## Section 3.6 | Surface Area of Revolved Objects

We have done quite a bit with integrals, but now that we know how to find lengths of curves, we have a very powerful tool in our toolkit. Namely, we have the differential of the curve length \[ds = \sqrt{1 + \left(\frac{d\alpha}{d\beta}\right)^2} d\beta ,\]where \(\alpha\) and \(\beta\) are placeholders for either \(x\) and \(y\), respectively, or \(y\) and \(x\), respectively.

Recall that this differential is derived from the formula \[\Delta s_i = \sqrt{1 + [f'(x_i^*)]^2} \Delta x ,\]which is the length of the curve along the function \(f\) on \([x_{i - 1},x_i]\). Now imagine that, prior to summing all these lengths, we decide to rotate this segment about the \(x\)-axis. The intent is

It may help with the visualization if you play with the following Wolfram Demonstration. Move the "Rotate" slider to the left. This will show the initial curve. Then slowing move the slider to the right and you will see that band being formed. You can also click and rotate the image to a side view to get a better glimpse of the concept.

Recall that this differential is derived from the formula \[\Delta s_i = \sqrt{1 + [f'(x_i^*)]^2} \Delta x ,\]which is the length of the curve along the function \(f\) on \([x_{i - 1},x_i]\). Now imagine that, prior to summing all these lengths, we decide to rotate this segment about the \(x\)-axis. The intent is

**not**to find the volume within this rotation, but to find the surface area of this rotation. This method is closely tied, in theory, to the "cylinder" or "shell" method we have learned previously. We will rotate this line segment about the \(x\)-axis, creating an infinitesimally thin band, or ring. Since the band has no thickness, it has no volume. It is technically a two-dimensional surface. This band has carved out a circumference and it has a "height" (which is actually the length of the line segment). Hence, the surface area of this band is \[A_i = 2\pi r \Delta s,\]where \(r\) is the distance from the axis of rotation to the line segment. If we are rotating about the \(x\)-axis, \(r\) would be \(y\). If, instead, we are rotating about the \(y\)-axis, \(r\) (the distance from the \(y\)-axis to the line segment being rotated) would be \(x\).It may help with the visualization if you play with the following Wolfram Demonstration. Move the "Rotate" slider to the left. This will show the initial curve. Then slowing move the slider to the right and you will see that band being formed. You can also click and rotate the image to a side view to get a better glimpse of the concept.

Surface Area of a Solid of Revolution from the Wolfram Demonstrations Project by Mito Are

**THEOREM | SURFACE AREA OF A REVOLVED OBJECT**

Let \(f(\lambda)\) be a function of either \(\lambda = x\) or \(\lambda = y\), where \(f'\) is continuous. Then the surface area of the object created when \(f\) is revolved about a line from \(\lambda = a\) to \(\lambda = b\) is \[A = \int_{\lambda = a}^{\lambda = b}{2 \pi r ds} ,\]where \(r\) is the distance from the line being rotated about to the function \(f\).

!WARNING!

You must make sure that the integrand, the differential, and the limits of integration are all with respect to the

You must make sure that the integrand, the differential, and the limits of integration are all with respect to the

**same**variable.**EXAMPLE 1 | Some quick setups with a nasty set of derivatives**

Suppose we are given the relation \(y^3 = (x - 3)^5 + 1\).

- Write an integral to give the surface area of the revolved object about the \(x\)-axis from \((3,1)\) to \((5,33^{1/3})\) two different ways.
- Write an integral to give the surface area of the revolved object about the \(y\)-axis from \((3,1)\) to \((5,33^{1/3})\) two different ways.
- Write an integral to give the surface area of the revolved object about the line \(y = 1\) from \((3,1)\) to \((5,33^{1/3})\) two different ways.
- Write an integral to give the surface area of the revolved object about the line \(x = -6\) from \((3,1)\) to \((5,33^{1/3})\) two different ways.

**ADVICE**

As you can see, surface area integrals can often be done using any combination of variables; however, try not to use a variable that will not make a function.

I should also note something important about the fact that I keep stating \(r\) is the

**distance**to the axis of rotation. This implies that \(r \geq 0\) which, in turn, implies that the entire surface integral is greater than 0.**EXAMPLE 2 [online #1] | A surprising result**

Find the area of the surface obtained by rotating \[x = \sqrt{a^2 - y^2}, 0 \leq y \leq \frac{a}{3}\] about the \(y\)-axis.

**EXAMPLE 3 [online #2] | Easy idea... tough integral**

Find the exact area of the surface obtained by rotating \[y = \sin{\left(\frac{\pi x}{2}\right)}, 0 \leq x \leq 2\] about the \(x\)-axis.

**EXAMPLE 4 [online #3] | Technology and the surface area integral**

The function \[f(t) = e^{-t^2}\] is famously known in mathematics to be the integrand of the Gauss error function. Use Simpson's Method with \(n = 10\) to approximate the area of the surface obtained by rotating this function from \(t = 0\) to \(t = 5\) about the \(t\)-axis and compare your result with the value given by your calculator or by Wolfram Alpha. (Round your answer to six decimal places.)

**EXAMPLE 5 [online #4] | A quandary**

"Gabriel's horn" is the surface defined by rotating \(y = \frac{1}{x}\) about the \(x\)-axis from \(x = 1\) to \(\infty\).

- Suppose you want to grasp this theoretical horn and fill it with your favorite refreshment. What is the volume of liquid you could pour into it?
- Suppose your infinitely-long horn needs to be cleaned up. In fact, it needs to be painted! How much paint would you need to coat the surface of Gabriel's horn?