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5.4: Area of a Surface of Revolution

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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

Multimedia Links

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)

and Math Video Tutorials by James Sousa, Surface Area of Revolution, Part 2 (5:43).

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}.

Review Answers

  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}

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Date Created:

Feb 23, 2012

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Aug 21, 2014
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