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Surface Area of Revolution · Tutorial 444 of 998

Numerically Estimating Surface Areas

Turn a surface-area integral into a numerical estimate, choose a quadrature rule and interval size, and assess the reliability of the result.

Intermediate 13 min read

What You'll Learn

  • Express a surface-area calculation as a numerical integral.
  • Apply the composite trapezoidal rule and composite Simpson’s rule.
  • Understand why the composite Simpson weights have their stated form.
  • Compare estimates made with different subinterval widths.
  • Use error bounds and refinement checks to assess numerical reliability.

When Surface-Area Integrals Need Numerical Methods

For a curve \(y=f(x)\), where \(f(x)\) is nonnegative on \([a,b]\), the surface area formed by revolving the graph about the \(x\)-axis is \(S=2\pi\int_a^b f(x)\sqrt{1+[f'(x)]^2}\,dx\). The integrand combines the radius of revolution with the curve’s arc-length factor.

Sometimes this integral has a convenient antiderivative. Often, however, the product involving a square root does not lead to a useful elementary antiderivative. In that case, numerical quadrature approximates the definite integral using values of its integrand at selected points.

It is useful to separate the geometry from the numerical calculation. Write the surface area as \(S=2\pi I\), where \(I\) is the integral of the remaining integrand. Approximate \(I\) first, then multiply the estimate by \(2\pi\). If lengths are measured in metres, the result for \(S\) is in square metres.

Definition: A quadrature rule approximates a definite integral by a weighted sum of function values at selected points in the interval. A composite rule applies a basic rule repeatedly across smaller subintervals.

The Composite Trapezoidal Rule

Divide \([a,b]\) into \(n\) equal subintervals, with width \(h=(b-a)/n\) and nodes \(x_i=a+ih\), for \(i=0,1,\ldots,n\). For a continuous integrand \(g\), the composite trapezoidal rule is

$$ T_n = h\left[ \frac{g(x_0)+g(x_n)}{2} +\sum_{i=1}^{n-1}g(x_i) \right]. $$

On each subinterval, the rule replaces the graph by the straight line joining the endpoint values. The area under that line is the area of a trapezoid. Adding the trapezoid areas gives the composite formula: each interior node is an endpoint of two neighbouring trapezoids, so its value is counted twice before the common factor \(h/2\) is applied. The two outer endpoints are counted only once.

This rule can be used for any positive integer \(n\). It does not require an even number of subintervals.

The Composite Simpson’s Rule

Simpson’s rule approximates the integrand by a quadratic polynomial across two subintervals at a time. Consequently, the number \(n\) of equal subintervals must be even. With the same node definitions and width \(h\), the composite rule is

$$ S_n = \frac{h}{3}\left[ g(x_0)+g(x_n) +4\sum_{\substack{i=1\\ i\ \mathrm{odd}}}^{n-1}g(x_i) +2\sum_{\substack{i=2\\ i\ \mathrm{even}}}^{n-2}g(x_i) \right]. $$

The first and last nodes have weight \(1\), odd-indexed interior nodes have weight \(4\), and even-indexed interior nodes have weight \(2\). For \(n=2\), the rule reduces to \(h[g(x_0)+4g(x_1)+g(x_2)]/3\).

Why the Simpson weights are \(1,4,1\): On one pair of subintervals, set \(x=x_1+hu\), so the three nodes correspond to \(u=-1,0,1\) and \(dx=h\,du\). The quadratic interpolation basis functions at those nodes are \(u(u-1)/2\), \(1-u^2\), and \(u(u+1)/2\). Integrating them from \(-1\) to \(1\) gives \(1/3\), \(4/3\), and \(1/3\), respectively. Multiplying by \(h\) gives \(h/3\), \(4h/3\), and \(h/3\). Applying this result to each consecutive pair of subintervals gives the composite weights: a shared even-indexed node receives \(1+1=2\), while each odd-indexed node receives \(4\).

Worked Example: Simpson’s Rule for a Revolved Parabola

Worked Example: Simpson’s Rule for a Revolved Parabola

Revolve \(y=x^2\) about the \(x\)-axis on \([0,1]\). Estimate the surface area using Simpson’s rule with \(n=2\) subintervals.

First, \(f'(x)=2x\), so the integral apart from the factor \(2\pi\) has integrand

$$ g(x)=x^2\sqrt{1+4x^2}, \qquad S=2\pi\int_0^1 g(x)\,dx. $$

Here \(h=(1-0)/2=0.5\). The three nodes and integrand values are

$$ \begin{aligned} x_0&=0, & g(x_0)&=0,\\ x_1&=0.5, & g(x_1)&=(0.5)^2\sqrt{1+4(0.5)^2} =\frac{\sqrt{2}}{4}\approx0.353553,\\ x_2&=1, & g(x_2)&=\sqrt{5}\approx2.236068. \end{aligned} $$

Substitute these values into the one-pair Simpson formula:

$$ \begin{aligned} I&\approx\frac{0.5}{3}\left[0+4(0.353553)+2.236068\right]\\ &=\frac{1}{6}(1.414212+2.236068)\\ &\approx0.608380. \end{aligned} $$

Thus the surface-area estimate is

$$ S\approx2\pi(0.608380)\approx3.822566. $$

The surface area is approximately \(3.822566\) square units. As a consistency check, dividing this estimate by \(2\pi\) recovers the Simpson estimate \(I\approx0.608380\) for the integral. This check verifies the factor conversion; it does not establish that the quadrature estimate is exact.

Worked Example: How Interval Width Changes a Trapezoidal Estimate

Worked Example: How Interval Width Changes a Trapezoidal Estimate

Use the same profile \(y=x^2\) and interval \([0,1]\), but compare composite trapezoidal estimates with \(n=2\) and \(n=4\). The integrand is again \(g(x)=x^2\sqrt{1+4x^2}\).

For \(n=2\), the nodes are \(0,0.5,1\), and the values are \(0\), \(0.353553\), and \(2.236068\). The trapezoidal estimate is

$$ \begin{aligned} T_2 &=0.5\left[\frac{0+2.236068}{2}+0.353553\right]\\ &=0.5(1.118034+0.353553)\\ &\approx0.735794. \end{aligned} $$

For \(n=4\), \(h=0.25\), and the nodes are \(0,0.25,0.5,0.75,1\). The needed values are \(g(0)=0\), \(g(0.25)\approx0.069877\), \(g(0.5)\approx0.353553\), \(g(0.75)\approx1.014061\), and \(g(1)\approx2.236068\). Therefore

$$ \begin{aligned} T_4 &=0.25\left[ \frac{0+2.236068}{2} +0.069877+0.353553+1.014061 \right]\\ &=0.25(1.118034+1.437491)\\ &\approx0.638881. \end{aligned} $$

Multiplying by \(2\pi\), the corresponding surface-area estimates are \(2\pi T_2\approx4.623127\) and \(2\pi T_4\approx4.014211\) square units. Halving the interval width has changed the estimate substantially. That is evidence that the coarser estimate may not be reliable to several decimal places; it is not, by itself, a proof that the finer estimate is accurate.

As a basic arithmetic check, dividing the two surface-area estimates by \(2\pi\) returns \(T_2\approx0.735794\) and \(T_4\approx0.638881\), respectively. The quadrature sums and the surface-area conversion are consistent.

Worked Example: A Different Profile with Simpson’s Rule

Worked Example: A Different Profile with Simpson’s Rule

Revolve \(y=1+x^3\) about the \(x\)-axis on \([0,1]\). Estimate the surface area using Simpson’s rule with \(n=2\).

Since \(f'(x)=3x^2\), the integrand is

$$ g(x)=(1+x^3)\sqrt{1+9x^4}. $$

The node width is \(h=0.5\). At the three nodes,

$$ \begin{aligned} g(0)&=1,\\ g(0.5)&=\left(1+\frac18\right)\sqrt{1+\frac{9}{16}} =\frac98\cdot\frac54 =\frac{45}{32}=1.40625,\\ g(1)&=2\sqrt{10}\approx6.324555. \end{aligned} $$

Applying Simpson’s rule gives

$$ \begin{aligned} I&\approx\frac{0.5}{3} \left[1+4(1.40625)+6.324555\right]\\ &=\frac{1}{6}(1+5.625+6.324555)\\ &\approx2.158259. \end{aligned} $$

Therefore

$$ S\approx2\pi(2.158259)\approx13.560742 $$

square units. Dividing by \(2\pi\) returns the integral estimate \(2.158259\), as required. This is a valid Simpson approximation, but one application of the rule alone cannot show how close it is to the true surface area.

Assessing Numerical Error

A finer partition often improves an approximation, but a numerical estimate should not be treated as exact merely because a calculator displays many digits. For a sufficiently smooth integrand, standard error bounds quantify how the error depends on the step width. If \(g\) has a continuous second derivative on \([a,b]\), the composite trapezoidal rule satisfies

$$ \left|\int_a^b g(x)\,dx-T_n\right| \leq \frac{(b-a)h^2}{12} \max_{a\leq x\leq b}|g''(x)|. $$

If \(g\) has a continuous fourth derivative on \([a,b]\), the composite Simpson’s rule, with even \(n\), satisfies

$$ \left|\int_a^b g(x)\,dx-S_n\right| \leq \frac{(b-a)h^4}{180} \max_{a\leq x\leq b}|g^{(4)}(x)|. $$

These bounds require the stated smoothness and a valid bound on the relevant derivative across the entire interval. If an integral is multiplied by a constant such as \(2\pi\), its absolute error bound is multiplied by the absolute value of that constant. For example, an error bound \(E\) for the integral gives a surface-area error bound \(2\pi E\).

In practice, finding a useful derivative bound may be difficult. A common diagnostic is to compute estimates at two resolutions and compare them. For example, compare \(T_n\) with \(T_{2n}\), or compare Simpson estimates \(S_n\) and \(S_{2n}\). A small difference supports the view that the estimates have stabilized, but it is not a guaranteed error bound: both estimates could still be inaccurate, particularly if the integrand changes rapidly in a region missed by the chosen nodes.

Choose the interval to match the actual curve segment being revolved. The integration limits determine which part of the profile contributes to the surface area. Changing those limits changes the geometric problem, not just the numerical accuracy. Once the intended interval is fixed, decreasing \(h\) increases the number of function evaluations and usually increases computational effort. The choice should balance the desired accuracy with the smoothness of the integrand and the cost of evaluating it.

A Common Pitfall: Confusing the Integral with the Surface Area

The quadrature rules approximate the integral \(I\), not automatically the whole surface area. For a profile revolved about the \(x\)-axis, the area is \(2\pi I\). For example, if a calculation gives \(I\approx0.608380\), reporting \(0.608380\) as the surface area omits the circumference factor. The correct conversion is \(S\approx2\pi(0.608380)\approx3.822566\) square units.

Another frequent error is using Simpson’s rule with an odd number of subintervals. Its composite form groups the intervals in pairs, so \(n\) must be even. If a chosen partition has an odd number of subintervals, use an even number for Simpson’s rule or use a rule that applies to that partition, such as the composite trapezoidal rule.

Practical Procedure

1
Write the surface-area integral.
Identify the radius and the curve-length factor, and include the appropriate constant for the axis of revolution.
2
Define the integrand and interval.
Keep the integration limits tied to the intended curve segment, and compute the integrand values at the quadrature nodes.
3
Choose a composite rule.
The trapezoidal rule allows any positive \(n\); Simpson’s rule requires even \(n\).
4
Calculate the integral estimate.
Show the step width, node values, weights, and weighted sum so the approximation can be checked.
5
Convert and assess.
Apply the surface-area factor, state the units and rounding, and compare with a refined estimate or a valid error bound when available.

Numerical quadrature makes surface areas accessible even when the defining integral has no convenient antiderivative. The key is to preserve the geometric meaning of the integral while treating the numerical result as an estimate whose reliability must be assessed.

Check Your Understanding

Try answering these questions before continuing.

  1. Why does a surface-area estimate for a curve revolved about the \(x\)-axis require multiplying the quadrature estimate by \(2\pi\)?
  2. How do the endpoint and interior weights arise in the composite trapezoidal rule?
  3. Why must composite Simpson’s rule use an even number of subintervals?
  4. What smoothness conditions are needed for the stated trapezoidal and Simpson error bounds?
  5. Why does agreement between estimates on two partitions not, by itself, prove that either estimate is accurate?

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