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Surface Area Of A Selected Curve Segment

Use geometric boundaries to select just the curve segment that generates the desired portion of a surface, then integrate over its corresponding bounds.

Intermediate 9 min read

What You'll Learn

  • Translate a requested band or cap into endpoints on the generating curve.
  • Find bounds by solving intersections with planes or other geometric boundaries.
  • Apply the appropriate surface-area integral to a selected graph segment.
  • Handle an endpoint where the graph derivative becomes unbounded.
  • Distinguish the curved surface generated by a segment from any flat end faces.
  • Check whether the chosen bounds describe only the requested portion.

Choosing Only the Curve Segment You Need

A surface of revolution is generated by rotating a curve around an axis. Often the task is not to find the area of the entire surface, but only one specified part: a band between two levels, a cap above a plane, or a truncated section between geometric boundaries. The key step is to identify which part of the curve generates that surface.

The integration bounds are not arbitrary. They are the coordinates of the endpoints of the selected curve segment. Those endpoints may be specified directly, or they may need to be found by intersecting the profile with a plane, another curve, or a geometric landmark such as a pole or equator.

Definition: A selected curve segment is the portion of a generating profile between two chosen endpoints. Rotating only that segment generates only the corresponding portion of the surface. Unless stated otherwise, surface area here means the curved surface generated by the segment; flat end faces are not included.

The Surface-Area Integral on a Selected Interval

For a graph \(y=f(x)\) on \(a\le x\le b\), revolved about the \(x\)-axis, the surface area is

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

For a graph \(x=g(y)\) on \(c\le y\le d\), revolved about the \(y\)-axis, it is

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

In each formula, the absolute value gives the distance from the profile to the axis of rotation. The interval must describe the selected curve segment. The usual formula assumes a smooth enough profile on the interval and that the generated surface is not counted repeatedly through overlap. If a derivative is unbounded at an endpoint, the integral is interpreted as an improper integral and must converge.

These formulas also explain why locating the endpoints matters. The integrand measures the surface generated by each small piece of arc, while the bounds decide which pieces of arc are included. Changing the bounds changes the portion of surface being measured.

Why the Bounds Select the Surface

For a graph \(y=f(x)\), a short horizontal change \(dx\) corresponds to an arc-length change approximately \(ds=\sqrt{1+(f'(x))^2}\,dx\). When this small arc is revolved about the \(x\)-axis, it sweeps out a narrow band whose radius is \(|f(x)|\). Its area is approximately circumference times slant width: \(2\pi |f(x)|\,ds\). Adding these areas along the selected segment and taking the limit of the sums gives the integral formula.

The same reasoning applies to \(x=g(y)\) revolved about the \(y\)-axis, with \(dy\) as the coordinate increment. Restricting the sum to the chosen segment restricts the resulting integral to the endpoints of that segment. This is why a correctly evaluated integral over the wrong interval still gives the wrong surface.

Find Endpoints Before Integrating

Translate the description of the requested surface into equations for its boundary. A horizontal plane \(y=h\), for example, meets a graph \(y=f(x)\) where \(f(x)=h\). A vertical plane \(x=k\) meets it where the profile has \(x=k\). If the profile is given implicitly, solve the curve equation together with the boundary equation.

There may be more than one intersection. Keep only the intersection points that bound the intended segment. Check the graph's domain and branch, and use the geometric description to decide which side of each boundary to include. A cap above a plane uses the portion above that plane; a band between two planes uses the portion between them.

1
Identify the axis and profile.
Choose the graph orientation that describes the generating curve and its distance from the axis clearly.
2
Translate the requested portion into boundaries.
Write the intersection or landmark conditions as equations.
3
Locate the endpoints and select the correct interval.
Check branches, domains, and which side of each boundary is wanted.
4
Integrate only over that segment.
Use the surface-area formula for the chosen orientation, then check that the result represents the intended curved portion.

Worked Examples

Worked Example: A Truncated Cone Band

Take the profile \(x=y\) and revolve it about the \(y\)-axis. Find the curved surface area between the horizontal planes \(y=1\) and \(y=3\).

The profile meets the planes at \(x=y=1\) and \(x=y=3\), so the selected segment has \(1\le y\le3\). As a graph of \(x\) in terms of \(y\), \(g(y)=y\) and \(g'(y)=1\). The radius is \(y\), which is positive throughout this interval.

$$ \begin{aligned} S &=2\pi\int_1^3 y\sqrt{1+(1)^2}\,dy\\ &=2\pi\sqrt{2}\int_1^3 y\,dy\\ &=2\pi\sqrt{2}\left[\frac{y^2}{2}\right]_1^3\\ &=\pi\sqrt{2}(9-1)\\ &=8\pi\sqrt{2}. \end{aligned} $$

Check the antiderivative: the derivative of \(\frac{y^2}{2}\) is \(y\), the integrand after factoring out \(\sqrt{2}\). The bounds correspond to the two specified planes, so this is the curved band only; the circular end faces are not included.

Worked Example: A Spherical Cap Above a Plane

A sphere of radius \(5\) is generated by revolving the right half of the circle \(x^2+y^2=25\) about the \(y\)-axis. Find the curved cap above the plane \(y=3\).

The right-hand profile is \(x=g(y)=\sqrt{25-y^2}\). Its top is the pole \(y=5\), and the plane \(y=3\) gives the other endpoint. Thus the selected interval is \(3\le y\le5\). For \(y<5\),

$$ g'(y)=-\frac{y}{\sqrt{25-y^2}}. $$

The derivative becomes unbounded as \(y\) approaches \(5\), so evaluate the integral as an improper integral. For \(3\le y<5\),

$$ \begin{aligned} g(y)\sqrt{1+\bigl(g'(y)\bigr)^2} &=\sqrt{25-y^2}\sqrt{1+\frac{y^2}{25-y^2}}\\ &=\sqrt{25-y^2}\sqrt{\frac{25}{25-y^2}}\\ &=5. \end{aligned} $$

Therefore the limiting integral is

$$ \begin{aligned} S &=\lim_{b\to5^-}2\pi\int_3^b 5\,dy\\ &=\lim_{b\to5^-}10\pi(b-3)\\ &=20\pi. \end{aligned} $$

The simplified integrand is constant on the open interval and has a finite integral up to \(y=5\), so the improper integral converges. As a check, the cap height is \(5-3=2\), and the result \(20\pi\) is \(2\pi\cdot5\cdot2\). The answer is the curved cap, not the disk in the plane \(y=3\).

Worked Example: A Spherical Band Between Two Vertical Planes

The top semicircle \(y=\sqrt{16-x^2}\) is revolved about the \(x\)-axis. Find the curved surface generated by the segment from \(x=0\) to the vertical plane \(x=3\).

The given planes directly specify the endpoints, so the selected interval is \(0\le x\le3\). On it, \(y\ge0\), and

$$ y'=-\frac{x}{\sqrt{16-x^2}}. $$

For this profile, the arc-length factor simplifies as follows:

$$ \begin{aligned} y\sqrt{1+(y')^2} &=\sqrt{16-x^2}\sqrt{1+\frac{x^2}{16-x^2}}\\ &=\sqrt{16-x^2}\sqrt{\frac{16}{16-x^2}}\\ &=4. \end{aligned} $$

Thus

$$ \begin{aligned} S &=2\pi\int_0^3 4\,dx\\ &=8\pi[x]_0^3\\ &=24\pi. \end{aligned} $$

Check the antiderivative: the derivative of \(4x\) is \(4\), the simplified integrand. The endpoints give a band from the equator at \(x=0\) to the circle where \(x=3\); they do not describe the whole sphere.

A Common Bounds Mistake

For the spherical cap in the second example, a tempting mistake is to use the full sphere's vertical range, \(-5\le y\le5\). That computes the entire sphere rather than the cap above \(y=3\): with the simplified integrand \(5\), it gives \(2\pi\int_{-5}^5 5\,dy=100\pi\).

The correction is to use the boundaries of the selected segment: the cutting plane \(y=3\) and the pole \(y=5\). Then \(2\pi\int_3^5 5\,dy=20\pi\). Always check the bounds against the geometric description before evaluating the integral.

Why Selected Segments Matter

Surface-area questions often describe a portion geometrically rather than giving integration bounds directly. Turning that description into endpoint equations is a central part of setting up the calculation. The integral then measures exactly the profile arc between those endpoints.

Keep the requested object clear. A band bounded by planes gives a curved band; it does not automatically include the planar disks that might close off a solid. Likewise, a cap cut by a plane includes the curved surface above the cut, not the disk at the cut, unless the problem explicitly asks for total surface area including that face.

Finally, bounds alone do not resolve every possible overlap when a profile crosses or meets the axis. The surface-area formula must be applied with its non-overlap conditions in mind; if different parts of the profile generate the same surface, counting each part separately may overcount. The selected-segment method identifies the intended arc, while the geometry determines whether its revolution traces the desired surface without duplication.

Check Your Understanding

For each question, identify the selected interval before deciding how to integrate.

  1. The line \(x=2y\) is revolved about the \(y\)-axis between the planes \(y=2\) and \(y=5\). What are the curve-segment endpoints?
  2. A graph \(y=f(x)\) is revolved about the \(x\)-axis. What equation locates its intersections with the horizontal plane \(y=h\)?
  3. For the circle \(x^2+y^2=R^2\), which \(y\)-bounds describe the top cap above \(y=h\), where \(0<h<R\), when using the right-hand profile about the \(y\)-axis?
  4. Does a surface-area integral over a selected segment include the flat disk at a cutting plane? Explain.
  5. For the top semicircle \(y=\sqrt{16-x^2}\) revolved about the \(x\)-axis, what bounds select the portion from the equator to the plane \(x=2\)?

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