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5.4: Area of a Surface of Revolution

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 $f$ is a smooth and non-negative function in the interval $[a, b],$ then the surface area $S$ generated by revolving the curve $y = f(x)$ between $x = a$ and $x = b$ about the $x-$axis is defined by

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

Equivalently, if the surface is generated by revolving the curve about the $y-$axis between $y = c$ and $y = d,$ then

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

Example 1:

Find the surface area that is generated by revolving $y = x^3$ on $[0, 2]$ about the $x-$axis (Figure 21).

Solution:

Figure 21

The surface area $S$ is

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

Using $u-$substitution by letting $u = 1 + 9x^4,$

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

Example 2:

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

Solution:

Figure 22

Since the curve is revolved about the $y-$axis, we apply

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

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

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

Simplifying,

$S = \pi \int_{0}^{3} \sqrt{4y + 1} dy.$

With the aid of $u-$substitution, let $u = 4y + 1,$

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

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 $x-$axis.

1. $y = 3x , 0 \le x \le 1$
2. $y = \sqrt{x}, 1 \le x \le 9$
3. $y = \sqrt{4 - x^2}, -1 \le x \le 1$

In problems #4–6 find the area of the surface generated by revolving the curve about the $y-$axis.

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

$S = \pi r \sqrt{r^2 + h^2}.$

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

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

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

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

Feb 23, 2012

Aug 21, 2014