6.4 Arc length of a curve and surface area (Page 3/8)

Calculus volume 1 Page 3 / 8

Area of a surface of revolution

The concepts we used to find the arc length of a curve can be extended to find the surface area of a surface of revolution. Surface area is the total area of the outer layer of an object. For objects such as cubes or bricks, the surface area of the object is the sum of the areas of all of its faces. For curved surfaces, the situation is a little more complex. Let $f (x)$ be a nonnegative smooth function over the interval $[a, b] .$ We wish to find the surface area of the surface of revolution created by revolving the graph of $y = f (x)$ around the $x -axis$ as shown in the following figure.

This figure has two graphs. The first graph is labeled “a” and is a curve in the first quadrant beginning at the y-axis. The curve is y=f(x). The second graph is labeled “b” and has the same curve y=f(x). There is also a solid surface formed by rotating the curve about the x-axis. — (a) A curve representing the function $f (x) .$ (b) The surface of revolution formed by revolving the graph of $f (x)$ around the $x -axis .$

As we have done many times before, we are going to partition the interval $[a, b]$ and approximate the surface area by calculating the surface area of simpler shapes. We start by using line segments to approximate the curve, as we did earlier in this section. For $i = 0, 1, 2,…, n,$ let $P = {x_{i}}$ be a regular partition of $[a, b] .$ Then, for $i = 1, 2,…, n,$ construct a line segment from the point $(x_{i - 1}, f (x_{i - 1}))$ to the point $(x_{i}, f (x_{i})) .$ Now, revolve these line segments around the $x -axis$ to generate an approximation of the surface of revolution as shown in the following figure.

This figure has two graphs. The first graph is labeled “a” and is a curve in the first quadrant beginning at the y-axis. The curve is y=f(x). The area under the curve above the x-axis has been divided into regions with vertical lines. The second graph is labeled “b” and has the same curve y=f(x). There is also a solid surface formed by rotating the curve about the x-axis. — (a) Approximating $f (x)$ with line segments. (b) The surface of revolution formed by revolving the line segments around the $x -axis .$

Notice that when each line segment is revolved around the axis, it produces a band. These bands are actually pieces of cones (think of an ice cream cone with the pointy end cut off). A piece of a cone like this is called a frustum of a cone.

To find the surface area of the band, we need to find the lateral surface area, $S,$ of the frustum (the area of just the slanted outside surface of the frustum, not including the areas of the top or bottom faces). Let $r_{1}$ and $r_{2}$ be the radii of the wide end and the narrow end of the frustum, respectively, and let $l$ be the slant height of the frustum as shown in the following figure.

This figure is a graph. It is a frustum of a cone above the x-axis with the y-axis in the center. The radius of the bottom of the frustum is rsub1 and the radius of the top is rsub2. The length of the side is labeled “l”. — A frustum of a cone can approximate a small part of surface area.

We know the lateral surface area of a cone is given by

Lateral Surface Area = π r s,

where $r$ is the radius of the base of the cone and $s$ is the slant height (see the following figure).

This figure is a cone. The cone has radius r, height h, and length of side s. — The lateral surface area of the cone is given by $π r s .$

Since a frustum can be thought of as a piece of a cone, the lateral surface area of the frustum is given by the lateral surface area of the whole cone less the lateral surface area of the smaller cone (the pointy tip) that was cut off (see the following figure).

This figure is a graph. It is a frustum of a cone. The radius of the bottom of the frustum is rsub1 and the radius of the top is rsub2. The length of the side is labeled “l”. There is also the top of a cone with broken lines above the frustum. It has length of side s. — Calculating the lateral surface area of a frustum of a cone.

The cross-sections of the small cone and the large cone are similar triangles, so we see that

\frac{r_{2}}{r_{1}} = \frac{s - l}{s} .

Solving for $s,$ we get

\begin{matrix} \frac{r_{2}}{r_{1}} & = & \frac{s - l}{s} \\ r_{2} s & = & r_{1} (s - l) \\ r_{2} s & = & r_{1} s - r_{1} l \\ r_{1} l & = & r_{1} s - r_{2} s \\ r_{1} l & = & (r_{1} - r_{2}) s \\ \frac{r_{1} l}{r_{1} - r_{2}} & = & s . \end{matrix}

Then the lateral surface area (SA) of the frustum is

\begin{matrix} S & = (Lateral SA of large cone) - (Lateral SA of small cone) \\ = π r_{1} s - π r_{2} (s - l) \\ = π r_{1} (\frac{r_{1} l}{r_{1} - r_{2}}) - π r_{2} (\frac{r_{1} l}{r_{1} - r_{2}} - l) \\ = \frac{π r_{1}^{2} l}{r_{1} - r_{2}} - \frac{π r_{1} r_{2} l}{r_{1} - r_{2}} + π r_{2} l \\ = \frac{π r_{1}^{2} l}{r_{1} - r_{2}} - \frac{π r_{1} r_{2} l}{r_{1} - r_{2}} + \frac{π r_{2} l (r_{1} - r_{2})}{r_{1} - r_{2}} \\ = \frac{π r_{1}^{2} l}{r_{1} - r_{2}} - \frac{π r_{1} r_{2} l}{r_{1} - r_{2}} + \frac{π r_{1} r_{2} l}{r_{1} - r_{2}} - \frac{π r_{2}^{2} l}{r_{1} - r_{2}} \\ = \frac{π (r_{1}^{2} - r_{2}^{2}) l}{r_{1} - r_{2}} = \frac{π (r_{1} - r_{2}) (r_{1} + r_{2}) l}{r_{1} - r_{2}} = π (r_{1} + r_{2}) l . \end{matrix}

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Source: OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2

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