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Surface Area of Revolution · Tutorial 434 of 998

Vertical Tangencies And Surface-Area Setups

Recognize endpoint vertical tangencies, choose a suitable graph orientation, and set up and evaluate matching surface-area integrals.

Intermediate 9 min read

What You'll Learn

  • Identify when a graph has a vertical tangent at an endpoint.
  • Choose between describing a profile as y=f(x) and as x=g(y).
  • Set correct bounds when changing the independent variable.
  • Handle an endpoint-singular surface-area integrand using an improper integral.
  • Check that two descriptions of the same revolved curve give the same area.

When a Vertical Tangent Affects the Setup

A curve can be smooth even when it is not a differentiable graph \(y=f(x)\) at every point of an interval. In particular, a graph may have a vertical tangent at an endpoint, so that \(dy/dx\) becomes unbounded there. The surface area can still be finite, but the formula written in terms of \(x\) may require an improper integral or may be less convenient than a different description of the same curve.

The practical choice is to describe the profile using the variable that gives a manageable derivative and bounds. For example, the curve \(y=\sqrt{x}\) for \(0\le x\le4\) is equivalently described by \(x=y^2\) for \(0\le y\le2\). The second description has a finite derivative at the endpoint where the first has a vertical tangent.

Key idea: A vertical tangent is a feature of the curve’s description as a graph, not necessarily a singularity of the curve itself. Reversing the dependent and independent variables can provide a regular surface-area setup.

Surface-Area Formulas for the Two Graph Orientations

For a nonnegative profile \(y=f(x)\) revolved about the \(x\)-axis, the radius of a surface band is \(y\), and the arc-length element is \(ds=\sqrt{1+(dy/dx)^2}\,dx\). Thus, when \(f\) is differentiable on the interval, the surface area is

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

If the same profile is written as \(x=g(y)\) and revolved about the \(x\)-axis, then \(ds=\sqrt{1+(dx/dy)^2}\,dy\). Its radius is still \(y\), so the formula becomes

$$ S=2\pi\int_c^d y\sqrt{1+\bigl(g'(y)\bigr)^2}\,dy. $$

The bounds \(c\) and \(d\) are the corresponding \(y\)-values on the curve. The same principle applies to revolution about the \(y\)-axis: the radius is the distance \(x\) from that axis. For a nonnegative profile, the formulas are \(2\pi\int x\sqrt{1+(dy/dx)^2}\,dx\) when using \(y=f(x)\), and \(2\pi\int g(y)\sqrt{1+(dg/dy)^2}\,dy\) when using \(x=g(y)\).

These formulas assume the profile can be described by the stated graph on the interval and that the arc-length integral exists. If a derivative is unbounded at an endpoint, interpret the integral as an improper integral: integrate away from the endpoint, then take the limit as the omitted portion shrinks to zero. A finite resulting limit gives a finite area.

Why Reversing the Variables Works

The two descriptions represent the same curve, so they must give the same surface area when the bounds correspond to the same endpoints. The change-of-variable relation explains why their integrands agree. Suppose \(x=g(y)\) is differentiable and has \(g'(y)>0\) on the interval, so it can be inverted to \(y=f(x)\). Then \(dy/dx=1/g'(y)\) and \(dx=g'(y)\,dy\).

Derivation: For revolution about the \(x\)-axis, substitute \(x=g(y)\) into the formula in \(x\). Using \(dy/dx=1/g'(y)\), and \(g'(y)>0\), gives
$$ \begin{aligned} 2\pi y\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx &=2\pi y\sqrt{1+\frac{1}{(g'(y))^2}}\,g'(y)\,dy\\ &=2\pi y\sqrt{(g'(y))^2+1}\,dy. \end{aligned} $$

This is exactly the surface-area element obtained directly from \(x=g(y)\). The calculation requires \(g'(y)>0\) for the displayed simplification; if the derivative changes sign, arc length uses \(\lvert g'(y)\rvert\) in the change of variable. At a vertical tangent, the inverse description may have a finite derivative even when \(dy/dx\) is unbounded, making the second form easier to use.

Worked Example: Revolving \(y=\sqrt{x}\) About the \(x\)-Axis

Worked Example: Revolving \(y=\sqrt{x}\) About the \(x\)-Axis

Find the surface area obtained by revolving \(y=\sqrt{x}\), \(0\le x\le4\), about the \(x\)-axis. At \(x=0\), the derivative \(dy/dx=1/(2\sqrt{x})\) is unbounded, so also calculate using \(x=y^2\).

First use \(y\) as a function of \(x\). For \(x>0\),

$$ \frac{dy}{dx}=\frac{1}{2\sqrt{x}}, \qquad y\sqrt{1+\left(\frac{dy}{dx}\right)^2} =\sqrt{x}\sqrt{1+\frac{1}{4x}} =\frac{1}{2}\sqrt{4x+1}. $$

The simplified integrand has a finite limit, \(1/2\), as \(x\to0^+\). The area is therefore the improper integral, whose value is obtained by integrating this continuous extension:

$$ \begin{aligned} S &=2\pi\lim_{\varepsilon\to0^+} \int_{\varepsilon}^{4} \frac{1}{2}\sqrt{4x+1}\,dx\\ &=\pi\lim_{\varepsilon\to0^+} \left[\frac{(4x+1)^{3/2}}{6}\right]_{\varepsilon}^{4}\\ &=\frac{\pi}{6}\left(17^{3/2}-1\right) =\frac{\pi}{6}\left(17\sqrt{17}-1\right). \end{aligned} $$

Now write the same curve as \(x=y^2\). Since \(x=0\) corresponds to \(y=0\), and \(x=4\) corresponds to \(y=2\), the bounds are \(0\le y\le2\). Here \(dx/dy=2y\), and the radius about the \(x\)-axis is \(y\):

$$ \begin{aligned} S &=2\pi\int_0^2 y\sqrt{1+4y^2}\,dy\\ &=2\pi\left[\frac{(1+4y^2)^{3/2}}{12}\right]_0^2\\ &=\frac{\pi}{6}\left(17\sqrt{17}-1\right). \end{aligned} $$

As a check, differentiating the antiderivative \((1+4y^2)^{3/2}/12\) gives \(y\sqrt{1+4y^2}\), the integrand without its factor \(2\pi\). Both graph orientations produce the same area. The \(y\)-description avoids the improper endpoint in this example.

Worked Example: The Same Curve Revolved About the \(y\)-Axis

Worked Example: The Same Curve Revolved About the \(y\)-Axis

Revolve \(y=\sqrt{x}\), \(0\le x\le4\), about the \(y\)-axis. The radius is now \(x\), not \(y\). Use the description \(x=y^2\), \(0\le y\le2\), so that \(dx/dy=2y\):

$$ \begin{aligned} S &=2\pi\int_0^2 y^2\sqrt{1+4y^2}\,dy. \end{aligned} $$

To evaluate, let \(t=2y\), so \(dy=dt/2\) and \(y^2=t^2/4\). The integral becomes

$$ \begin{aligned} \int_0^2 y^2\sqrt{1+4y^2}\,dy &=\frac18\int_0^4 t^2\sqrt{1+t^2}\,dt\\ &=\frac{1}{64}\left[t\sqrt{1+t^2}(2t^2+1)-\operatorname{arsinh}(t)\right]_0^4\\ &=\frac{132\sqrt{17}-\operatorname{arsinh}(4)}{64}. \end{aligned} $$

Therefore,

$$ S=\frac{\pi}{32}\left(132\sqrt{17}-\operatorname{arsinh}(4)\right). $$

To check the antiderivative, differentiate \(\bigl[t\sqrt{1+t^2}(2t^2+1)-\operatorname{arsinh}(t)\bigr]/8\); the result is \(t^2\sqrt{1+t^2}\), as required. The factor \(1/8\) from the substitution then gives the displayed value of the original integral. This example also shows why the axis must be considered when choosing the radius: the radius is \(y^2\) here, whereas it was \(y\) in the first example.

Worked Example: Matching Bounds After a Shift

Worked Example: Matching Bounds After a Shift

Consider \(y=\sqrt{x-1}\) on \(1\le x\le5\), revolved about the \(x\)-axis. The tangent is vertical at the left endpoint \(x=1\). Rewriting the profile gives \(x=y^2+1\). The endpoint values \(x=1\) and \(x=5\) correspond to \(y=0\) and \(y=2\), respectively.

Using \(x\) first, for \(x>1\) we have \(dy/dx=1/(2\sqrt{x-1})\). The radius-times-arc-length factor simplifies to

$$ y\sqrt{1+\left(\frac{dy}{dx}\right)^2} =\sqrt{x-1}\sqrt{1+\frac{1}{4(x-1)}} =\frac{1}{2}\sqrt{4x-3}. $$

Thus, treating the endpoint as improper,

$$ \begin{aligned} S &=2\pi\lim_{\varepsilon\to0^+} \int_{1+\varepsilon}^{5}\frac12\sqrt{4x-3}\,dx\\ &=\frac{\pi}{6}\left(17^{3/2}-1\right). \end{aligned} $$

For the reversed description, \(dx/dy=2y\), so

$$ \begin{aligned} S &=2\pi\int_0^2 y\sqrt{1+4y^2}\,dy\\ &=\frac{\pi}{6}\left(17\sqrt{17}-1\right). \end{aligned} $$

The answers agree. Differentiating \((4x-3)^{3/2}/6\) gives \(\sqrt{4x-3}\), and differentiating \((1+4y^2)^{3/2}/12\) gives \(y\sqrt{1+4y^2}\); these verify the antiderivatives used. The bounds also match the same curve segment, rather than merely having the same numerical length.

A Common Setup Error

A frequent mistake is to change from \(y=f(x)\) to \(x=g(y)\) but keep the original \(x\)-bounds. For the curve \(y=\sqrt{x}\) from \(x=0\) to \(x=4\), the corresponding \(y\)-bounds are \(0\) and \(2\), not \(0\) and \(4\).

Pitfall and correction: The incorrect setup \(2\pi\int_0^4 y\sqrt{1+4y^2}\,dy\) uses the \(x\)-endpoint \(4\) as though it were a \(y\)-endpoint. Its value is \(\frac{\pi}{6}(65\sqrt{65}-1)\), not the area of the stated curve segment. The corrected bounds come from \(y=\sqrt{x}\): \(y(0)=0\) and \(y(4)=2\). Therefore use \(2\pi\int_0^2 y\sqrt{1+4y^2}\,dy\).

Another error is to conclude that the surface area is infinite just because \(dy/dx\) is unbounded at one endpoint. The full integrand, including the radius, determines whether the improper integral converges. In the first example, the product simplifies to \(\frac12\sqrt{4x+1}\), which has a finite endpoint limit.

Choosing a Useful Description

Before integrating, check whether the curve is naturally a graph in the chosen variable across the whole interval. If \(dy/dx\) becomes unbounded at an endpoint, test whether writing \(x\) as a function of \(y\) gives finite derivatives and straightforward bounds. Then identify the radius from the axis of revolution and compute the corresponding arc-length element. The choice of variable can simplify the calculation, but it does not change the curve segment or the surface being generated.

1
Identify the tangent issue.
Check whether the derivative for the chosen graph is unbounded at an endpoint.
2
Try the other graph orientation.
Rewrite the profile as \(x=g(y)\) if that gives a simpler derivative.
3
Match the bounds.
Convert both endpoints to the new independent variable.
4
Use the correct radius.
Measure distance from the specified axis of revolution.
5
Check convergence and the result.
If necessary, take an endpoint limit, and verify the antiderivative or compare equivalent setups.

Check Your Understanding

Use the formulas and checks in this tutorial to answer the following.

  1. For \(y=\sqrt{x}\) on \(0\le x\le9\), what interval for \(y\) corresponds to the same curve segment when written as \(x=y^2\)?
  2. Why is the surface-area integral in \(x\) interpreted as improper for \(y=\sqrt{x}\) at \(x=0\)?
  3. When the profile \(x=y^2\) is revolved about the \(y\)-axis, what is the radius of a surface band?
  4. State the condition on \(g'(y)\) used in the displayed change-of-variable derivation.
  5. Explain why an unbounded graph derivative at one endpoint does not by itself establish that the surface area is infinite.

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