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Surface Area of Revolution · Tutorial 430 of 998

Revolution About A Shifted Horizontal Axis

Set up and interpret surface-area integrals for graphs revolved about a shifted horizontal axis, using the curve’s perpendicular distance from that axis.

Intermediate 9 min read

What You'll Learn

  • Identify the radius of revolution as the perpendicular distance from the graph to \(y=c\).
  • Write the surface-area integral for \(y=f(x)\) over a specified interval.
  • Choose the correct radius expression when the graph lies above or below the axis.
  • Use arc length to explain why the radius and slope factor both appear.
  • Recognize why a signed radius can produce an invalid surface area.

Why the Position of the Axis Matters

When a curve is revolved around a horizontal line, the distance from the curve to that line determines the radius of each circular band of the surface. If the axis is the \(x\)-axis, the radius for a graph \(y=f(x)\) is \(|f(x)|\). Moving the axis to \(y=c\) changes that distance to \(|f(x)-c|\).

This adjustment is simple to state, but important to make correctly. The expression \(f(x)-c\) is a signed vertical difference; a radius is a distance and cannot be negative. Whether the graph lies above or below the axis determines how that distance is written without absolute-value notation.

Key idea: For a graph \(y=f(x)\) revolved around the horizontal line \(y=c\), the perpendicular radius is \(|f(x)-c|\). Under the conditions below, the surface-area element is \(2\pi |f(x)-c|\,ds\), where \(ds\) is an element of arc length.

The Surface-Area Formula

Suppose \(f\) is continuously differentiable on \([a,b]\), and the graph stays on one side of the axis \(y=c\) throughout the interval. The graph may touch the axis, but it does not cross from one side to the other. Its arc-length element, when \(x\) is the parameter, is

$$ ds=\sqrt{1+\bigl(f'(x)\bigr)^2}\,dx. $$

A thin band of the surface has circumference approximately \(2\pi r\), where \(r\) is the distance to the axis, and width approximately \(ds\). Its area is therefore approximately \(2\pi r\,ds\). Taking the limiting sum of these band areas gives the surface-area integral:

$$ S=2\pi\int_a^b |f(x)-c|\sqrt{1+\bigl(f'(x)\bigr)^2}\,dx. $$

The factor \(\sqrt{1+(f'(x))^2}\) accounts for the slant length of the curve, rather than its horizontal projection \(dx\). The factor \(|f(x)-c|\) is the radius of the circular band. Both factors are needed: using only the radius and \(dx\) would treat a sloping curve as though it were horizontal.

Conditions for this setup: The curve is given as \(y=f(x)\) on \([a,b]\), has a well-defined arc length there, and stays on one side of \(y=c\). For a continuously differentiable graph, the displayed formula applies directly. If the graph crosses the axis or has points where the graph description is unsuitable, the curve must be analyzed in suitable pieces; the setup here does not resolve possible overlap in the resulting surface.

Choosing the Radius from the Curve’s Position

If the graph is above the axis, then \(f(x)\ge c\), so the distance is \(f(x)-c\). If it is below, then \(f(x)\le c\), so the distance is \(c-f(x)\). In either case, the radius is nonnegative.

Position of graph on the interval Radius \(r(x)\) Surface-area setup
\(f(x)\ge c\) \(f(x)-c\) \(2\pi\int_a^b (f(x)-c)\sqrt{1+(f'(x))^2}\,dx\)
\(f(x)\le c\) \(c-f(x)\) \(2\pi\int_a^b (c-f(x))\sqrt{1+(f'(x))^2}\,dx\)

These are not different surface-area principles. They are two ways to write the same distance, \(|f(x)-c|\), after establishing the graph’s position relative to the axis.

Worked Example: A Line Above a Shifted Axis

Worked Example: A Line Above a Shifted Axis

Find the surface area formed by revolving \(y=x+1\), for \(0\le x\le2\), around \(y=-1\).

First compare the graph with the axis:

$$ f(x)-c=(x+1)-(-1)=x+2. $$

On \(0\le x\le2\), \(x+2>0\), so the graph lies above the axis and the radius is \(r(x)=x+2\). Also, \(f'(x)=1\), and therefore

$$ \sqrt{1+\bigl(f'(x)\bigr)^2} =\sqrt{1+1^2} =\sqrt{2}. $$

Substitute the radius and arc-length factor into the formula:

$$ \begin{aligned} S &=2\pi\int_0^2 (x+2)\sqrt{2}\,dx\\ &=2\pi\sqrt{2}\left[\frac{x^2}{2}+2x\right]_0^2\\ &=2\pi\sqrt{2}\left(\left(\frac{4}{2}+4\right)-0\right)\\ &=2\pi\sqrt{2}(6)\\ &=12\pi\sqrt{2}. \end{aligned} $$

Check the setup: at \(x=0\), the point on the curve is one unit above \(y=0\), while the axis \(y=-1\) is one unit farther down, so the radius is \(2\), matching \(0+2\). At \(x=2\), the curve is at \(y=3\), so its distance from \(y=-1\) is \(4\), matching \(2+2\). The radius expression agrees at both endpoints and throughout the interval.

Worked Example: A Graph Below the Axis

Worked Example: A Graph Below the Axis

Find the surface area formed by revolving \(y=x-4\), for \(0\le x\le2\), around \(y=1\).

Here \(f(x)-c=(x-4)-1=x-5\), which is negative throughout the interval. Thus the curve is below the axis, and its radius is

$$ r(x)=c-f(x)=1-(x-4)=5-x. $$

The derivative is \(f'(x)=1\), so \(\sqrt{1+(f'(x))^2}=\sqrt{2}\). The surface area is

$$ \begin{aligned} S &=2\pi\int_0^2 (5-x)\sqrt{2}\,dx\\ &=2\pi\sqrt{2}\left[5x-\frac{x^2}{2}\right]_0^2\\ &=2\pi\sqrt{2}\left((10-2)-0\right)\\ &=16\pi\sqrt{2}. \end{aligned} $$

Check the radius at the endpoints. At \(x=0\), the curve is at \(y=-4\), which is \(5\) units below \(y=1\); the formula gives \(5-0=5\). At \(x=2\), the curve is at \(y=-2\), which is \(3\) units below the axis; the formula gives \(5-2=3\). The radius decreases as the graph approaches the axis, as expected.

Worked Example: A Curved Profile with a Simple Arc-Length Factor

Worked Example: A Curved Profile with a Simple Arc-Length Factor

Find the surface area formed by revolving \(y=\sqrt{1-x^2}\), for \(0\le x\le \tfrac12\), around \(y=0\).

The graph lies above the axis, so the radius is \(r(x)=\sqrt{1-x^2}\). Differentiate:

$$ f'(x)=-\frac{x}{\sqrt{1-x^2}}. $$

The arc-length factor simplifies as follows:

$$ \begin{aligned} \sqrt{1+\bigl(f'(x)\bigr)^2} &=\sqrt{1+\frac{x^2}{1-x^2}}\\ &=\sqrt{\frac{1-x^2+x^2}{1-x^2}}\\ &=\frac{1}{\sqrt{1-x^2}}. \end{aligned} $$

Multiplying by the radius cancels the square-root factors. Therefore,

$$ \begin{aligned} S &=2\pi\int_0^{1/2}\sqrt{1-x^2}\, \frac{1}{\sqrt{1-x^2}}\,dx\\ &=2\pi\int_0^{1/2}1\,dx\\ &=2\pi\left[x\right]_0^{1/2}\\ &=2\pi\left(\frac12-0\right)\\ &=\pi. \end{aligned} $$

Check the simplification: for every \(x\) in this interval, \(1-x^2>0\), so the square roots are defined and their product is \(1\). The radius and arc-length factor therefore give exactly the constant integrand used above.

Why the Absolute Value Cannot Be Dropped

A common error is to use \(f(x)-c\) as the radius even when the graph lies below the axis. A signed vertical difference can be negative, but revolving a curve produces a nonnegative area. For the graph \(y=x-4\) around \(y=1\), for example, using the signed difference gives \(f(x)-c=x-5\). At \(x=0\), that expression is \(-5\), although the actual radius is \(5\).

$$ \text{Incorrect: }\quad 2\pi\int_0^2 (x-5)\sqrt{2}\,dx =2\pi\sqrt{2}(2-10) =-16\pi\sqrt{2}. $$

A negative result cannot be a surface area. Correcting the radius to the distance \(c-f(x)=5-x\) gives

$$ \text{Correct: }\quad 2\pi\int_0^2 (5-x)\sqrt{2}\,dx =16\pi\sqrt{2}. $$

The error is not in the arc-length factor. It is in treating a signed difference as a distance. The absolute value in \(|f(x)-c|\), or the appropriate above-axis or below-axis expression, prevents this mistake.

A Reliable Setup Procedure

Before evaluating a surface-area integral, determine the axis, the radius, and the arc-length factor separately. This separates the geometric decision from the calculus and makes it easier to catch sign errors.

1
Record the axis and interval.
For this topic, the axis is \(y=c\), and the graph is \(y=f(x)\) on \([a,b]\).
2
Determine which side contains the graph.
Compare \(f(x)\) with \(c\) across the interval, not just at one point.
3
Write the perpendicular radius.
Use \(f(x)-c\) above the axis, \(c-f(x)\) below it, or \(|f(x)-c|\) to express the distance directly.
4
Include the arc-length factor.
Multiply the radius by \(\sqrt{1+(f'(x))^2}\), then by \(2\pi\).
5
Check the radius before evaluating.
Test a point or endpoint to confirm that the radius is a nonnegative distance from the curve to \(y=c\).

Check Your Understanding

Use the graph’s position relative to the horizontal axis of revolution to answer each question.

  1. For \(y=f(x)\) revolved around \(y=3\), what is the radius if \(f(x)\ge3\) throughout the interval?
  2. For the same axis, what is the radius if \(f(x)\le3\) throughout the interval?
  3. Write the surface-area integral for \(y=2x+1\), \(0\le x\le1\), revolved around \(y=0\). Do not evaluate it.
  4. What geometric quantity does \(\sqrt{1+(f'(x))^2}\,dx\) represent?
  5. Why is \(f(x)-c\) alone not always a valid radius?

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