Surface-Area Problems With Coordinate Axes

Surface Area About a Coordinate Axis

A curve revolved about an axis sweeps out a surface. To calculate its area, we combine two ingredients: the circumference traced by a point on the curve and the small length of curve that is being rotated. The key setup decision is the radius: it is the distance from the curve to the axis of rotation.

For a curve \(y=f(x)\), the arc-length differential is \(ds=\sqrt{1+(f'(x))^2}\,dx\). For a curve written as \(x=g(y)\), it is \(ds=\sqrt{1+(g'(y))^2}\,dy\). The choice between these forms depends on how the curve is described and which variable gives a convenient integral.

Definition: If a smooth curve segment is revolved once about an axis without overlapping itself, its surface area is the integral of the circumference of revolution times the arc-length differential. About the \(x\)-axis, the radius is the distance \(|y|\); about the \(y\)-axis, it is \(|x|\).

When the curve lies above the \(x\)-axis, the radius is \(y\), and when it lies to the right of the \(y\)-axis, the radius is \(x\). In these common cases, the formulas are:

\begin{aligned} \text{About the }x\text{-axis:}\qquad S&=2\pi\int_a^b y\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx,\\[4pt] \text{About the }y\text{-axis:}\qquad S&=2\pi\int_c^d x\sqrt{1+\left(\frac{dx}{dy}\right)^2}\,dy. \end{aligned}

Each formula requires the appropriate graph description and a well-defined arc length on the interval. If a curve is below or left of the axis, its radius is a distance, so use \(|y|\) or \(|x|\), not a signed coordinate. The examples here use curves that stay on one side of the axis and generate a single surface.

Why the Formula Uses Circumference and Arc Length

Consider a small piece of a curve \(y=f(x)\) revolved about the \(x\)-axis. Its distance from the axis is \(y\), so its circular path has circumference \(2\pi y\). A short curve piece has length approximately \(ds\). Rotating that piece produces a narrow band whose area is approximately circumference times slant length, or \(2\pi y\,ds\).

For a smooth graph, the arc-length differential is \(ds=\sqrt{1+(f'(x))^2}\,dx\). Adding the areas of the narrow bands along the interval and taking the limit of increasingly fine partitions gives the integral. The same reasoning applies about the \(y\)-axis, replacing the radius \(y\) with \(x\) and using the arc length written in terms of \(y\).

Identify the axis and radius.
Measure the perpendicular distance from the curve to the axis: \(y\) for a curve above the \(x\)-axis, or \(x\) for a curve right of the \(y\)-axis.

Choose the arc-length variable.
Use \(ds=\sqrt{1+(dy/dx)^2}\,dx\) for \(y=f(x)\), or \(ds=\sqrt{1+(dx/dy)^2}\,dy\) for \(x=g(y)\).

Integrate circumference times arc length.
Multiply the circumference \(2\pi r\) by \(ds\), then integrate over the interval that traces the specified curve segment.

A Polynomial Profile About the \(x\)-Axis

Worked Example: A Line Revolved About the \(x\)-Axis

Find the surface area generated by revolving \(y=x+1\), for \(0\le x\le 1\), about the \(x\)-axis.

The curve is above the \(x\)-axis, so the radius is \(r=y=x+1\). Its derivative is \(dy/dx=1\), hence \(ds=\sqrt{1+1^2}\,dx=\sqrt{2}\,dx\). Therefore,

\begin{aligned} S &=2\pi\int_0^1 (x+1)\sqrt{2}\,dx\\ &=2\pi\sqrt{2}\left[\frac{x^2}{2}+x\right]_0^1\\ &=2\pi\sqrt{2}\left(\frac{1}{2}+1-0\right)\\ &=3\pi\sqrt{2}. \end{aligned}

For a check, an antiderivative of the integrand \(2\pi\sqrt{2}(x+1)\) is \(\pi\sqrt{2}(x^2+2x)\), since its derivative is \(2\pi\sqrt{2}(x+1)\). Evaluating this antiderivative at \(1\) and \(0\) gives \(3\pi\sqrt{2}\), as above.

A Polynomial Profile About the \(y\)-Axis

For rotation about the \(y\)-axis, it can be convenient to describe the curve as \(x=g(y)\). Then the radius is \(x\), and the differential is measured along the curve in the \(y\)-direction. The next example uses the same type of straight profile but makes this choice explicit.

Worked Example: A Line Revolved About the \(y\)-Axis

Find the surface area generated by revolving \(x=y+1\), for \(0\le y\le 2\), about the \(y\)-axis.

At height \(y\), the radius is the horizontal distance \(x=y+1\). Since \(dx/dy=1\), the arc-length differential is \(ds=\sqrt{1+1^2}\,dy=\sqrt{2}\,dy\). Thus,

\begin{aligned} S &=2\pi\int_0^2 (y+1)\sqrt{2}\,dy\\ &=2\pi\sqrt{2}\left[\frac{y^2}{2}+y\right]_0^2\\ &=2\pi\sqrt{2}\left((2+2)-0\right)\\ &=8\pi\sqrt{2}. \end{aligned}

To check the evaluation, the derivative of \(2\pi\sqrt{2}\left(y^2/2+y\right)\) is \(2\pi\sqrt{2}(y+1)\), the integrand. Its values at \(2\) and \(0\) differ by \(8\pi\sqrt{2}\).

A Circular Profile About the \(x\)-Axis

A circle can make the arc-length factor look complicated, but the circle’s equation may simplify the product of radius and arc-length factor. The interval matters: the calculation below uses only a selected arc, not the entire circle.

Worked Example: A Circular Arc Revolved About the \(x\)-Axis

Find the surface area generated by revolving the upper circular arc \(y=\sqrt{25-x^2}\), for \(3\le x\le 4\), about the \(x\)-axis.

On this interval, \(y>0\), so the radius is \(y\). Differentiating the profile gives

\frac{dy}{dx} =-\frac{x}{\sqrt{25-x^2}} =-\frac{x}{y}.

Therefore, using \(x^2+y^2=25\),

\sqrt{1+\left(\frac{dy}{dx}\right)^2} =\sqrt{1+\frac{x^2}{y^2}} =\sqrt{\frac{x^2+y^2}{y^2}} =\frac{5}{y}.

The radius \(y\) cancels the \(1/y\) in the arc-length factor, so the surface area is

\begin{aligned} S &=2\pi\int_3^4 y\cdot\frac{5}{y}\,dx\\ &=10\pi\int_3^4 1\,dx\\ &=10\pi(4-3)\\ &=10\pi. \end{aligned}

As a check, the simplified area integrand is the constant \(10\pi\), whose antiderivative \(10\pi x\) has derivative \(10\pi\). Its endpoint difference is \(40\pi-30\pi=10\pi\).

An Exponential Profile About the \(x\)-Axis

For an exponential curve, the arc-length factor is generally not constant. A substitution can make the integral manageable, but the radius and the arc-length differential must still be identified before integrating.

Worked Example: An Exponential Curve Revolved About the \(x\)-Axis

Find the surface area generated by revolving \(y=e^x\), for \(0\le x\le\ln 2\), about the \(x\)-axis.

The radius is \(y=e^x\), and \(dy/dx=e^x\). Thus,

\begin{aligned} S &=2\pi\int_0^{\ln 2} e^x\sqrt{1+e^{2x}}\,dx. \end{aligned}

Set \(u=e^x\), so \(du=e^x\,dx\). The endpoints become \(u=1\) when \(x=0\), and \(u=2\) when \(x=\ln 2\). Therefore,

\begin{aligned} S &=2\pi\int_1^2\sqrt{1+u^2}\,du\\ &=2\pi\left[\frac{u\sqrt{1+u^2}+\ln\left(u+\sqrt{1+u^2}\right)}{2}\right]_1^2\\ &=\pi\left(2\sqrt{5}+\ln(2+\sqrt{5})-\sqrt{2}-\ln(1+\sqrt{2})\right). \end{aligned}

For a check, the derivative of \(\frac{1}{2}\left(u\sqrt{1+u^2}+\ln(u+\sqrt{1+u^2})\right)\) is \(\sqrt{1+u^2}\). The substitution \(u=e^x\), \(du=e^x\,dx\), therefore reproduces the original integrand, and the transformed endpoints are \(1\) and \(2\) as used above.

A Common Radius Mistake

The radius is not automatically the variable that appears in the integral. It is the distance from the curve to the axis. This distinction is especially easy to miss when the curve is written as \(y=f(x)\) but rotated about the \(x\)-axis.

Common mistake: For \(y=x+1\) revolved about the \(x\)-axis on \(0\le x\le1\), using \(x\) as the radius gives the incorrect setup \(2\pi\int_0^1 x\sqrt{2}\,dx=\pi\sqrt{2}\). The correct radius is the curve’s height \(y=x+1\), so the correct result is \(2\pi\int_0^1(x+1)\sqrt{2}\,dx=3\pi\sqrt{2}\).

The mistake changes the surface area because \(x\) measures horizontal distance from the \(y\)-axis, not vertical distance from the \(x\)-axis. Before integrating, state the axis, the perpendicular distance to it, and the appropriate arc-length differential. This simple check catches many setup errors.

Choosing and Checking a Setup

For a coordinate-axis problem, the most reliable procedure is to translate the geometry into the integral before doing any integration. First identify the curve segment and its interval. Next identify the radius from the chosen axis. Then compute the derivative needed for arc length and write \(ds\) in the same variable used for the bounds. Finally, integrate and check the antiderivative or substitution.

Axis of rotation	Curve description	Radius	Arc-length differential
\(x\)-axis	\(y=f(x)\)	\(\|y\|\)	\(\sqrt{1+(dy/dx)^2}\,dx\)
\(y\)-axis	\(x=g(y)\)	\(\|x\|\)	\(\sqrt{1+(dx/dy)^2}\,dy\)

These formulas calculate the area of the surface traced by the specified curve segment. They do not automatically include end caps, which are separate surfaces, and they assume the revolution traces the surface without unintended overlap. When a problem specifies only a curve segment, use that segment’s bounds rather than replacing them with the full domain of a related curve.

Check Your Understanding

For each question, identify the radius and arc-length differential before evaluating or simplifying.

Write the surface-area integral for \(y=2x+1\), \(0\le x\le 1\), revolved about the \(x\)-axis.
For \(x=3-y\), \(0\le y\le 1\), revolved about the \(y\)-axis, what is the radius as a function of \(y\)?
For \(y=e^x\) revolved about the \(x\)-axis, what are \(y\) and \(dy/dx\) in the surface-area integrand?
On the upper arc \(y=\sqrt{25-x^2}\), why does the factor \(y\sqrt{1+(dy/dx)^2}\) simplify to \(5\)?
Explain why \(x\) is not the radius when a curve \(y=f(x)\) is revolved about the \(x\)-axis.