key Point 8 write down the integral:

Answer

Use the substitution $u=1+9x^4$ so $\frac{du}{dx}=36x^3$ to write down the integral in terms of $u$ :

Answer

Perform the integration:

Answer

Apply the limits of integration to find the area:

Answer

Exercises

1. The line $y=x$ between $x=0$ and $x=1$ is rotated around the $x$ axis.
   1. Find the area of the surface generated.
   2. Verify this result by finding the curved surface area of the corresponding cone. (The curved surface area of a cone of radius $r$ and slant height $\ell$ is $\pi r\ell$ .)
2. Find the area of the surface generated when $y=\sqrt{x}$ in the interval $1\le x\le 2$ is rotated about the $x$ axis.

Answer