Have you ever noticed that flower vases tend to have extravagant shapes? Pottery is the art of forming vases, jars, and many more objects using ceramic materials like clay. How are these shapes achieved? By means of rotation. Vessels are made by spinning around the clay in a potter's wheel until it reaches the desired shape and then put in an oven to harden up.
It's time to connect this idea with Calculus. Imagine that, rather than spinning around clay in a potter's wheel, you spin around a function. By doing this you will obtain what is known as a surface of revolution. Keep reading this article if you wish to know more about surfaces of revolution and how to find their area.
Meaning of the Surface Area of Revolution
You have seen that by means of integration you can find the area below a curve described by a function. These shapes are not usually related to the areas of any geometric figures, so integration is a big help!
Integration can also be used to find the surface area of three-dimensional objects with extravagant shapes. Take for instance a parabolic antenna, which is a device used for telecommunications, like TV and the internet. The shape of a parabolic antenna can be obtained by revolving a parabola around its axis of symmetry.
Figure 1. A parabolic antenna has the shape of the surface of revolution obtained by revolving a parabola around its axis of symmetry
Remember that the axis of symmetry of a parabola is the axis that passes through its vertex and its focus.
So what is a surface of revolution?
A surface of revolution is a surface obtained by revolving a curve around a fixed axis.
In general, you can obtain surfaces of revolution by revolving any kind of functions with respect to any straight line. Surfaces of revolution have area, but have no volume, as they are completely hollow.
The difference between a surface of revolution and a solid of revolution is that the solid is filled up, so it has a volume. For this reason a solid of revolution is also called a volume of revolution. Check out more in our article about Solids of Revolution!
Area of a Surface of Revolution in Calculus
As mentioned before, you can obtain a surface of revolution by revolving a function around a fixed axis. Consider the parabolic antenna example. You need to start with a segment of a parabola, which you need to picture in the three-dimensional space, as revolving it will require an additional dimension.
Figure 2. The segment of a parabola graphed on the \(xy-\)plane of the three-dimensional space.
Now that you have a segment of a curve graphed on the \(xy-\)plane, you need to revolve it around a fixed axis. This axis is known as the axis of revolution.
An axis of revolution is the axis around which a function is revolved to obtain a surface or a solid of revolution.
Usually, in Calculus, the axis of revolution is either the \(x-\)axis or the \(y-\)axis, but it can be any straight line.
Going back into the parabolic antenna example, you should revolve the parabola segment around the \(y-\)axis to obtain this surface of revolution.
Figure 3. Rotation of the segment using the \(y-\)axis as axis of revolution.
This process produces the surface of revolution. This parabolic antenna is known as a paraboloid of revolution.
Figure 4. A paraboloid of revolution
Please note that, generally, by revolving a curve around a different axis, you will obtain a different surface as well. For example, by revolving the same curve around the \(x-\)axis instead, you obtain a trumpet!
Figure 5. A trumpet obtained by revolving a segment of a parabola around the \(x-\)axis
A Gabriel's horn is a surface of revolution obtained by revolving the curve
\[ f(x) = \frac{1}{x} \quad \text{for}\quad 1 \leq x\]
around the \(x-\)axis.
Figure 6. A Gabriel's horn
This surface of revolution is involved in an interesting Paradox. Since the curve extends to infinity you might be thinking that it has an infinite surface area, and if you were to fill it, it would have an infinite volume.
Turns out that while it is true that Gabriel's horn has an infinite surface area, it has a finite volume. The paradox comes from the fact that this would imply that you can fill Gabriel's horn with paint, but you would never have enough paint to paint its surface!
Formula for finding the Area of a Surface of Revolution
Suppose you obtain a surface of revolution by revolving a function around the \(x-\)axis. You can find the area of this surface of revolution by using the formula
\[ S = 2\pi \int_a^b f(x)\sqrt{1+f'(x)^2}\,\mathrm{d}x.\]
For this formula to work you must assume that the function is positive in the interval \( [a,b]\), and that its derivative \( f'(x)\) exists and is continuous in the same interval.
The formula for the area of a surface of revolution can be used to prove the formula for the surface area of a sphere.
Begin by noting that you can obtain a sphere of radius R by revolving the curve
\[ f(x)=\sqrt{R^2-x^2} \quad \text{for} \quad -R\leq x \leq R\]
around the \(x-\) axis.
Figure 7. A sphere with radius R obtained by revolving \( f(x) = \sqrt{R^2-x^2}\) around the \(x-\)axis
To use the formula for the area of a surface of revolution you will need to find the derivative \( f'(x)\). You can achieve this with the help of the Chain Rule and the Power Rule, that is
\[ \begin{align} f'(x) &= \frac{1}{2}\left(\frac{1}{\sqrt{R^2-x^2}}\right)(-2x) \\ &= \frac{-x}{\sqrt{R^2-x^2}}. \end{align}\]
The formula involves
\[\sqrt{1+\left( f'(x) \right) ^2},\]
so square the derivative,
\[ \left(f'(x) \right) ^2 = \frac{x^2}{R^2-x^2},\]
add \(1\) while simplifying,
\[ \begin{align} 1 + \left( f'(x) \right) ^2 &= 1+\frac{x^2}{R^2-x^2} \\ &= \frac{R^2-x^2+x^2}{R^2-x^2} \\ &= \frac{R^2}{R^2-x^2}, \end{align}\]
and take its square root,
\[ \begin{align} \sqrt{1+\left( f'(x) \right)^2} &= \sqrt{\frac{R^2}{R^2-x^2}} \\ &= \frac{R}{\sqrt{R^2-x^2}}. \end{align}\]
Note that, when you multiply by \( f(x) \) this will simplify, giving you
\[ \begin{align} f(x)\sqrt{1+\left( f'(x)\right)^2} &= \cancel{\sqrt{R^2-x^2}}\left(\frac{R}{\cancel{\sqrt{R^2-x^2}}}\right) \\ &= R\end{align} . \]
This means that the integration will be fairly simple!
\[ \begin{align} A &= 2\pi \int_{-R}^R R \,\mathrm{d}x \\ &= 2\pi R \int_{-R}^R \mathrm{d}x \\ &= 2\pi R\left( R-(-R) \right) \\ &= 4\pi R^2. \end{align}\]
By using the formula for the area of a surface of revolution you obtained the formula for the surface area of a sphere! Neat!
While this was a really nice example, please note that the integrals involved in finding the area of a surface of revolution are usually rather complicated, so the use of Computer Algebra Systems (CAS) is recommended for the evaluation of such integrals.
Derivation of the Formula for the Area of a Surface of Revolution
Just like with solids of revolution and the arc length of a curve, the best way to calculate the area of a surface of revolution usually starts by splitting the whole surface into simpler shapes. By revolving a small portion of a curve you will get a surface that looks like a truncated cone. This surface is called a frustum of a cone.
Figure 8. A frustum of a cone obtained by revolving a small segment of a curve
Now suppose you know how to find the surface area of a frustum of a cone. All you need to do is add the surface area of all the frustums, and you are done! The sum will be an approximation of the area of the surface of revolution. As usual, by taking the limit as the number of frustums tends to infinity, you will get an exact match.
The task becomes finding the surface area of a frustum of a cone. Start by recalling that the lateral surface area of a cone, which will be denoted as \( L_S\), is given by
\[ L_S = \pi R s,\]
where \( R \) is the radius of the base of the cone, and \( s \) is its slant height.
Figure 9. Dimensions of a cone
To build the frustum you can simply take off the tip of the cone, so the area of the frustum, which will be named \(A_f\) becomes
\[ A_f=\pi R s - \pi r (s-\ell),\]
where \( r \) is the radius of the tip of the cone, and \( s-\ell \) is its slant height.
Figure 10. Dimensions of a frustum
This time, you can relate \(\ell\) to the same segment that you would use to find the arc length of a curve, and the radii of the frustum can be found by evaluating the function at two consecutive values as shown in the following figure.
Figure 11. Dimensions of the frustum in terms of the function \(f(x)\)
All what is left to know is \( s \). This can be found with some algebra and similar triangles, that is
\[ \begin{align} \frac{r}{R} &= \frac{s-\ell}{s} \\ rs &= Rs-R\ell \\ rs-Rs &= -R\ell \\ s(R-r) &=R\ell \\ s &= \frac{R\ell}{R-r}. \end{align}\]
With this, it is possible to write and simplify the expression for the area of a frustum, so
\[ \begin{align} A_f &= \pi R s - \pi r (s-\ell) \\ &= \pi (Rs-rs+r\ell). \end{align} \]
Now substitute the expression you just found for \(s\), that is
\[ \begin{align} A_f &= \pi \left( R\frac{R\ell}{R-r}-r\frac{R\ell}{R-r}+r\ell \right) \\ &= \pi \left( \frac{R^2\ell}{R-r}-\frac{Rr\ell}{R-r} +r\ell\right) \\ &= \pi\ell \left( \frac{R^2-Rr+Rr-r^2}{R-r}\right) \\ &= \pi\ell\left( \frac{R^2-r^2}{R-r} \right). \end{align}\]
The expression can be further simplified by factoring the difference of squares, so
\[ \begin{align} A_f &= \pi\ell\left( \frac{(R+r)\cancel{(R-r)}}{\cancel{R-r}}\right) \\ &= \pi\ell(R+r). \end{align}\]
You can now use the above expression for the area of the frustum of the cone and write it in terms of quantities related to \( f(x)\), that is
\[ A_{f_i} = \pi\left[ f(x_{i-1})+f(x_i) \right] \sqrt{(\Delta x)^2+(\Delta y)^2},\]
which you can rewrite as
\[ A_{f_i} = \pi \left[ f(x_{i-1})+f(x_i) \right] \Delta x\sqrt{1+\left( \frac{\Delta y}{\Delta x}\right)},\]
where the index \(i \) was added to denote that this is the \(i^{th}\) piece of the whole surface. Now you can use the Intermediate Value Theorem to rewrite
\[ \frac{\Delta y}{\Delta x}\]
in terms of the derivative of \( f\), so
\[ A_{f_i} = \pi \left[ f(x_{i-1})+f(x_i) \right] \Delta x \sqrt{1+\left(f'(x_i^*\right)^2},\]
where \( x_i^*\) is some value between \(x_{i-1}\) and \(x_i\).
By adding all the frustums, you get an approximation for the area of the surface of revolution, that is
\[ A \approx \sum_{i=1}^N \pi \left[ f(x_{i-1})+f(x_i) \right] \Delta x \sqrt{1+\left(f'(x_i^*\right)^2}. \]
Now take the limit as \( N \) goes to infinity, this will:
- Turn the sum, \( \Sigma \), into an integral, \( \int\).
- The values \( x_{i-1}\), \( x_i \), and \( x_i^* \) squeeze up, so all coincide, so these will be just \(x\).
- This implies that \( f(x_{i-1}) + f(x_i) = 2f(x) \).
- Turn the length of the interval, \(\Delta x\), into a differential, \( \mathrm{d}x\).
- You will get an exact match of the area.
This way you obtain the formula for the area of a surface of revolution, which is
\[ A = 2 \pi \int_a^b f(x)\sqrt{1+\left(f'(x)\right)^2}\,\mathrm{d}x.\]
Area of a Surface of Revolution Examples
Here are some examples on finding the area of a surface of revolution.
Find the area of the surface of revolution obtained from revolving
\[ f(x) = \frac{1}{3}x^3 \quad \text{for} \quad 0\leq x \leq 1\]
around the \(x-\)axis.
Solution:
You should begin by finding the derivative of \( f(x) \) with the help of the Power Rule, this will give you
\[ f'(x)= x^2.\]
Next, you will need to square the above derivative, add it to \(1\), and take its square root. That is
\[ \begin{align} \sqrt{1+(f'(x))^2} &= \sqrt{1 + (x^2)^2} \\ &= \sqrt{1+x^4}. \end{align}\]
The area of the surface of revolution is given by
\[ \begin{align} A &= 2\pi \int_0^1 \frac{1}{3}x^3\sqrt{1+x^4}\,\mathrm{d}x \\ &= \frac{2\pi}{3} \int_0^1 x^3 \sqrt{1+x^4}\,\mathrm{d}x. \end{align}\]
Since \( x^3\) is proportional to the derivative of \(x^4\) you can use Integration by Substitution by letting
\[ u=x^4,\]
so
\[\mathrm{d}u = 4x^3 \mathrm{d}x.\]
This way you can first evaluate the indefinite integral
\[ \begin{align} \int x^3\sqrt{1+x^4}\,\mathrm{d}x &= \frac{1}{4} \int \sqrt{1+u}\,\mathrm{d}u \\ &= \frac{1}{4} \int (1+u)^{^1/_2} \, \mathrm{d}u \end{align}\]
with the help of the Power Rule, obtaining
\[ \begin{align} \int x^3\sqrt{1+x^4}\,\mathrm{d}x &= \frac{1}{4}\left(\frac{2}{3}(1+u)^{^3/_2}\right) \\ &= \frac{1}{6}(1+x^4)^{^3/_2}. \end{align}\]
Finally, use the above result to evaluate the definite integral and simplify, that is
\[ \begin{align} A &= 2\pi\left[\frac{1}{6}(1+(1)^4)^{^3/_2}-\frac{1}{6}(1+(0)^4)^{^3/_2} \right] \\ &= \frac{\pi}{3}(2^{^3/_2}-1) \\ &= \frac{\pi}{3}(2\sqrt{2}-1), \end{align}\]
and you can use a calculator to find the numeric value of the above expression, which is
\[ A \approx 1.914724 \, \, \text{square units.}\]
Sometimes you will need assistance from a CAS to evaluate the integrals!
Find the area of the surface of revolution obtained from revolving
\[ g(x) = x^2 \quad \text{for} \quad 1 \leq x \leq 3\]
around the \(x-\)axis.
Solution:
You might think that since the function is of a lower degree finding the involved surface of revolution will be easier. Unfortunately, this is not the case. As usual, begin by finding the derivative of the function, which is
\[ g'(x)=2x.\]
Now you need square it, add it to \(1\), and take its square root, giving you
\[ \begin{align} \sqrt{1+(g'(x))^2} &= \sqrt{1+(2x)^2} \\ &= \sqrt{1+4x^2}. \end{align}\]
This means that the area of the surface of revolution is given by
\[ A = 2\pi\int_1^3 x^2\sqrt{1+4x^2}\,\mathrm{d}x.\]
The evaluation of the integral
\[ \int_1^3 x^2\sqrt{1+4x^2}\,\mathrm{d}x\]
involves inverse hyperbolic functions, so it might be better to use a Computer Algebra System for its evaluation. This will give you
\[ \int_1^3 x^2\sqrt{1+4x^2}\,\mathrm{d}x \approx 40.984,\]
so you can plug in this value into the area of the surface of revolution and obtain
\[ \begin{align} A &= 2\pi \int_1^3 x^2\sqrt{1+4x^2}\,\mathrm{d}x \\ &\approx 257.51\, \, \text{square units.} \end{align}\]
Surface Area of Revolution - Key takeaways
- A surface of revolution is a surface obtained by revolving a curve around a fixed axis.
- Surfaces of revolution have no volume, as they are completely hollow.
- An axis of revolution is the axis around which a function is revolved to obtain a surface or a solid of revolution.
- You can obtain different surfaces of revolution from the same curve by revolving it around different axis of revolution.
- The formula for finding the area of a surface of revolution is given by\[ S = 2\pi \int_a^b f(x)\sqrt{1+f'(x)^2}\,\mathrm{d}x.\]
- Usually, the definite integral involved in this calculation is rather complex, so the use of a Computer Algebra System is strongly advised.
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