Arc Length and Surface Area

In this section, we'll study a few more geometric objects using the definite integral. First, we'll ask how to determine the distance we've travelled if we drive along a curvy graph. Then we'll tackle roughly the same question for surfaces of revolution by computing the surface area.


Arc Length

Consider the graph of a function $  y = f(x)  $ and imagine that it is a road along which we drive our car. We would like to compute the distance it would take to travel along this road from $  x = a  $ to $  x = b  $ .

We know how to find the distance if the road were straight: in that case, we could just use the Pythagorean Theorem. However, when the road is winding, we do not have a clear way to find the distance. The idea will be to approximate the graph by a sequence of straight lines and add up the distance along those line segments. This is demonstrated below:

Here, you can see that as the approximating line segments are made shorter, they provide a more accurate approximation of the graph. This means that the distance along the line segments more accurately reflects the distance along the graph.

We will then break the interval $  [a,b]  $ into smaller pieces of width $  dx  $ . When $  dx  $ is very small, the graph looks very much like a straight line and so we can form the triangle in the picture. The length of this hypotenuse $  ds  $ is approximately the length along the curve for this little segment. From the picture, we have

\[  ds = \sqrt{dx^2 + dy^2} = \sqrt{1 + (\frac{dy}{dx})^2}~dx 
 \]

Notice that the ratio that appears in this expression, $  \frac{dy}{dx}  $ , is very suggestive: it is measuring the slope of this line which is approximating the graph. When this interval is very small, this line is the tangent line and $  \frac{dy}{dx}  $ is its slope. This means that $  \frac{dy}{dx} = f^\prime(x)  $ as the notation suggests.

Now to find the total length along the curve s, we simply add up all of the lengths of the very short line segments:

\[  
s = \int ds = \int_a^b \sqrt{1 + f^\prime(x)^2} ~dx 
 \]


Some examples

  1. First, we'll find the circumference of a circle of radius R. Of course, we expect to find the result that $  C = 2\pi R  $ , but here we can verify the result.

    The circle can be described as $  x^2 + y^2 = R^2  $ which means that, as a graph, the upper semi-circle is

    \[ 
y = f(x) = \sqrt{R^2 - x^2} 
 \]

    We will find the length along this upper semi-circle which is half the circumference of the circle. Notice that

    
\begin{eqnarray*} 
f^\prime(x) & = & -\frac{x}{\sqrt{R^2 - x^2}}\\ 
1+f^\prime(x)^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} \\ 
\sqrt{1+f^\prime(x)^2} & = & \frac{R}{\sqrt{R^2-x^2}} 
\end{eqnarray*}

    Then we can evaluate the arc length along the upper semi-circle as

    
\begin{eqnarray*} 
s & = & \int_{-R}^R \sqrt{1+f^\prime(x)^2}~dx \\ 
& = & 2 \int_0^R \frac{R}{\sqrt{R^2-x^2}} ~dx \\ 
& = & 2R\sin^{-1}(\frac xR)|_0^R \\ 
& = & \pi R 
\end{eqnarray*}

    This means that the circumference of the circle, which is twice the length along the upper semi-circle, is $  C = 2\pi R  $ as expected.

  2. Now we'll compute the length along the parabola $  y = x^2  $ between $  x = 0  $ and $  x = 1  $ . This is, in fact, the curve shown in the demonstration above. Through our approximation there, we expect that the arc length is approximately 1.479...

    Let's begin:

    
\begin{eqnarray*} 
f(x) & = & x^2 \\ 
f^\prime(x) & = & 2x \\ 
1 + f^\prime(x) & = & 1 + 4x^2 
\end{eqnarray*}

    This means that the arc length is given by $  s = \int_0^1 
\sqrt{1+4x^2}~dx  $ . This is a difficult integral to evaluate using the Fundamental Theorem of Calculus: an antiderivative for the integrand is not immediately clear through one of the techniques we've discussed. However, if we look on a table of integrals, we can really save ourselves some work. In particular, there you will notice that

    \[ 
\int \sqrt{a^2+u^2}~du = \frac u2\sqrt{a^2+u^2} + \frac{a^2}2 
\ln(u+\sqrt{a^2+u^2}) + C 
 \]

    This is not exactly the integral we want to evaluate, but it is pretty close. In fact, if we use the substitution $  u = 2x  $ , we find that

    
\begin{eqnarray*} 
s & = & \int_0^1 \sqrt{1+4x^2}~dx \\ 
& = & \frac 12 \int_0^2 \sqrt{1+u^2}~du \\ 
& = & \frac 12 \big[\frac u2\sqrt{1+u^2} + \frac 12\ln(u+\sqrt{1+u^2})\big]|_0^2 \\ 
& = & \frac12\big[\sqrt{5} + \frac 12 \ln(2+\sqrt{5})\big] \\ 
& = & 1.479 
\end{eqnarray*}

    This is the same result that we found in the demonstration above.


Surface Area

Now we would like to find the surface area of a surface obtained by revolving the portion of the graph $  y = f(x)  $ between x = a and x = b about the x axis. Again, this is difficult to understand directly, but we can approximate the surface area by small pieces whose surface area can be computed.

To that end, let's again approximate the curve $  y = f(x)  $ by dividing up the interval $  [a,b]  $ into little pieces of width $  dx  $ . The picture is the same one we saw above.

When we revolve this straight line about the x axis, we have a portion of a cone whose base radius is $  f(x)  $ .
Now imagine that we cut this cone and lay it out flat. What we would have would almost be rectangular and so the area would be approximately $  2\pi f(x)~ds 
= 2\pi f(x) \sqrt{1 + f^\prime(x)^2}~dx  $ .

Again, if we add up all the areas, we find that the surface area is

\[ 
A = \int_a^b 2\pi f(x) \sqrt{1 + f^\prime(x)^2}~dx 
 \]


Examples

  1. First, we'll find the surface area of the sphere of radius $  R  $ which can be found by revolving the graph $  y = f(x) = \sqrt{R^2 - x^2}  $ about the x axis. We have already computed above that $ 
\sqrt{1+f^\prime(x)^2} = \frac{R}{\sqrt{R^2-x^2}}  $ .

    This means that

    \[ 
A = \int_{-R}^R 2\pi \sqrt{R^2 -x^2}\frac{R}{R^2-x^2}~dx 
= 2\pi \int_{-R}^R R~dx = 4\pi R^2 
 \]

    This may be a familiar formula but now you verified it for yourself.

  2. Now we'll compute the surface area obtained by revolving the graph of $  y = \sqrt{x}  $ between x = 0 and x = 1 about the y axis.

    
\begin{eqnarray*} 
A & = & 2\pi \int_0^1 \sqrt{x} \sqrt{1 + \frac 14 \frac 1x} ~dx \\ 
& = & 2\pi \int_0^1 \sqrt{x + \frac 14}~dx \\ 
& = & 2\pi \frac 23 (x+\frac 14)^{3/2}|_0^1 \\ 
& = & \frac {4\pi}{3}((\frac 74)^{3/2} - (\frac 14)^{3/2}) 
\end{eqnarray*}