Arclength and Surfaces of Revolution

Given a function \(y=f(x),\ a \le x \le b\), we see how to find the length of the curve, along the graph, from \((a,f(a))\) to \((b,f(b))\). If this curve is revolved about an axis, it generates a surface of revolution, and we see how to find the resulting surface area.

Arclength

The arclength along \(y=f(x)\) from \((a,f(a))\) to \((b,f(b))\) is determined from the formula

\(\displaystyle \int_a^b \sqrt{1 + \left(f'(x)\right)^2}\ dx.\)

Except for in carefully chosen cases, integrals of this form are impossible to calculate without using numerical methods (approximations). This makes computational software especially useful for computing arclength. We’ll begin with an example where the function \(f(x)\) is “carefully chosen,” so that \(\sqrt{1 + \left(f'(x)\right)^2}\) has an antiderivative in terms of elementary functions.

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Example 1. Find the arclength of \(f(x) = \frac23 x^{3/2}\) from the point on the graph where \(x=0\) to the point where \(x=1\). Since \(f'(x) = x^{1/2}\), the integral to be calculated is

\(\int_0^1 \sqrt{1 + x}\ dx.\)

Within SageMath the antiderivative can be calculated by the command integral(sqrt(1+x), x). In order to get the definite integral that is the arclength¹ the SageMath command is integral(sqrt(1+x), x, 0, 1), which produces the answer \(\dfrac{4}{3}\sqrt{2}\ – \dfrac{2}{3}\).

Example 2. Say that you want to find the arclength along the sine curve \(y=\sin(x)\) from \((0,0)\) to \((\pi,0)\). In SageMath, once a function f(x) has been defined (by putting a line f(x)=... that defines it), you can find the derivative by the command diff(f, x). This means that the following would give the arclength of the sine curve above:

f(x) = sin(x)
integral(sqrt( 1 + (diff(f,x))^2 ), x, 0, pi)

If you’ve gone and attempted to compute the above, you may confused when the output is not any number. Instead, it looks a lot like the line you typed in (especially if you determined the derivative, \(\cos(x)\), yourself; your output should be something like integrate(sqrt( (cos(x))^2 + 1 ), x, 0, pi) .)

So…what is going on? It is not just an issue in SageMath; if you were working in another computational software system, like Mathematica, you would also get a weird answer: \(\sqrt{2}\ \texttt{EllipticE}\left[\texttt{x}, \frac12\right]\). The reason for these answers is that the function \(\sqrt{\cos^2(x) + 1}\) does not have an anti-derivative that is in terms of elementary functions; one cannot use an antiderivative in the Fundamental Theorem of Calculus to find the definite integral.

However, you can get an approximate numerical value of the integral. To do so in SageMath, you could put the following²

f(x) = sin(x) 
numerical_integral(sqrt( 1 + (diff(f,x))^2 ), 0, pi)

 

and get an ordered pair (3.820197789027712, 4.241271544000682e-14). The first number is the approximate value of the integral. The second number is an upper bound on the error; that is, a maximum amount by which the true value of the integral might differ from the approximation.

Exercises

For each of the following curves \(y=f(x)\), if the function being integrated in the formula for arclength has an anti-derivative in terms of elementary functions then write down that anti-derivative. Otherwise, say it does not exist.  In both cases, have a computation in your notebook that supports this.  Finally, find a numerical value for the arclength.

1. 1. 1. \(f(x) = \frac{x^3}{3} + \frac{1}{4x},\ a=1, b=2.\)
      2. \(f(x) = \arctan(x),\ a=0, b=1.\)
      3. \(f(x) = \ln(x),\ a=1, b=e.\)
      4. \(f(x) = \sqrt{1 – x^2},\ a=0, b=1.\)

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