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# 5.4: Area of a Surface of Revolution

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

A student will be able to:

• Learn how to find the area of a surface that is generated by revolving a curve about an axis or a line.

In this section we will deal with the problem of finding the area of a surface that is generated by revolving a curve about an axis or a line. For example, the surface of a sphere can be generated by revolving a semicircle about its diameter (Figure 19) and the circular cylinder can be generated by revolving a line segment about any axis that is parallel to it (Figure 20).

Figure 19

Figure 20

Area of a Surface of Revolution

If \begin{align*}f\end{align*} is a smooth and non-negative function in the interval \begin{align*}[a, b],\end{align*} then the surface area \begin{align*}S\end{align*} generated by revolving the curve \begin{align*}y = f(x)\end{align*} between \begin{align*}x = a\end{align*} and \begin{align*}x = b\end{align*} about the \begin{align*}x-\end{align*}axis is defined by

\begin{align*} S = \int_{a}^{b} 2 \pi f(x) \sqrt{1 + [f'(x)]^2} dx = \int_{a}^{b} 2 \pi y \sqrt{1 + \left (\frac{dy} {dx}\right )^2} dx.\end{align*}

Equivalently, if the surface is generated by revolving the curve about the \begin{align*}y-\end{align*}axis between \begin{align*}y = c\end{align*} and \begin{align*}y = d,\end{align*} then

\begin{align*} S = \int_{c}^{d} 2 \pi g (y) \sqrt{1 + [g'(y)]^2} dy = \int_{c}^{d} 2 \pi x \sqrt{1 + \left (\frac{dx} {dy}\right )^2}dy.\end{align*}

Example 1:

Find the surface area that is generated by revolving \begin{align*}y = x^3\end{align*} on \begin{align*}[0, 2]\end{align*} about the \begin{align*}x-\end{align*}axis (Figure 21).

Solution:

Figure 21

The surface area \begin{align*}S\end{align*} is

\begin{align*}S &= \int_{a}^{b} 2 \pi y \sqrt{1 + \left (\frac{dy} {dx}\right )^2} dx \\ &= \int_{0}^{2} 2 \pi x^3 \sqrt{1 + (3x^2)^2} dx \\ &= 2\pi \int_{0}^{2} x^3 (1 + 9x^4)^{1/2} dx.\end{align*}

Using \begin{align*}u-\end{align*}substitution by letting \begin{align*}u = 1 + 9x^4,\end{align*}

\begin{align*}S &= 2 \pi \int_{1}^{145} u^{1/2} \frac{du}{36} \\ &= \frac{2 \pi}{36} \left [\frac{2} {3} u^{3/2}\right ]^{145}_{1} \\ &= \frac{2 \pi}{36}\cdot \frac{2} {3} \left [(145)^{3/2} - 1\right ]\\ &\approx \frac{4\pi} {108} [1745]\\ &\approx 203\end{align*}

Example 2:

Find the area of the surface generated by revolving the graph of \begin{align*}f(x) = x^2\end{align*} on the interval \begin{align*}[0, \sqrt{3}]\end{align*} about the \begin{align*}y-\end{align*}axis (Figure 22).

Solution:

Figure 22

Since the curve is revolved about the \begin{align*}y-\end{align*}axis, we apply

\begin{align*} S = \int_{c}^{d} 2 \pi x \sqrt{1 + \left (\frac{dx} {dy}\right )^2} dy.\end{align*}

So we write \begin{align*}y = x^2\end{align*} as \begin{align*} x = \sqrt{y}\end{align*}. In addition, the interval on the \begin{align*}x-\end{align*}axis \begin{align*} [0, \sqrt{3}]\end{align*} becomes \begin{align*}[0, 3].\end{align*} Thus

\begin{align*} S = \int_{0}^{3} 2 \pi \sqrt{y} \sqrt{1 + \left (\frac{1} {2\sqrt{y}}\right )^2} dy.\end{align*}

Simplifying,

\begin{align*} S = \pi \int_{0}^{3} \sqrt{4y + 1} dy.\end{align*}

With the aid of \begin{align*}u-\end{align*}substitution, let \begin{align*}u = 4y + 1,\end{align*}

\begin{align*}S &= \frac{\pi} {4} \int_{1}^{13} u^{1/2} du \\ &= \frac{\pi} {6} \left [(13)^{3/2} - 1\right ] \\ &= \frac{\pi} {6} [46.88 - 1]\\ &\approx 24\end{align*}

For video presentations of finding the surface area of revolution (16.0), see Math Video Tutorials by James Sousa, Surface Area of Revolution, Part 1 (9:47)

## Review Questions

In problems #1 - 3 find the area of the surface generated by revolving the curve about the \begin{align*}x-\end{align*}axis.

1. \begin{align*} y = 3x , 0 \le x \le 1\end{align*}
2. \begin{align*} y = \sqrt{x}, 1 \le x \le 9\end{align*}
3. \begin{align*} y = \sqrt{4 - x^2}, -1 \le x \le 1\end{align*}

In problems #4–6 find the area of the surface generated by revolving the curve about the \begin{align*}y-\end{align*}axis.

1. \begin{align*} x = 7y + 2, 0 \le y \le 3\end{align*}
2. \begin{align*} x = y^3, 0 \le y \le 8\end{align*}
3. \begin{align*} x = \sqrt{9 - y^2}, -2 \le y \le 2\end{align*}
4. Show that the surface area of a sphere of radius \begin{align*}r\end{align*} is \begin{align*} 4\pi r^2\end{align*}.
5. Show that the lateral area \begin{align*}S\end{align*} of a right circular cone of height \begin{align*}h\end{align*} and base radius \begin{align*}r\end{align*} is

\begin{align*} S = \pi r \sqrt{r^2 + h^2}.\end{align*}

1. \begin{align*} 3 \pi \sqrt{10}\end{align*}
2. \begin{align*} \approx 112\end{align*}
3. \begin{align*} 8 \pi\end{align*}
4. \begin{align*} 75 \pi \sqrt{50}\end{align*}
5. \begin{align*} \approx 823583\end{align*}
6. \begin{align*} 24\pi\end{align*}
7. We can create a sphere of radius \begin{align*}r\end{align*} by rotating a semicircle of radius \begin{align*}r\end{align*} around the \begin{align*}x-\end{align*}axis. The formula for a circle of radius \begin{align*}r\end{align*} is \begin{align*}x^2+y^2=r^2\end{align*}, and we can solve for \begin{align*}y\end{align*} to get the equation for the semicircle: \begin{align*}y=\sqrt{r^2-x^2}= (r^2-x^2)^{\frac{1}{2}}\end{align*}. Then we can also solve for: \begin{align*}\frac{dy}{dx} = \frac{1}{2} (-2x)(r^2-x^2)^{\frac{-1}{2}} = -x (r^2-x^2)^{\frac{-1}{2}}\end{align*} and so \begin{align*}\left(\frac{dy}{dx}\right)^2 = x^2 (r^2-x^2)^{-1}\end{align*}. Now we can plug all this into the integral that gives us the surface area, integrating from \begin{align*}-r \le x \le r\end{align*}:

\begin{align*}A &= \int_{-r}^{r} 2 \pi y \sqrt{1+[y^{\prime}]^2}dx\\ A &= \int_{-r}^{r} 2 \pi (r^2-x^2)^{\frac{1}{2}} \sqrt{1+(x^2(r^2-x^2)^{-1})}dx\\ A &= \int_{-r}^{r} 2 \pi \sqrt{(r^2-x^2) (1+(x^2(r^2-x^2)^{-1}))}dx\\ A &= \int_{-r}^{r} 2 \pi \sqrt{r^2-x^2+x^2}dx\\ A &= 2 \pi r \int_{-r}^{r} dx\\ A &= 2 \pi r (x) \Big |_{-r}^r\\ A &= 4 \pi r^2\end{align*}

1. We can create this right circular cone by rotating a line around the \begin{align*}x-\end{align*}axis over the interval \begin{align*}0 \le x \le h\end{align*}. Two points we know that have to be on the line are \begin{align*}(0, r)\end{align*} and \begin{align*}(h, 0)\end{align*}, which then gives us an equation for the line of \begin{align*}y = - \frac{r}{h}x+r\end{align*}. Then we can determine the derivative of \begin{align*}y\end{align*} with respect to \begin{align*}x: y^{\prime} = - \frac{r}{h}\end{align*}, and so \begin{align*}(y^{\prime})^2 = \frac{r^2}{h^2}\end{align*}. And now we are ready to calculate the integral that gives the lateral surface area:

\begin{align*}S &= \int_0^h 2 \pi y \sqrt{1+[y^{\prime}]^2}dx\\ S &= \int_0^h 2 \pi \left(- \frac{r}{h}x + r \right) \sqrt{1+ \frac{r^2}{h^2}}dx\\ S &= 2 \pi \sqrt{\frac{h^2}{h^2} + \frac{r^2}{h^2}} \left( - \frac{r}{h} \left(\frac{x^2}{2}\right)+rx\right) \Bigg |_0^h\\ S &= \frac{2 \pi}{h} \sqrt{h^2+r^2} \left(\frac{-rh}{2}+rh\right)\\ S &= \pi r \sqrt{h^2+r^2}\end{align*}

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Feb 23, 2012
Jun 10, 2016
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CK.MAT.ENG.SE.1.Calculus.5.4