Title: Default for 14.4 Lengths of Curves and Surfaces of Revolution
Course: HELM Workbooks 2023-2024

The length of a curve

To find the length of a curve in the \(xy\) plane we first divide the curve into a large number of pieces. We measure (or, at least, approximate) the length of each piece and then by an obvious summation process obtain an estimate for the length of the curve. Theoretically, we allow the number of pieces to increase without bound, implying that the length of each piece will tend to zero. In this limit the summation process becomes an integration process.

[Image: See text below]

The Figure above shows the portion of the curve \(y(x)\) between \(x = a\) and \(x = b\). A small piece of this curve has been selected and can be considered as the hypotenuse of a triangle with base \(\delta x\) and height \(\delta y\). (Here \(\delta x\) and \(\delta y\) are intended to be ‘small’ so that the curved segment can be regarded as a straight segment.)

Using Pythagoras’ theorem, the length of the hypotenuse is: \[\sqrt{\delta x^2 + \delta y^2}= \sqrt{1+\left(\frac{\delta y}{\delta x}\right)^2} \: \delta x \]

By summing all such contributions between \(x=a\) and \(x=b\), and letting \(\delta x \to 0\) we obtain an expression for the total length of the curve: \[ \lim_{\delta x \to 0} \sum_{x=a}^{x=b} \sqrt{1+\left(\frac{\delta y}{\delta x}\right)^2} \: \delta x \]

But we already know how to write such an expression in terms of an integral. We obtain the following result:

Key Point

Given a curve with equation \(y = f(x)\), then the length of the curve between the points where \(x = a\) and \(x = b\) is given by the formula: \[\int_a^b \sqrt{1+\left(\frac{\mathrm{d} y}{\mathrm{d} x}\right)^2}\ \mathrm{d}x\]

Because of the complicated form of the integrand, and in particular the square root, integrals of this type are often difficult to calculate. In practice, approximate numerical methods rather than exact methods are normally needed to perform the integration. We shall first illustrate the application of the formula in the Key Point above by a problem which could be calculated in a much simpler way, before looking at some harder problems.

Example

Find the length of the curve \(y = 3x + 2\) between \(x = 1\) and \(x = 5\)

Solution

In this Example, the curve is in fact a straight line segment, and its length could be obtained using Pythagoras’ theorem without the need for integration.

Notice from the formula in the Key Point that it is necessary to find \(\frac{\mathrm{d}y}{\mathrm{d}x}\), which in this case is \(3\).

Applying the formula we find \[\begin{aligned}\text{length of curve} & \int_1^5 \sqrt{1+(3)^2} \mathrm{d}x \\ & = \int_1^5 \sqrt{10} \mathrm{d}x \\ & = \left[\sqrt{10}x\right]^5_1 \\ & = (5-1)\sqrt{10} = 4 \sqrt{10} = 12.65 \; \text{to} \: 2 \text{ d.p} \end{aligned} \]

Thus the length of the curve \(y = 3x + 2\) between the points where \(x = 1\) and \(x = 5\) is \(12.65\) units.

Exercise

Find the length of the curve \(y = \cosh x\) between \(x = 0\) and \(x = 2\) shown in the diagram.

[Image: Graph of y=cosh(x) between x=0 and x=2.]

First write down \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\)

\(\dfrac{\mathrm{d}y}{\mathrm{d}x}=\)

Hence write down the required integral:

\(\displaystyle{\int_0^2}\)  \(\mathrm{d}x\)

This integral can be evaluated by making use of the hyperbolic identity \(\cosh^2 x - \sinh^2 x \equiv 1\).

Write down the integral which results after applying this identity:

\(\displaystyle{\int_0^2}\) \(\mathrm{d}x\)

Perform the integration to find the required length, giving your answer to 2 decimal places.
Feedback: The formula in the Key Point tells us that, given a curve with equation \(y=f(x)\), the length of the curve between the points where \(x=a\) and \(x=b\) is given by the formula \[\int_a^b \sqrt{1+\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2} \mathrm{d}x\] If \(y=\cosh x\) then \(\frac{\mathrm{d}y}{\mathrm{d}x}=\sinh x\), therefore the required integral is \[\int_0^2 \sqrt{1+\sinh^2 x} \mathrm{d}x\] Making use of the hyperbolic identity \(\cosh^2 x - \sinh^2 x \equiv 1\), the integral can be written as \[\int_0^2 \cosh x \mathrm{d}x\] Therefore, the length of the curve \(y=\cosh x\) between \(x=0\) and \(x=2\) is \[\big[\sinh x\big]^2_0=\sinh(2)-\sinh(0)=3.62686=3.63 \]

Exercise

Find the length of the curve \(y = x^2\) between \(x = [value]\) and \(x = [value]\).

Given \(y=x^2\) then \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=2x\). Use this result and apply the formula in the first Key Point to obtain the integral required:

\(\displaystyle{\int_{[value]}^{[value]}}\) \(\mathrm{d}x\)

Make the substitution \(x=\dfrac{1}{2}\sinh u\), giving \(\dfrac{\mathrm{d}x}{\mathrm{d}u}=\dfrac{1}{2}\cosh u\), to obtain an integral in terms of \(u\):

\(\displaystyle{\int_{0}^{\sinh^{-1} [value]}}\) \(\mathrm{d}u\)

Use the hyperbolic identity \(\cosh^2 u - \sinh^2 u \equiv 1\) to eliminate \(\sinh^2 u\):

\(\displaystyle{\int_{0}^{\sinh^{-1} [value]}}\) \(\mathrm{d}u\)

Use the hyperbolic identity \(\cosh^2 u = \frac{1}{2}(\cosh 2u +1)\) to rewrite the integrand in terms of \(\cosh 2u\):

\(\displaystyle{\int_{0}^{\sinh^{-1} [value]}}\) \(\mathrm{d}u\)

Finally, perform the integration to find the required length, giving your answer to 2 decimal places.
Feedback: The formula in the Key Point tells us that, given a curve with equation \(y=f(x)\), the length of the curve between the points where \(x=a\) and \(x=b\) is given by the formula \[\int_a^b \sqrt{1+\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2} \: \mathrm{d}x\] Therefore, our required integral is \[\int_{0}^{\sinh^{-1} [value]} \sqrt{1+4 x^2} \: \mathrm{d}x\] Using the substitution \(x=\frac{1}{2}\sinh u\), we calculate that the upper limit is \(u=\sinh^{-1}(2\cdot [value])=\sinh^{-1}([value])\), and the lower limit is \(u=0.\) Therefore, the integral becomes \[\int_{0}^{\sinh^{-1} [value]} \sqrt{1+\sinh^2 u}\frac{1}{2}\cosh u \: \mathrm{d}u\] Using the hyperbolic identity \(\cosh^2 u - \sinh^2 u \equiv 1\), we eliminate \(\sinh^2 u\), to obtain \[\frac{1}{2}\int_{0}^{\sinh^{-1} [value]}\cosh^2 u \: \mathrm{d}u\] Using the hyperbolic identity \(\cosh^2 u = \frac{1}{2}(\cosh 2u +1)\), the integrand can be written as \[\frac{1}{4}\int_{0}^{\sinh^{-1} [value]} (\cosh 2u + 1) \: \mathrm{d}u\] Now we can perform the integration: \( \begin{aligned} \frac{1}{4} \int_{0}^{\sinh^{-1} [value]} (\cosh 2u + 1) \: \mathrm{d}u & = \frac{1}{4} \left[ \frac{\sinh 2 u}{2} + u \right]_{0}^{\sinh^{-1} [value]} \\ & = \frac{1}{4} \left(\left( \frac{\sinh(2\cdot \sinh^{-1}([value]))}{2}+\sinh^{-1}([value]) \right)-\left( \frac{\sinh(2\cdot 0)}{2}+0\right)\right) \\ & = [value] \end{aligned} \)

Exercise

Find the length of the line \(y = [value]\) between \(x = [value]\) and \(x = [value]\) using the technique of this Section. Verify your result from your knowledge of the straight line. Give your answer in exact form.

The length is .
Feedback: The distance is from \(([value],[value])\) to \(([value],[value])\) along the line. This is given using Pythagoras' theorem as \(\sqrt{([value]-[value])^2+([value]-[value])^2}=[value] \) Using the formula \[\text{length of curve}=\int_a^b \sqrt{1+\left(\frac{dy}{dx}\right)^2}\ \mathrm{d}x,\] the length of the line is calculated to be \[\int_{[value]}^{[value]} \sqrt{1+([value])^2} \ \mathrm{d}x=[value].\]

Exercise

Find the length of \(y = [value]x^{\frac{3}{2}}\) between \(x = [value]\) and \(x = [value]\). Give your answer in exact form.

The length is
Feedback: Given \( y = [value] \) then \( \dfrac{dy}{dx} = [value].\) Using the formula \[\text{length of curve}=\int_a^b \sqrt{1+\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2} \ \mathrm{d}x,\] the length of the line is calculated to be: \[\ell = \int_{[value]}^{[value]} \sqrt{1+[value]} \ \mathrm{d}x = \left[ [value] \right]_{[value]}^{[value]}=[value].\] For this integral you can use the substitution \( u= 1+[value].\)

Exercise

Calculate the length of the curve \(y^2 = [value]\) between \(x = [value]\) and \(x = [value]\), in the first quadrant. Give your answer in exact form.

The length is
Feedback: We are given \( y^2 = [value] \) then \( y = [value]\)on the first quadrant. The derivative \( \dfrac{\mathrm{d}y}{\mathrm{d}x} = [value].\) Using the formula \[\text{length of curve}=\int_a^b \sqrt{1+\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2} \mathrm{d}x,\] the length of the line is calculated to be: \[ \ell = \int_{[value]}^{[value]} \sqrt{1+[value]} \, \mathrm{d}x = \left[ [value] \right]_{[value]}^{[value]} = [value].\] For this integral you can use the substitution \( u= 1+[value].\)

The area of a surface of revolution

In previous workbooks we found an expression for the volume of a solid of revolution. Here we consider the more complicated problem of formulating an expression for the surface area of a solid of revolution.

[Image: See text below.]

The Figure above shows the portion of the curve \(y(x)\) between \(x = a\) and \(x = b\) which is rotated around the \(x\) axis through \(360^{\circ}\). A small disc, of thickness \(\delta x\), of the solid of revolution has been selected. Its radius is \(y\) and so its circumference has length \(2 \pi y\). (As usual we assume \(\delta x\) is ‘small’ so that the curved part of \(y(x)\) representing the hypotenuse of the highlighted ‘triangle’ can be regarded as straight). This surface ‘ribbon’, shown shaded, has a length \(2 \pi y\) and a width \(\sqrt{(\delta x)^2 + (\delta y)^2}\) and so its area is, to a good approximation, \(2 \pi y \sqrt{(\delta x)^2 + (\delta y)^2}\). We now let \(\delta x \to 0\) to obtain the result in the Key Point below:

Key Point

Given a curve with equation \(y = f(x)\), then the surface area of the solid generated by rotating that part of the curve between the points where \(x = a\) and \(x = b\) around the \(x\) axis is given by the formula: \[\text{area of surface} = \int_a^b 2 \pi y \sqrt{1+\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2} \ \mathrm{d}x\]

Exercise

Find the area of the surface generated when the part of the curve \(y=x^3\) between \(x=0\) and \(x=4\) is rotated around the \(x\) axis.

Using the previous Key Point, write down the integral:

\(\displaystyle \int\) |  | \(\mathrm{d}x\)

Use the substitution \(u=1+9x^4\) so \(\dfrac{\mathrm{d}u}{\mathrm{d}x}=36 x^3\) to write down the integral in terms of \(u\):

\(\displaystyle \int\) |  | \(\mathrm{d}u\)

Perform the integration and apply the limits of integration to find the area, giving your answer in exact form.
Feedback: \[\text{area}=\int_a^b 2\pi y \sqrt{1+\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2} \ \mathrm{d}x= \int_0^4 2 \pi x^3 \sqrt{1+(3x^2)^2} \ \mathrm{d}x = \int_0^4 2 \pi x^3 \sqrt{1+9x^4} \ \mathrm{d}x\] Using the substitution \(u=1+9x^4\), we calculate that the upper limit is \(u=1+9(4)^4=2305\), and the lower limit is \(u=1+9(0)^4=1\). Therefore, the integral becomes \[A = \int_1^{2305} 2 \pi x^3 \sqrt{u} \frac{\mathrm{d}u}{36x^3} = \frac{\pi}{18}\int_1^{2305} \sqrt{u}\ \mathrm{d}u\] Performing the integration, we obtain \[\frac{\pi}{18} \left[\frac{2 u^{\frac{3}{3}}}{3}\right]^{2305}_1\] Applying the limits of integration, the area is calculated to be: \(A = \dfrac{\pi}{27}\left((2305)^{\frac{3}{2}}-1\right).\)

Exercise

The line \(y =[value]\) between \(x = [value]\) and \(x = [value]\) is rotated around the \(x\) axis.

Find the area of the surface generated, giving your answer in exact form.

The area is
Feedback: Given \(y =[value]\) than \( \dfrac{\mathrm{d}y}{\mathrm{d}x} = [value].\) Using the formula \[\text{area}=\int_a^b 2\pi y \sqrt{1+\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2} \ \mathrm{d}x,\] the area of the surface generated is calculated to be: \[A = \int_{[value]}^{[value]} 2\pi ([value]) \sqrt{1+[value]} \, \mathrm{d}x= \left[ [value] \right]_{[value]}^{[value]}= ([value]).\]

Exercise

Find the area of the surface generated when \(y =[value]\) in the interval \([value]\leq x \leq [value]\) is rotated about the \(x\)-axis. Give your answer in exact form.

The area is
Feedback: Given \(y =[value]\) then \( \dfrac{\mathrm{d}y}{\mathrm{d}y}= \displaystyle [value].\) Using the formula \[\text{area}=\int_a^b 2\pi y \sqrt{1+\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2} \ \mathrm{d}x,\] the area of the surface generated is calculated to be: \[A = \int_{[value]}^{[value]} 2\pi ([value]) \sqrt{1+[value]} \ \mathrm{d}x = \left[ [value] \right]_{[value]}^{[value]}= [value].\]