## Calculus Of One Real Variable – By Pheng Kim Ving Chapter 12: Applications Of The Integral – Section 12.7: Areas Of Surfaces Of Revolution

12.7
Areas Of Surfaces Of Revolution

 1. Surfaces Of Revolution

Consider a sphere of radius r. See Fig. 1.1. We wish to find the surface area of this sphere. For this purpose we set up
an xy-coordinate system in such a way that its origin is at the centre of the sphere. In Fig. 1.1 we add the z-axis to get # Fig. 1.1

Surface Area Of A Sphere.

the xyz-coordinate system to make it clear that the sphere is in the 3-D space. The sphere intersects the xy-plane in a
circle. Clearly we can consider the surface of the sphere as generated by revolving the upper semi-circle about the x-axis.
We can also consider it as generated by revolving the lower semi-circle about the x-axis. Its area is calculated in Example
5.1.
The upper and lower semi-circles are graphs of functions.

When the graph of a function is revolved (rotated) about the x-axis, it generates a surface, called a surface of revolution.
See Fig. 1.2. In general, when a plane curve is revolved about a line in the plane of the curve, it generates a surface
called a surface of revolution. In this section we'll find areas of surfaces of  revolution. # Graph of f, when revolved about x-axis, generates a surface of revolution.  # Area of element of surface of revolution swept out by a chord.

the circumference of revolution at xi–1) and |Pi–1 Pi| is the length of chord Pi–1 Pi. Now r(xi–1) = | f(xi–1)|; the absolute
value is necessary because a radius must be positive or 0 and we don't require the restriction that f(x) be positive or 0 on
[a, b], so that general f(x) may be sometimes negative as well as sometimes positive. Wherever f(x) is negative, we
take its absolute value for the radius of the circle generated by the point (x, f(x)); see Examples 2.1 and 2.2 below. By
Section 12.6 Part 2, there's ci in [xi–1, xi] such that: In Section 12.6 Parts 1 and 2, we approximated the length of the graph by the sum |P0 P1| +|P1 P2| + ... +
|Pn–1 Pn| of the chords. So here we of course approximate the area S generated by the graph over [a, b] by the sum S1 +
S2 + ... + Sn of the areas generated by the chords. Consequently: We've just shown that:

 The area S of the surface of revolution for y = f(x) from x = a to x = b about the x-axis is: ### Areas As Integrals Of Differential Areas

Let s be the arc length of the graph of f over [a, b]. As discussed in Section 12.6 Part 3, the arc length element or
differential of arc length for f at x is the length ds, as shown in Fig. 2.2, of a short tangent line segment at x, given by: So Eq. [2.1] can be written as:  # Fig. 2.2 The area S can be regarded as the integral of the differential area dS: The fact that S is the integral of dS is consistent with the fact that s is the integral of ds. If you forget Eq. [2.1], try to
obtain it by going thru Eqs. [2.2], [2.3], and [2.4].

### Example 2.1

Find the area of the surface of revolution obtained by revolving the graph of y = f(x) = 2x from x = –3 to x = –1 about
the x-axis.

Incorrect Solution

Let S be the required area. Then: EOS

Of course the solution above is incorrect, since an area can't be negative. The culprit is the incorrect absolute value. It
must be that |2x| = –2x, as 2x < 0 on [–3, –1]. The correct solution is given below.

Correct Solution

Let S be the required area. Then: EOS

A sketch of the graph of f for help doesn't have to be drawn if not asked for. Simply recall the general-situation Fig. 2.2
mentally, apply Eq. [2.1], and get the correct absolute values.

### Example 2.2

Calculate the area of the surface generated by revolving the arc of y = x3 from x = –1 to x = 2 about the x-axis.

## Solution

Let S be the desired area. Then:  EOS

If f(x) (or y) changes sign over [a, b], we must consider sub-intervals of [a, b] where it's positive and where it's negative
in order to get its correct absolute values, then add up the integrals over those sub-intervals.

Again a sketch of the graph of f for help doesn't have to be drawn if not asked for. Simply recall the general-situation Fig.
2.2 mentally, apply Eq. [2.1], and get the correct absolute values.

Suppose the graph of a continuously differentiable function f over [a, b] is revolved about the y-axis. See Fig. 3.1. The
graph generates a surface of revolution. We're going to find the area S of this surface. Let x be an arbitrary point in [a,
b]. The tangent line segment at x whose length is the differential arc length ds describes the differential area dS of S.
The area S is the integral of the differential area dS. We have: the absolute value being required since there are x < 0 if a < 0. So:

 The area S of the surface of revolution for y = f(x) from x = a to x = b about the y-axis is:  # Differential surface area dS is described by differential arc length ds.

Remark that in Eq. [3.1], the integration is still along the x-axis, not along the y-axis, although the revolution is this time

### Example 3.1 ## Solution

Let S be the required area. Then: ## EOS

A sketch of the graph of f for help doesn't have to be drawn if not asked for. Simply recall the general-situation Fig. 3.1
mentally, apply Eq. [3.1], and get the correct absolute values.

 4. General Case  # Area of revolution is integral of differential area. Radius of revolution at x is r(x) = |f(x) – k|. Here r(x) = k – f(x) for all x in [a, b]. # Area of revolution is integral of differential area. Radius of revolution at x is r(x) = |x – k|. Here r(x) = k – x for all x in [a, b].

In summary:

 The area S of the surface of revolution for f(x) from x = a to x = b: ### Example 4.1

Find the area of the surface generated by revolving the line segment y = x + 2 from y = 2 to y = 5 about the line y = 4.

## Solution

The x-coordinate of the point of intersection of y = x + 2 and y = 4 is x where x + 2 = 4 or x = 2. From y = x + 2 we
have x = y – 2. When y increases from 2 to 5 along the line y = x + 2, x increases from 2 – 2 = 0 to 5 – 2 = 3. Let S be
the desired area. Then: EOS

Because the formulas for areas of surfaces of revolution presented in this section correspond to integration along the
x-axis, we must change any y-interval to its corresponding x-interval in order to get the x-limits of integration to be
utilized in these formulas.

A sketch of the graph of f for help doesn't have to be drawn if not asked for. According to the case at hand, simply recall
the general-situation either Fig. 4.1 or Fig. 4.2 mentally, apply either Eq. [4.1] or Eq. [4.2], and get the correct absolute
values. Eq. [4.3] can be used in any case.

 5. Applications

### Example 5.1 ## Solution # Fig. 5.1

Graph For Example 5.1. EOS

 Problems & Solutions

1. Find the area of the surface of revolution obtained by revolving the graph of: from x = 1 to x = 4 about the x-axis.

## Solution

Let S be the required area. Then:   2. Calculate the area of the surface of revolution obtained by revolving the curve: ### Solution

Let S be the desired area. Then:  3. Compute the area of the surface generated by revolving the arc y = ln x from x = 1 to x = e about the y-axis.

Solution

Let S be the required area. Then:    4. Find the area of the surface obtained by revolving the line segment x = y + 3 from y = –5 to y = 1 about the line
x = 2.

### Solution

From x = y + 3 we have y = x – 3. When y increases from –5 to 1 along the line x = y + 3, x increases from – 5 + 3 =
–2 to 1 + 3 = 4. Let S be the desired area. Then:   ### Solution The equation of the line segment (0, h)–( r, 0) in the xy-coordinate system with its origin at the center of the circular
base of the cone is: 