10 Curves in the Plane Chapter Introduction 10.2 Parametric Equations

10.1 Arc Length and Surface Area

^†^†margin: (a) (b) Figure 10.1.1: Graphing

y=\sin x

[0,\pi]

and approximating the curve with line segments.

In previous sections we have used integration to answer the following questions:

1.

Given a region, what is its area?
2.

Given a solid, what is its volume?

In this section, we address a related question: Given a curve, what is its length? This is often referred to as arc length.

Consider the graph of $y=\sin x$ on $[0,\pi]$ given in Figure 10.1.1 (a). How long is this curve? That is, if we were to use a piece of string to exactly match the shape of this curve, how long would the string be?

As we have done in the past, we start by approximating; later, we will refine our answer using limits to get an exact solution.

The length of straight-line segments is easy to compute using the Distance Formula. We can approximate the length of the given curve by approximating the curve with straight lines and measuring their lengths.

In Figure 10.1.1 (b), the curve $y=\sin x$ has been approximated with 4 line segments (the interval $[0,\pi]$ has been divided into 4 equally-lengthed subintervals). It is clear that these four line segments approximate $y=\sin x$ very well on the first and last subinterval, though not so well in the middle. Regardless, the sum of the lengths of the line segments is $3.79$ , so we approximate the arc length of $y=\sin x$ on $[0,\pi]$ to be $3.79$ .

In general, we can approximate the arc length of $y=f(x)$ on $[a,b]$ in the following manner. Let $a=x_{0}<x_{1}<\dotso<x_{n-1}<x_{n}=b$ be a partition of $[a,b]$ into $n$ subintervals. Let $\Delta x_{i}$ represent the length of the $i\,^{\text{th}}$ subinterval $[x_{i-1},x_{i}]$ .

^†^†margin: Figure 10.1.2: Zooming in on the

i^{\text{th}}

subinterval

[x_{i-1},x_{i}]

of a partition of

[a,b]

Figure 10.1.2 zooms in on the $i^{\text{th}}$ subinterval where $y=f(x)$ is approximated by a straight line segment. The dashed lines show that we can view this line segment as the hypotenuse of a right triangle whose sides have length $\Delta x_{i}$ and $\Delta y_{i}$ . Using the Pythagorean Theorem, the length of this line segment is $\displaystyle\sqrt{(\Delta x_{i})^{2}+(\Delta y_{i})^{2}}.$ Summing over all subintervals gives an arc length approximation

L\approx\sum_{i=1}^{n}\sqrt{(\Delta x_{i})^{2}+(\Delta y_{i})^{2}}.

As it is written, this is not a Riemann Sum. While we could conclude that taking a limit as the subinterval length goes to zero gives the exact arc length, we would not be able to compute the answer with a definite integral. We need first to do a little algebra.

In the above expression factor out a $\Delta x_{i}^{2}$ term:

\displaystyle\sum_{i=1}^{n}\sqrt{(\Delta x_{i})^{2}+(\Delta y_{i})^{2}}

\displaystyle=\sum_{i=1}^{n}\sqrt{(\Delta x_{i})^{2}\left(1+\frac{(\Delta y_{i% })^{2}}{(\Delta x_{i})^{2}}\right)}.

Now pull the $(\Delta x_{i})^{2}$ term out of the square root:

L\approx\sum_{i=1}^{n}\sqrt{1+\frac{(\Delta y_{i})^{2}}{(\Delta x_{i})^{2}}}\ % \Delta x_{i}.

This is nearly a Riemann Sum. Consider the $(\Delta y_{i})^{2}/(\Delta x_{i})^{2}$ term. The expression $\Delta y_{i}/\Delta x_{i}$ measures the “change in $y$ /change in $x$ ,” that is, the “rise over run” of $f$ on the $i\,^{\text{th}}$ subinterval. The Mean Value Theorem of Differentiation (Theorem 3.2.1) states that there is a $c_{i}$ in the $i\,^{\text{th}}$ subinterval where $f\,^{\prime}(c_{i})=\Delta y_{i}/\Delta x_{i}$ . Thus we can rewrite our above expression as:

L\approx\sum_{i=1}^{n}\sqrt{1+[f\,^{\prime}(c_{i})]^{2}}\ \Delta x_{i}.

This is a Riemann Sum. As long as $f\,^{\prime}$ is continuous on $[a,b]$ , we can invoke Theorem 5.3.2 and conclude

L=\int_{a}^{b}\sqrt{1+[f\,^{\prime}(x)]^{2}}\ dx.

Key Idea 10.1.1 Arc Length

Let $f$ be differentiable on an open interval containing $[a,b]$ , where $f\,^{\prime}$ is also continuous on $[a,b]$ . Then the arc length of $f$ from $x=a$ to $x=b$ is

L=\int_{a}^{b}\sqrt{1+[f\,^{\prime}(x)]^{2}}\ dx.

Watch the video:
Arc Length from https://youtu.be/PwmCZAWeRNE

As the integrand contains a square root, it is often difficult to use the formula in Key Idea 10.1.1 to find the length exactly. When exact answers are difficult to come by, we resort to using numerical methods of approximating definite integrals. The following examples will demonstrate this.

^†^†margin: Figure 10.1.3: A graph of

f(x)=x^{3/2}

from Example 10.1.1.

Example 10.1.1 Finding arc length

Find the arc length of $f(x)=x^{3/2}$ from $x=0$ to $x=4$ .

SolutionA graph of $f$ is given in Figure 10.1.3. We begin by finding $f\,^{\prime}(x)=\frac{3}{2}x^{1/2}$ . Using the formula, we find the arc length $L$ as

	$\displaystyle L$	$\displaystyle=\int_{0}^{4}\sqrt{1+\left(\frac{3}{2}x^{1/2}\right)^{2}}\ dx$
		$\displaystyle=\int_{0}^{4}\sqrt{1+\frac{9}{4}x}\ dx$
		$\displaystyle=\int_{0}^{4}\left(1+\frac{9}{4}x\right)^{1/2}\ dx$
		$\displaystyle=\frac{2}{3}\cdot\frac{4}{9}\left(1+\frac{9}{4}x\right)^{3/2}\Big% {\|}_{0}^{4}$
		$\displaystyle=\frac{8}{27}\left(10^{3/2}-1\right)\text{units}.$

Example 10.1.2 Finding arc length

Find the arc length of $\displaystyle f(x)=\frac{1}{8}x^{2}-\ln x$ from $x=1$ to $x=2$ .

SolutionA graph of $f$ is given in Figure 10.1.4; the portion of the curve measured in this problem is in bold. ^†^†margin: Figure 10.1.4: A graph of $f(x)=\frac{1}{8}x^{2}-\ln x$ from Example 10.1.2. This function was chosen specifically because the resulting integral can be evaluated exactly. We begin by finding $f\,^{\prime}(x)=x/4-1/x$ . The arc length is

	$\displaystyle L$	$\displaystyle=\int_{1}^{2}\sqrt{1+\left(\frac{x}{4}-\frac{1}{x}\right)^{2}}\ dx$
		$\displaystyle=\int_{1}^{2}\sqrt{1+\frac{x^{2}}{16}-\frac{1}{2}+\frac{1}{x^{2}}% }\ dx$
		$\displaystyle=\int_{1}^{2}\sqrt{\frac{x^{2}}{16}+\frac{1}{2}+\frac{1}{x^{2}}}% \ dx$
		$\displaystyle=\int_{1}^{2}\sqrt{\left(\frac{x}{4}+\frac{1}{x}\right)^{2}}\ dx$
		$\displaystyle=\int_{1}^{2}\left(\frac{x}{4}+\frac{1}{x}\right)\ dx$
		$\displaystyle=\left(\frac{x^{2}}{8}+\ln x\right)\Bigg{\|}_{1}^{2}$
		$\displaystyle=\frac{3}{8}+\ln 2\ \text{units}.$

The previous examples found the arc length exactly through careful choice of the functions. In general, exact answers are much more difficult to come by and numerical approximations are necessary.

Example 10.1.3 Approximating arc length numerically

Find the length of the sine curve from $x=0$ to $x=\pi$ .

SolutionThis is somewhat of a mathematical curiosity; in Example 5.4.3 we found the area under one “hump” of the sine curve is 2 square units; now we are measuring its arc length.

^†^†margin:

\begin{array}[]{cc}x&\sqrt{1+\cos^{2}x}\\ \hline 0&\sqrt{2}\\ \pi/4&\sqrt{3/2}\\ \pi/2&1\\ 3\pi/4&\sqrt{3/2}\\ \pi&\sqrt{2}\\ \end{array}

Figure 10.1.5: A table of values of

y=\sqrt{1+\cos^{2}x}

to evaluate a definite integral in Example 10.1.3.

The setup is straightforward: $f(x)=\sin x$ and $f\,^{\prime}(x)=\cos x$ . Thus

L=\int_{0}^{\pi}\sqrt{1+\cos^{2}x}\ dx.

This integral cannot be evaluated in terms of elementary functions so we will approximate it with Simpson’s Method with $n=4$ . Figure 10.1.5 gives $\sqrt{1+\cos^{2}x}$ evaluated at 5 evenly spaced points in $[0,\pi]$ . Simpson’s Rule then states that

	$\displaystyle\int_{0}^{\pi}\sqrt{1+\cos^{2}x}\ dx$	$\displaystyle\approx\frac{\pi-0}{4\cdot 3}\left(\sqrt{2}+4\sqrt{3/2}+2(1)+4% \sqrt{3/2}+\sqrt{2}\right)$
		$\displaystyle\approx 3.82918.$

Using a computer with $n=100$ the approximation is $L\approx 3.8202$ ; our approximation with $n=4$ is quite good. Our approximation of $3.79$ from the beginning of this section isn’t as close.

Surface Area of Solids of Revolution

^†^†margin:

(a)

(b) Figure 10.1.6: Establishing the formula for surface area.

We have already seen how a curve $y=f(x)$ on $[a,b]$ can be revolved around an axis to form a solid. Instead of computing its volume, we now consider its surface area.

We begin as we have in the previous sections: we partition the interval $[a,b]$ with $n$ subintervals, where the $i\,^{\text{th}}$ subinterval is $[x_{i},x_{i+1}]$ . On each subinterval, we can approximate the curve $y=f(x)$ with a straight line that connects $f(x_{i})$ and $f(x_{i+1})$ as shown in Figure 10.1.6(a). Revolving this line segment about the $x$ -axis creates part of a cone (called a frustum of a cone) as shown in Figure 10.1.6(b). The surface area of a frustum of a cone is

A=2\pi r_{\text{avg}}L,

where $r_{\text{avg}}$ is the average of $R_{1}$ and $R_{2}$ . The length is given by $L$ ; we use the material just covered by arc length to state that

L\approx\sqrt{1+[f\,^{\prime}(c_{i})]^{2}}\Delta x_{i}

for some $c_{i}$ in the $i\,^{\text{th}}$ subinterval. The radii are just the function evaluated at the endpoints of the interval: $f(x_{i-1})$ and $f(x_{i})$ . Thus the surface area of this sample frustum of the cone is approximately

2\pi\frac{f(x_{i-1})+f(x_{i})}{2}\sqrt{1+[f\,^{\prime}(c_{i})]^{2}}\Delta x_{i}.

Since $f$ is a continuous function, the Intermediate Value Theorem states there is some $d_{i}$ in $[x_{i-1},x_{i}]$ such that $f(d_{i})=\dfrac{f(x_{i-1})+f(x_{i})}{2}$ ; we can use this to rewrite the above equation as

2\pi f(d_{i})\sqrt{1+[f\,^{\prime}(c_{i})]^{2}}\Delta x_{i}.

Summing over all the subintervals we get the total surface area to be approximately

\text{Surface Area}\approx\sum_{i=1}^{n}2\pi f(d_{i})\sqrt{1+[f\,^{\prime}(c_{% i})]^{2}}\Delta x_{i},

which is a Riemann Sum. Taking the limit as the subinterval lengths go to zero gives us the exact surface area, given in the upcoming Key Idea.

If instead we revolve $y=f(x)$ about the $y$ -axis, the radii of the resulting frustum are $x_{i-1}$ and $x_{i}$ ; their average value is simply the midpoint of the interval. In the limit, this midpoint is just $x$ . This gives the second part of Key Idea 10.1.2.

Key Idea 10.1.2 Surface Area of a Solid of Revolution

Let $f$ be differentiable on an open interval containing $[a,b]$ where $f\,^{\prime}$ is also continuous on $[a,b]$ .

1.

The surface area of the solid formed by revolving the graph of $y=f(x)$ , where $f(x)\geq 0$ , about the $x$ -axis is

$\text{Surface Area}=2\pi\int_{a}^{b}f(x)\sqrt{1+[f\,^{\prime}(x)]^{2}}\ dx.$
2.

The surface area of the solid formed by revolving the graph of $y=f(x)$ about the $y$ -axis, where $a,b\geq 0$ , is

$\text{Surface Area}=2\pi\int_{a}^{b}x\sqrt{1+[f\,^{\prime}(x)]^{2}}\ dx.$

Example 10.1.4 Finding surface area of a solid of revolution

Find the surface area of the solid formed by revolving $y=\sin x$ on $[0,\pi]$ around the $x$ -axis, as shown in Figure 10.1.7.

SolutionThe setup turns out to be easier than the resulting integral. Using Key Idea 10.1.2, we have the surface area $S A$ is: ^†^†margin: Figure 10.1.7: Revolving $y=\sin x$ on $[0,\pi]$ about the $x$ -axis.

	SA	$\displaystyle=2\pi\int_{0}^{\pi}\sin x\sqrt{1+\cos^{2}x}\ dx$
		$\displaystyle=-2\pi\int_{1}^{-1}\sqrt{1+u^{2}}\ du\qquad\qquad\text{substitute% $u=\cos x$}$
		$\displaystyle=2\pi\int_{-\pi/4}^{\pi/4}\sec^{3}\theta\ d\theta\qquad\qquad% \text{substitute $u=\tan\theta$}$
		$\displaystyle=\left.\pi\left(\sec\theta\tan\theta+\ln\left\lvert\sec\theta+% \tan\theta\right\rvert\right)\right\|_{-\pi/4}^{\pi/4}\qquad\text{by \autoref{% ex_trigint6}}$
		$\displaystyle=\pi\left(\sqrt{2}+\ln\left(\sqrt{2}+1\right)-\left(-\sqrt{2}+\ln% \left(\sqrt{2}-1\right)\right)\right)$
		$\displaystyle=\pi\left(2\sqrt{2}+\ln\left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right)\right)$
		$\displaystyle=2\pi\left(\sqrt{2}+\ln\left(\sqrt{2}+1\right)\right)\ \text{% units}^{2}\qquad\qquad\text{rationalize the denominator.}$

It is interesting to see that the surface area of a solid, whose shape is defined by a trigonometric function, involves both a square root and a natural logarithm.

^†^†margin:

(a)

(b) Figure 10.1.8: The solids used in Example 10.1.5.

Example 10.1.5 Finding surface area of a solid of revolution

Find the surface area of the solid formed by revolving the curve $y=x^{2}$ on $[0,1]$ about:
1. the $x$ -axis 2. the $y$ -axis.

Solution

1.

The solid formed by revolving $y=x^{2}$ around the $x$ -axis is graphed in Figure 10.1.8(a). Like the integral in Example 10.1.4, this integral is easier to setup than to actually integrate. While it is possible to use a trigonometric substitution to evaluate this integral, it is significantly more difficult than a solution employing the hyperbolic sine:

$\displaystyle SA$ $\displaystyle=2\pi\int_{0}^{1}x^{2}\sqrt{1+(2x)^{2}}\ dx.$

$\displaystyle=\left.\frac{\pi}{32}\left(2(8x^{3}+x)\sqrt{1+4x^{2}}-\sinh^{-1}(% 2x)\right)\right|_{0}^{1}$

$\displaystyle=\frac{\pi}{32}\left(18\sqrt{5}-\sinh^{-1}2\right)\ \text{units}^% {2}.$
2.

Since we are revolving around the $y$ -axis, the “radius” of the solid is not $f(x)$ but rather $x$ . Thus the integral to compute the surface area is:

$\displaystyle SA$ $\displaystyle=2\pi\int_{0}^{1}x\sqrt{1+(2x)^{2}}\ dx$

$\displaystyle=\frac{\pi}{4}\int_{1}^{5}\sqrt{u}\ du\qquad\text{substitute $u=1% +4x^{2}$}$

$\displaystyle=\left.\frac{\pi}{4}\frac{2}{3}u^{3/2}\right|_{1}^{5}$

$\displaystyle=\frac{\pi}{6}\left(5\sqrt{5}-1\right)\ \text{units}^{2}.$

The solid formed by revolving $y=x^{2}$ about the $y$ -axis is graphed in Figure 10.1.8 (b).

Our final example is a famous mathematical “paradox.”

Example 10.1.6 The surface area and volume of Gabriel’s Horn

Consider the solid formed by revolving $y=1/x$ about the $x$ -axis on $[1,\infty)$ . Find the volume and surface area of this solid. (This shape, as graphed in Figure 10.1.9, is known as “Gabriel’s Horn” since it looks like a very long horn that only a supernatural person, such as an angel, could play.) ^†^†margin: Figure 10.1.9: A graph of Gabriel’s Horn.

SolutionTo compute the volume it is natural to use the Disk Method. We have:

	$\displaystyle V$	$\displaystyle=\pi\int_{1}^{\infty}\frac{1}{x^{2}}\ dx$
		$\displaystyle=\lim_{b\to\infty}\pi\int_{1}^{b}\frac{1}{x^{2}}\ dx$
		$\displaystyle=\lim_{b\to\infty}\left.\pi\left(\frac{-1}{x}\right)\right\|_{1}^{b}$
		$\displaystyle=\lim_{b\to\infty}\pi\left(1-\frac{1}{b}\right)$
		$\displaystyle=\pi\ \text{units}^{3}.$

Gabriel’s Horn has a finite volume of $\pi$ cubic units. Since we have already seen that regions with infinite length can have a finite area, this is not too difficult to accept.

We now consider its surface area. The integral is straightforward to setup:

	$\displaystyle SA$	$\displaystyle=2\pi\int_{1}^{\infty}\frac{1}{x}\sqrt{1+1/x^{4}}\ dx.$
Integrating this expression is not trivial. We can, however, compare it to other improper integrals. Since $1<\sqrt{1+1/x^{4}}$ on $[1,\infty)$ , we can state that
	$\displaystyle 2\pi\int_{1}^{\infty}\frac{1}{x}\ dx$	$\displaystyle<2\pi\int_{1}^{\infty}\frac{1}{x}\sqrt{1+1/x^{4}}\ dx.$

By Key Idea 8.6.1, the improper integral on the left diverges. Since the integral on the right is larger, we conclude it also diverges, meaning Gabriel’s Horn has infinite surface area.

Hence the “paradox”: we can fill Gabriel’s Horn with a finite amount of paint, but since it has infinite surface area, we can never paint it.

Somehow this paradox is striking when we think about it in terms of volume and area. However, we have seen a similar paradox before, as referenced above. We know that the area under the curve $y=1/x^{2}$ on $[1,\infty)$ is finite, yet the shape has an infinite perimeter. Strange things can occur when we deal with the infinite.

Exercises 10.1

Terms and Concepts

1.

T/F: The integral formula for computing Arc Length was found by first approximating arc length with straight line segments.
2.

T/F: The integral formula for computing Arc Length includes a square-root, meaning the integration is probably easy.

Problems

In Exercises 3–12, find the arc length of the function on the given interval.

3.

$\displaystyle f(x)=x$ on $[0,1]$ .
4.

$\displaystyle f(x)=\sqrt{8}x$ on $[-1,1]$ .
5.

$\displaystyle f(x)=\frac{1}{3}x^{3/2}-x^{1/2}$ on $[0,1]$ .
6.

$\displaystyle f(x)=\frac{1}{12}x^{3}+\frac{1}{x}$ on $[1,4]$ .
7.

$\displaystyle f(x)=2x^{3/2}-\frac{1}{6}\sqrt{x}$ on $[0,9]$ .
8.

$\displaystyle f(x)=\cosh x$ on $[-\ln 2,\ln 2]$ .
9.

$\displaystyle f(x)=\frac{1}{2}\bigl{(}e^{x}+e^{-x}\bigr{)}$ on $[0,\ln 5]$ .
10.

$\displaystyle f(x)=\frac{1}{12}x^{5}+\frac{1}{5x^{3}}$ on $[.1,1]$ .
11.

$\displaystyle f(x)=\ln\bigl{(}\sin x\bigr{)}$ on $[\pi/6,\pi/2]$ .
12.

$\displaystyle f(x)=\ln\bigl{(}\cos x\bigr{)}$ on $[0,\pi/4]$ .

In Exercises 13–20, set up the integral to compute the arc length of the function on the given interval. Do not evaluate the integral.

13.

$\displaystyle f(x)=x^{2}$ on $[0,1]$ .
14.

$\displaystyle f(x)=x^{10}$ on $[0,1]$ .
15.

$\displaystyle f(x)=\sqrt{x}$ on $[0,1]$ .
16.

$\displaystyle f(x)=\ln x$ on $[1,e]$ .
17.

$\displaystyle f(x)=\sqrt{1-x^{2}}$ on $[-1,1]$ . (Note: this describes the top half of a circle with radius 1.)
18.

$\displaystyle f(x)=\sqrt{1-x^{2}/9}$ on $[-3,3]$ . (Note: this describes the top half of an ellipse with a major axis of length 6 and a minor axis of length 2.)
19.

$\displaystyle f(x)=\frac{1}{x}$ on $[1,2]$ .
20.

$\displaystyle f(x)=\sec x$ on $[-\pi/4,\pi/4]$ .

In Exercises 21–28, use Simpson’s Rule, with $n=4$ , to approximate the arc length of the function on the given interval. Note: these are the same problems as in Exercises 13–20.

21.

$\displaystyle f(x)=x^{2}$ on $[0,1]$ .
22.

$\displaystyle f(x)=x^{10}$ on $[0,1]$ .
23.

$\displaystyle f(x)=\sqrt{x}$ on $[0,1]$ . (Note: $f\,^{\prime}(x)$ is not defined at $x=0$ .)
24.

$\displaystyle f(x)=\ln x$ on $[1,e]$ .
25.

$\displaystyle f(x)=\sqrt{1-x^{2}}$ on $[-1,1]$ . (Note: $f\,^{\prime}(x)$ is not defined at the endpoints.)
26.

$\displaystyle f(x)=\sqrt{1-x^{2}/9}$ on $[-3,3]$ . (Note: $f\,^{\prime}(x)$ is not defined at the endpoints.)
27.

$\displaystyle f(x)=\frac{1}{x}$ on $[1,2]$ .
28.

$\displaystyle f(x)=\sec x$ on $[-\pi/4,\pi/4]$ .

In Exercises 29–32, find the surface area of the described solid of revolution.

29.

The solid formed by revolving $y=2x$ on $[0,1]$ about the $x$ -axis.
30.

The solid formed by revolving $y=x^{3}$ on $[0,1]$ about the $x$ -axis.
31.

The solid formed by revolving $y=\sqrt{x}$ on $[0,1]$ about the $x$ -axis.
32.

The sphere formed by revolving $y=\sqrt{1-x^{2}}$ on $[-1,1]$ about the $x$ -axis.

	$\displaystyle SA$	$\displaystyle=2\pi\int_{0}^{1}x^{2}\sqrt{1+(2x)^{2}}\ dx.$
		$\displaystyle=\left.\frac{\pi}{32}\left(2(8x^{3}+x)\sqrt{1+4x^{2}}-\sinh^{-1}(% 2x)\right)\right\|_{0}^{1}$
		$\displaystyle=\frac{\pi}{32}\left(18\sqrt{5}-\sinh^{-1}2\right)\ \text{units}^% {2}.$

	$\displaystyle SA$	$\displaystyle=2\pi\int_{0}^{1}x\sqrt{1+(2x)^{2}}\ dx$
		$\displaystyle=\frac{\pi}{4}\int_{1}^{5}\sqrt{u}\ du\qquad\text{substitute $u=1% +4x^{2}$}$
		$\displaystyle=\left.\frac{\pi}{4}\frac{2}{3}u^{3/2}\right\|_{1}^{5}$
		$\displaystyle=\frac{\pi}{6}\left(5\sqrt{5}-1\right)\ \text{units}^{2}.$