## Area of a Surface of Revolution (Part 1)

1. Suppose we consider some function $y=f(x)$ on the domain $[a,b]$. For example:

We revolve this about the $x$ axis by 360 degrees. So this looks like

Question: what is the area occupied by this surface?
2. The trick is to write the area as the sum of the area of smaller strips. So for example, one such strip might be:

What is its area? We approximate this as a cylinder with radius
(1)$\displaystyle r=f(x)$
and height is the segment’s arclength
(2)$\displaystyle h=\sqrt{\left(\frac{\mathrm{d}x}{\mathrm{d}x}\right)^{2}+\left(\frac{\mathrm{d}f(x)}{\mathrm{d}x}\right)^{2}}\,\mathrm{d}x$.
Thus the cylinder’s surface area is
(3)$\displaystyle A_{cyl} = 2\pi r\cdot h = 2\pi f(x)\sqrt{\left(\frac{\mathrm{d}x}{\mathrm{d}x}\right)^{2}+\left(\frac{\mathrm{d}f(x)}{\mathrm{d}x}\right)^{2}}\,\mathrm{d}x$.
We have many cylinders, and we add them all up. This sum is a very natural integral:
(4)$\displaystyle A_{surface} = \int 2\pi f(x)\sqrt{\left(\frac{\mathrm{d}x}{\mathrm{d}x}\right)^{2}+\left(\frac{\mathrm{d}f(x)}{\mathrm{d}x}\right)^{2}}\,\mathrm{d}x$.
This can be somewhat simplified as
(4′)$\displaystyle A_{surface} = \int 2\pi f(x)\sqrt{1+\left(\frac{\mathrm{d}f(x)}{\mathrm{d}x}\right)^{2}}\,\mathrm{d}x$.

Example 1. Consider the curve $f(x)=x^{3}/3$ on the domain $[0,1]$. What is the area for its surface of revolution?

Solution. We see the arc-length differential for this function is
(5)$\displaystyle \mathrm{d}s = \sqrt{1+\bigl(f'(x)\bigr)^{2}}\,\mathrm{d}x=\sqrt{1+(x^{2})^{2}}\,\mathrm{d}x$.
The differential area for the surface is
(6)$\displaystyle \mathrm{d}A=\frac{2\pi}{3} x^{3}\sqrt{1+x^{4}}\,\mathrm{d}x$.
Integrating gives us the area
(7)$\displaystyle A=\int^{1}_{0}\frac{2\pi}{3} x^{3}\sqrt{1+x^{4}}\,\mathrm{d}x$.
But now we have to perform this integral!

We let $u = x^{4}$ and $\mathrm{d}u=4x^{3}\,\mathrm{d}x$. Thus our integral (7) becomes
(8)$\displaystyle \int^{1}_{0}\frac{2\pi}{3} x^{3}\sqrt{1+x^{4}}\,\mathrm{d}x=\frac{2\pi}{3}\int^{x=1}_{x=0}\sqrt{1+u}\frac{\mathrm{d}u}{4}$.
This can be solved directly as
(9)$\displaystyle \frac{2\pi}{3}\int^{x=1}_{x=0}\sqrt{1+u}\frac{\mathrm{d}u}{4}=\frac{2\pi}{12}\left.\frac{(1+u)^{3/2}}{3/2}\right|^{x=1}_{x=0}$.
Now we unroll our substitution, transforming (9) into
(10)$\displaystyle \frac{2\pi}{12}\left.\frac{(1+u)^{3/2}}{3/2}\right|^{x=1}_{x=0} = \left.\frac{2\pi}{12}\frac{2}{3}(1+x^{4})\right|^{x=1}_{x=0}=\frac{4\pi}{36}(2-1)=\frac{\pi}{9}$.
That concludes our example.

Example 2 [Variation of Math.SE]. Consider the curve
(11)$\displaystyle y=\frac{x^{2}}{4}-\frac{\ln(x)}{2}$ for $1\leq x\leq 2$
What is the area for the surface of revolution about the $x$ axis?

Solution. We have the surface’s area be described by the integral
(12)$\displaystyle A=\int^{2}_{1}2\pi y\sqrt{1+\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^{2}}\,\mathrm{d}x$
We need to express the integrand as an explicit function of $x$. So we begin with
(13)\displaystyle\begin{aligned}\frac{\mathrm{d}y}{\mathrm{d}x}&=\frac{x}{2}-\frac{1}{2x}\\ &=\frac{x^{2}-1}{2x}\end{aligned}
Great, so now we can consider its square
(14)\displaystyle\begin{aligned}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^{2}&=\left(\frac{x^{2}-1}{2x}\right)^{2}\\ &=\frac{(x^{2}-1)}{4x^{2}}\\ &=\frac{x^{4}-2x^{2}+1}{4x^{2}}\end{aligned}
Now we can consider the arc-length portion of the integrand
(15)\displaystyle\begin{aligned}\sqrt{1+\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^{2}}&=\sqrt{1+\frac{x^{4}-2x^{2}+1}{4x^{2}}}\\ &=\sqrt{\frac{4x^{2}+x^{4}-2x^{2}+1}{4x^{2}}}\\ &=\frac{\sqrt{4x^{2}+x^{4}-2x^{2}+1}}{2x}\\ &=\frac{\sqrt{x^{4}+2x^{2}+1}}{2x}=\frac{x^{2}+1}{2x} \end{aligned}
Great, what now?

We can set up the integral in Equation (12) as
(16)$\displaystyle \int^{2}_{1}2\pi y\sqrt{1+\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^{2}}\,\mathrm{d}x = \int^{2}_{1}2\pi\left(\frac{x^{2}}{4}-\frac{\ln(x)}{2}\right)\cdot\left(\frac{x^{2}+1}{2x}\right)\,\mathrm{d}x$
which we will consider as the sum of two integrals:
(17a)$\displaystyle A_{1} = \int^{2}_{1}2\pi\frac{x^{2}}{4}\left(\frac{x^{2}+1}{2x}\right)\,\mathrm{d}x$
and
(17b)$\displaystyle A_{2} = \int^{2}_{1}2\pi\left(-\frac{\ln(x)}{2}\right)\cdot\left(\frac{x^{2}+1}{2x}\right)\,\mathrm{d}x$.
Now we have to evaluate each integral in turn.

We use substitution for (17a), take
(18)$\displaystyle u=x^{2}$ and $\displaystyle \mathrm{d}u = 2x\,\mathrm{d}x$.
Then we have
(19)\displaystyle \begin{aligned}\int^{2}_{1}\pi\frac{x}{8x}\left(x^{2}+1\right)2x\,\mathrm{d}x &= \frac{\pi}{8}\int^{2}_{1}(x^{2}+1)2x\,\mathrm{d}x\\ &=\frac{\pi}{8}\int^{x=2}_{x=1}(u+1)\,\mathrm{d}u\end{aligned}.

This is a trivial integral to evaluate
(20)$\displaystyle \frac{\pi}{8}\int^{x=2}_{x=1}(u+1)\,\mathrm{d}u =\left.\frac{\pi}{8}\frac{u^{2}}{2}+u\right|^{x=2}_{x=1}$.

We unroll our substitution, letting $u = x^{2}$ and we get
(21)\displaystyle \begin{aligned} \frac{\pi}{8}\int^{x=2}_{x=1}(u+1)\,\mathrm{d}u &=\left.\frac{\pi}{8}\left(\frac{u^{2}}{2}+u\right)\right|^{x=2}_{x=1}\\ &=\left.\frac{\pi}{8}\left(\frac{x^{4}}{2}+x^{2}\right)\right|^{x=2}_{x=1}\\ &=\frac{\pi}{8}\left[12-\frac{3}{2}\right]=\frac{21\pi}{16}\end{aligned}.

Great this gives us $A_{1}$, but what about our other integral in Equation (17b)?

So we want to evaluate the integral
(17b)$\displaystyle A_{2} = \int^{2}_{1}2\pi\left(-\frac{\ln(x)}{2}\right)\cdot\left(\frac{x^{2}+1}{2x}\right)\,\mathrm{d}x$.
Lets use the substitution
(22)$\displaystyle v=\ln(x)$ and $\displaystyle \mathrm{d}v = \frac{\mathrm{d}x}{x}$.
Note this means that $x=\exp(v)$. So our integral is
(23)$\displaystyle \int^{2}_{1}2\pi\left(-\frac{\ln(x)}{2}\right)\cdot\left(\frac{x^{2}+1}{2}\right)\frac{\mathrm{d}x}{x} = \frac{2\pi}{4}\int^{x=2}_{x=1}v(\mathrm{e}^{2v}+1)\,\mathrm{d}v$.
Lets break this up into more manageable problems.

We see that we have one term be
(24)$\displaystyle \int^{x=2}_{x=1}v\mathrm{e}^{2v}\,\mathrm{d}v$.
Using integration by parts we have, letting $f(v)=v$ and $g(v)=\exp(2v)/2$, that
(24′)$\displaystyle \int^{x=2}_{x=1}v\mathrm{e}^{2v}\,\mathrm{d}v=\int f(v)g'(v)\,\mathrm{d}v$.
Thus integration by parts yields
(25)$\displaystyle \int^{x=2}_{x=1}v\mathrm{e}^{2v}\,\mathrm{d}v=\left.\frac{v\mathrm{e}^{2v}}{2}\right|^{x=2}_{x=1}-\int \frac{\mathrm{e}^{2v}}{2}\,\mathrm{d}v$.
But these are trivial integrals, we find
(26)\displaystyle \begin{aligned}\left.\frac{v\mathrm{e}^{2v}}{2}\right|^{x=2}_{x=1}-\int \frac{\mathrm{e}^{2v}}{2}\,\mathrm{d}v &= \left.\frac{v\mathrm{e}^{2v}}{2}\right|^{x=2}_{x=1}-\left.\frac{\mathrm{e}^{2v}}{4}\right|^{x=2}_{x=1}\\ &=\left.\frac{(2v-\mathrm{e}^{v})\mathrm{e}^{v}}{4}\right|^{x=2}_{x=1}\end{aligned}.
This is one term for Equation (23). The other is
(27)$\displaystyle frac{2\pi}{4}\int^{x=2}_{x=1}v\,\mathrm{d}v=\left.\frac{v^{2}}{2}\right|^{x=2}_{x=1}$.
Combining Equations (26) and (27) gives us
(28)$\displaystyle \frac{2\pi}{4}\int^{x=2}_{x=1}v(\mathrm{e}^{2v}+1)\,\mathrm{d}v = \left.\frac{2\pi}{4}\left(\frac{(2v-1)\mathrm{e}^{v}}{4}+\frac{v^{2}}{2}\right)\right|^{x=2}_{x=1}$.
Now we must unroll our substitution, again. So we plug in $v=\ln(x)$, giving us
(29)\displaystyle \begin{aligned} \frac{2\pi}{4}\int^{x=2}_{x=1}v(\mathrm{e}^{2v}+1)\,\mathrm{d}v &= \left.\frac{2\pi}{4}\left(\frac{(2v-1)\mathrm{e}^{2v}}{4}+\frac{v^{2}}{2}\right)\right|^{x=2}_{x=1}\\ &=\left.\frac{2\pi}{4}\left(\frac{\bigl(2\ln(x)-1\bigr)x^{2}}{4}+\frac{\ln(x)^{2}}{2}\right)\right|^{x=2}_{x=1}\end{aligned}.
We then evaluate this to find
(30)\displaystyle \begin{aligned} A_{2}&=\frac{2\pi}{4}\int^{x=2}_{x=1}v(\mathrm{e}^{2v}+1)\,\mathrm{d}v \\ &=\left.\frac{2\pi}{4}\left(\frac{\bigl(2\ln(x)-1\bigr)x^{2}}{4}+\frac{\ln(x)^{2}}{2}\right)\right|^{x=2}_{x=1}\\ &=\frac{\pi}{2}\left[\left(\frac{2\ln(2)-1}{4}\right)4+\frac{\ln(2)^{2}}{2}-\left(\frac{-1}{4}\right)\right]\\ &=\frac{\pi}{2}\left[2\ln(2)-1+\frac{\ln(2)^{2}}{2}+\frac{1}{4}\right] \end{aligned}.
This is the other half of our problem.

Now Equation (21) gave us $A_{1}$ and Equation (30) gave us $A_{2}$. We can now combine these together to get the area
(31)$\displaystyle A = A_{1}+A_{2} = \frac{21\pi}{16} + \frac{\pi}{2}\left[2\ln(2)-1+\frac{\ln(2)^{2}}{2}+\frac{1}{4}\right]$.
That concludes our problem.

Exercise 1. Let $y = \sqrt{4x+1}$. Consider the surface of revolution about the $x$ axis. Calculate the area for the surface of revolution restricted along the interval $[0,2]$. [from Math.SE].