Revolving A Graph About The X-Axis

From a Curve to a Surface

A curve revolved about an axis sweeps out a surface. To find its area, each small piece of the curve contributes a narrow band around the axis. The band’s circumference depends on how far the curve is from the axis, while its width depends on the length of the curve itself.

For a graph \(y=f(x)\) revolved about the \(x\)-axis, the distance from the graph to the axis is \(|f(x)|\). The small arc-length element is \(ds=\sqrt{1+[f'(x)]^2}\,dx\). Combining these quantities gives the central formula for this topic.

Surface-area formula: If \(f\) is continuously differentiable on \([a,b]\), the surface area generated by revolving its graph about the \(x\)-axis is

S=2\pi\int_a^b |f(x)|\sqrt{1+[f'(x)]^2}\,dx.

The factor \(|f(x)|\) is the radius, and \(\sqrt{1+[f'(x)]^2}\,dx\) is the arc-length element.

Why the Formula Has This Form

The circumference of a circle with radius \(r\) is \(2\pi r\). A short curve segment of length \(ds\), revolved around the \(x\)-axis, therefore sweeps out a narrow band whose area is approximately its circumference multiplied by its slant width:

dS\approx 2\pi r\,ds.

Here the radius is the distance to the axis, \(r=|f(x)|\), not necessarily \(f(x)\). The arc length of a graph segment comes from the right triangle whose horizontal side is \(dx\) and whose vertical side is \(dy=f'(x)\,dx\):

ds=\sqrt{(dx)^2+(dy)^2} =\sqrt{1+[f'(x)]^2}\,dx.

Substituting these expressions into the band-area relation gives the surface-area element. Integrating that element along the graph gives the formula.

Approximate the curve by short segments.
On a small interval, use a straight segment of length \(\Delta s\) to approximate the graph.

Revolve a segment around the axis.
When the graph stays on one side of the axis over the segment, the rotated piece is a frustum band. Its lateral area is \(\pi(r_1+r_2)\Delta s\), which is \(2\pi\) times the average endpoint radius times \(\Delta s\).

Refine the partition.
As the segments become shorter, the average radius approaches the radius at each point and the sum of the band areas approaches the integral of \(2\pi|f(x)|\,ds\).

Use the graph’s arc-length element.
Replacing \(ds\) by \(\sqrt{1+[f'(x)]^2}\,dx\) yields the stated formula.

This derivation explains why the factor involving the derivative is essential: the bands are measured along the slanted curve, not along the horizontal \(x\)-axis. If \(f\) changes sign, the radius remains nonnegative because it is a distance. One can split the interval at zeros when convenient; the same formula uses \(|f(x)|\) throughout.

A Practical Setup

For a particular problem, identify the function and interval, find the derivative, determine the radius, and substitute all three into the integral. Before evaluating, check that the interval follows the specified curve segment and that the derivative belongs to the original function. If the graph lies below the axis, its radius is still positive.

Worked Example: A Quadratic Profile

Find the surface area when \(y=x^2\), \(0\le x\le 1\), is revolved about the \(x\)-axis.

The graph is nonnegative on this interval, so its radius is \(x^2\). Its derivative is \(f'(x)=2x\), giving

S=2\pi\int_0^1x^2\sqrt{1+4x^2}\,dx.

To evaluate, let \(u=2x\). Then \(x=u/2\), \(dx=du/2\), and the limits \(x=0,1\) become \(u=0,2\):

\int_0^1x^2\sqrt{1+4x^2}\,dx =\frac18\int_0^2u^2\sqrt{1+u^2}\,du.

An antiderivative of \(u^2\sqrt{1+u^2}\) is

G(u)=\frac18\left(u\sqrt{1+u^2}(2u^2+1)-\operatorname{arsinh}(u)\right).

Differentiating verifies it: the derivative of the expression inside parentheses is \(8u^2\sqrt{1+u^2}\), so \(G'(u)=u^2\sqrt{1+u^2}\). Thus

\int_0^1x^2\sqrt{1+4x^2}\,dx =\frac18\bigl(G(2)-G(0)\bigr) =\frac{18\sqrt5-\operatorname{arsinh}(2)}{64}.

Since \(\operatorname{arsinh}(2)=\ln(2+\sqrt5)\), the surface area is

S=\frac{\pi}{32}\left(18\sqrt5-\ln(2+\sqrt5)\right).

The antiderivative check and endpoint substitution verify the integral evaluation; multiplying that integral by \(2\pi\) gives the stated surface area.

When the Graph Lies Below the Axis

A negative function value does not mean a negative radius. The radius is the distance from the point on the graph to the \(x\)-axis, so it is \(|f(x)|\). For a graph that stays below the axis on the whole interval, this means the radius is \(-f(x)\).

Worked Example: A Linear Profile Below the Axis

Find the area when \(y=x-2\), \(0\le x\le 1\), is revolved about the \(x\)-axis.

On this interval, \(x-2<0\), so the radius is \(|x-2|=2-x\). The derivative is \(f'(x)=1\). Therefore

S=2\pi\int_0^1(2-x)\sqrt{1+1^2}\,dx =2\pi\sqrt2\int_0^1(2-x)\,dx.

An antiderivative of \(2-x\) is \(2x-x^2/2\). Evaluating at the endpoints gives

S=2\pi\sqrt2\left[2x-\frac{x^2}{2}\right]_0^1 =2\pi\sqrt2\left(\frac32-0\right) =3\pi\sqrt2.

Differentiating \(2x-x^2/2\) gives \(2-x\), and substituting the endpoints gives the integral value \(3/2\). The area is positive, as required.

A Trigonometric Profile

For trigonometric functions, the derivative and the resulting square root should be written out carefully. Familiar identities can sometimes simplify the calculation, but the arc-length factor is still \(\sqrt{1+[f'(x)]^2}\), not just a trigonometric expression chosen from memory.

Worked Example: Revolving a Cosine Profile

Find the surface area when \(y=\cos x\), \(0\le x\le \pi/2\), is revolved about the \(x\)-axis.

On this interval, \(\cos x\ge0\), so the radius is \(\cos x\). Since \(f'(x)=-\sin x\), the setup is

S=2\pi\int_0^{\pi/2}\cos x\sqrt{1+\sin^2x}\,dx.

Use \(u=\sin x\), so \(du=\cos x\,dx\), with endpoints \(u=0\) and \(u=1\). Then

S=2\pi\int_0^1\sqrt{1+u^2}\,du.

An antiderivative is

H(u)=\frac12\left(u\sqrt{1+u^2}+\operatorname{arsinh}(u)\right).

Indeed, differentiating gives

H'(u)=\frac12\left(\sqrt{1+u^2}+\frac{u^2}{\sqrt{1+u^2}}+\frac1{\sqrt{1+u^2}}\right) =\sqrt{1+u^2}.

Therefore

S=2\pi\bigl(H(1)-H(0)\bigr) =\pi\left(\sqrt2+\operatorname{arsinh}(1)\right) =\pi\left(\sqrt2+\ln(1+\sqrt2)\right).

The derivative check confirms the antiderivative, and the transformed endpoints give the displayed value.

Common Setup Error: Using \(dx\) Instead of \(ds\)

A frequent mistake is to multiply the circumference by horizontal width \(dx\) rather than by the curve’s arc-length element \(ds\). This ignores the slant of the graph and generally underestimates the area.

Incorrect calculation: For \(y=x^2\) on \([0,1]\), using \(2\pi\int_0^1x^2\,dx=2\pi/3\) treats the curve as if its length element were \(dx\). The correct setup includes \(f'(x)=2x\), so it is \(2\pi\int_0^1x^2\sqrt{1+4x^2}\,dx\). The missing square-root factor is the correction.

Another error is to use \(f(x)\) as the radius even when it is negative. The radius is always a distance, so the formula uses \(|f(x)|\). In the linear example, using \(x-2\) instead of \(2-x\) would produce a negative integral, which cannot represent an area.

Choosing and Checking the Integral

The formula gives a reliable setup when the profile is a continuously differentiable graph on a finite interval. Its parts have distinct geometric roles: \(2\pi|f(x)|\) is the circumference at a point, and the square-root factor converts horizontal change into distance along the curve. A useful final check is to ask whether both factors match those roles.

The interval should describe exactly the curve segment being revolved.
The radius must be nonnegative, so use \(|f(x)|\).
Differentiate the given function before forming the arc-length factor.
The integrand should combine the radius and the arc-length factor.
The resulting area should not be negative.

Some surface-area integrals have elementary antiderivatives, while others do not. The setup remains meaningful either way: it precisely describes the area, even when the integral must be evaluated by another method.

Check Your Understanding

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

For \(y=3-x\) on \([0,1]\), what integral gives the surface area when the graph is revolved about the \(x\)-axis?
Why does revolving \(y=-f(x)\) use the radius \(|f(x)|\), rather than a negative radius?
For \(y=\sin x\) on \([0,\pi/2]\), write the surface-area integral without evaluating it.
What geometric length does the factor \(\sqrt{1+[f'(x)]^2}\,dx\) represent?
Which factor is missing from the incorrect setup \(2\pi\int_a^b |f(x)|\,dx\), and when would that omission change the answer?