# Topic 9 Further Applications of Integration

## 9.1 Arc length

Suppose \(f\) is a function such that \(f'\) is continuous. Locally, that is, over a small interval of \(x\), the curve defined by \(y=f(x)\) is approximately a straight line segment. The length of the small line segment is \[ \sqrt{(\mathrm{d}x)^2+(\mathrm{d}y)^2}=\sqrt{1+(f'(x))^2} \mathrm{d} x. \]

*Remark*. The equality can be obtained using the Mean Value Theorem.

Using this approximation, the length of the curve defined between \((x_a, y_a)\) and \(x_b, y_b\) can be calculated as the definite integral \[ \int_{x_a}^{x_b}\sqrt{1+(f'(x))^2} \mathrm{d} x. \]

*Remark*. If the curve is define by \(x=g(y)\), we may switch \(x\) and \(y\). The length of the curve is
\[
\int_{y_a}^{y_b}\sqrt{1+(g\prime(y))^2} \mathrm{d} y.
\]

If the curve is parametrized by \(t\), that is a point on the curve is given by \((x(t), y(t))\), where \(x(t)\) and \(y(t)\) are functions of \(t\), then \[ \sqrt{(\mathrm{d}x)^2+(\mathrm{d}y)^2}= \sqrt{(\frac{\mathrm{d}x}{\mathrm{d}t})^2+(\frac{\mathrm{d}y}{\mathrm{d}t})^2}\mathrm{d} t=\sqrt{(x'(t))^2+(y'(t))^2}\mathrm{d} t \] and the length of the curve parametrized by \(t\) in \([a, b]\) is \[ \int_a^b \sqrt{(x'(t))^2+(y'(t))^2}\mathrm{d} t. \]

If we fixed an initial point \((a, b)\) on the curve \(y=f(x)\), then the arc length of the curve from \((a, b)\) to \((x, y)\) defines a function \[ s(x)=\int_a^x \sqrt{1+(f'(t))^2} \mathrm{d} t \] which is called the arc length function.

The differential \(\mathrm{d} s=\sqrt{1+(f'(x))^2} \mathrm{d} x\) will be used to calculate the surface area of the revolution of a curve.

**Exercise 9.1 **Find the exact length of the curve
\[
y^{2}=4(x+4)^{3}, \quad 0 \le x \le 2, \quad y>0
\]

*Solution*.

Since \(y>0\), solve for \(y\), we get \(y=2(x+4)^{\frac32}\).

Then \[ \frac{\mathrm{d} y}{\mathrm{d} x}=3(x+4)^{\frac12} \]

The arc length of the curve is given by the integral \[ \begin{aligned} &\int_0^2 \sqrt{1+\left(\frac{\mathrm{d} y}{\mathrm{d} x}\right)^2}\mathrm{d} x\\ =&\int_0^2 \sqrt{1+9(x+4)}\mathrm{d} x\\ =&\int_0^2 \sqrt{9x+37}\mathrm{d} x\\ =&\frac{2}{27}(9x+37)^{\frac32}\big|_0^2\\ =&\frac{2}{27}(55\sqrt{55}-37\sqrt{37}). \end{aligned} \]

**Exercise 9.2 **Find the exact length of the curve
\[
x=\frac{y^{4}}{8}+\frac{1}{4 y^{2}}, \quad 1 \le y \le 2
\]

*Solution*.

It is easier to find \(\frac{\mathrm{d} x}{\mathrm{d} y}\) which is \[ \frac{\mathrm{d} x}{\mathrm{d} y}=\frac{y^3}{2}-\frac{y^{-3}}{2}. \]

The arc length of the curve is given by the integral \[ \begin{aligned} &\int_1^2 \sqrt{1+\left(\frac{\mathrm{d} x}{\mathrm{d} y}\right)^2}\mathrm{d} y\\ =&\int_1^2 \sqrt{1+\left(\frac{y^3}{2}-\frac{y^{-3}}{2}\right)^2}\mathrm{d} x\\ =&\int_1^2\sqrt{\left(\frac{y^3}{2}+\frac{y^{-3}}{2}\right)^2}\mathrm{d} x\\ =&\int_1^2\left(\frac{y^3}{2}+\frac{y^{-3}}{2}\right)\mathrm{d} x\\ =&\left(\frac{y^4}{8}-\frac{y^{-2}}{4}\right)\bigg|_1^2\\ =&\frac{33}{16}. \end{aligned} \]

**Exercise 9.3 **Find the exact length of the curve
\[
y=\ln (\cos x), \quad 0 \le x \le \pi / 3
\]

*Solution*.

We first find the derivative \[ \frac{\mathrm{d} y}{\mathrm{d} x}=-\frac{\sin x}{\cos x}=-\tan x. \]

The arc length of the curve is given by the integral \[ \begin{aligned} &\int_0^{\pi/3} \sqrt{1+\left(\frac{\mathrm{d} y}{\mathrm{d} x}\right)^2}\mathrm{d} x\\ =&\int_0^{\pi/3} \sqrt{1+\left(-\tan x\right)^2}\mathrm{d} x\\ =&\int_0^{\pi/3} \sec x \mathrm{d} x\\ =&\ln(\sec x+\tan x)\big|_0^{\pi/3}\\ % =&\ln\left(\sec\frac{\pi}{3}+\tan\frac{\pi}{3}\right)-\ln(\sec0+\tan0)\\ =&\ln(2+\sqrt{3}) \end{aligned} \]

**Exercise 9.4 **Find the exact length of the curve
\[
y=\frac{1}{4} x^{2}-\frac{1}{2} \ln x, \quad 1 \le x \le 2
\]

*Solution*.

First find the derivative \[ \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{x}{2}-\frac{x^{-1}}{2}. \]

The arc length of the curve is given by the integral \[ \begin{aligned} &\int_1^2 \sqrt{1+\left(\frac{\mathrm{d} y}{\mathrm{d} x}\right)^2}\mathrm{d} x\\ =&\int_1^2 \sqrt{1+\left(\frac{x}{2}-\frac{1}{2x}\right)^2}\mathrm{d} x\\ =&\int_1^2 \frac{x}{2}+\frac{1}{2x} \mathrm{d} x\\ =&\left(\frac{x^2}{4}+\frac12\ln x\right)\bigg|_1^2\\ =&\frac34+\frac12\ln 2 \end{aligned} \]

**Exercise 9.5 **Use trigonometric substitution to parametrize the the curve
\[
x^{2/3}+y^{2/3}=1
\]
and find its length.

*Solution*.

The curve can be parametrized by \(x=\sin^3t\) and \(y=\cos^3t\) with \(0\le t\le 2\pi\).

By the symmetries of the sine and cosine functions, the curve has four pieces with equal length.

We will find the arc length of the piece with \(0\le t \le \pi/2\).

For \(t \in [0,\pi/2]\), we have \[ \begin{aligned} &\sqrt{(\mathrm{d} x)^2 + (\mathrm{d} y)^2}\\ =&\sqrt{(3\sin^2t\cos t)^2+(-3\cos^2t\sin t)^2} \mathrm{d} t\\ =&3\sin t\cos t \mathrm{d} t\\ =&\frac32\sin2t\mathrm{d}t. \end{aligned} \]

The arc length of the curve for \(0\le t \le \pi/2\) is given by the integral \[ \begin{aligned} &\int_0^{\pi/2} \frac32\sin2t\mathrm{d}t\\ =&-\frac34\cos2t|_0^{\pi/2}\\ =&\frac32. \end{aligned} \]

So the total length of the curve is \(4\cdot \frac32=6\).

## 9.2 Area of a surface of revolution

When rotate a curve along an axis, locally, the surface generated by a small piece of the curve is approximately a band of a cone. Suppose the radius (roughly speaking the distance from a point in the piece of curve to the rotation axis) is \(r\). Then the area of the band is approximately \[ 2\pi r\mathrm{d} s \] where \(s\) is the arc length function.

Using integral, the area of the surface of revolution can be calculated by \[ \int 2\pi r\mathrm{d} s. \]

The explicit form of \(r\) and \(\mathrm{d} s\) may be expressed according to the rotation axis.

**Exercise 9.6 **Find the exact area of the surface obtained by rotating the curve about the \(x\)-axis.
\[
9 x=y^{2}+18, \quad 2 \le x \le 6
\]

*Solution*.

Since the rotation axis is the \(x\)-axis, the radius of a band on the surface is parametrized by \(y\). An arc above an interval near \((x, y)\) is about \[ \begin{aligned} &\sqrt{1+\left(\frac{\mathrm{d} x}{\mathrm{d} y}\right)^2}\\ =&\sqrt{1+\left(\frac29 y\right)^2} \mathrm{d} y\\ =&\frac19\sqrt{81+4y^2}\mathrm{d} y \end{aligned} \]

As the curve \(9x=y^2+18\) is symmetric with respect to the \(x\)-axis, the surface of revolution can be obtained by only the upper half part of the whole curve, that is, we may assume that \(y\ge 0\).

Solve for \(y\) from the inequality \(9\cdot 2\le y^2+18\le 9\cdot 6\) together with \(y\ge 0\), we get \(0\le y\le 6\)

Then, the are of the surface of revolution is given by \[ \begin{aligned} &2\pi \int_0^6 \frac19y\sqrt{81+4y^2}\mathrm{d} y\\ =&\frac{2\pi}{9}\int_0^6y\sqrt{81+4y^2}\mathrm{d} y\\ =&\frac{2\pi}{72}\int_0^6\sqrt{81+4y^2}\mathrm{d}(81+4y^2)\\ =&\frac{\pi}{54}(81+4y^2)^{\frac32}\big|_0^6\\ =&49\pi. \end{aligned} \]

**Exercise 9.7 **Find the exact area of the surface obtained by rotating the curve about the \(x\)-axis.
\[
y=\sin(\pi x), \quad 0 \le x \le 1
\]

*Solution*.

Since the rotation axis is the \(x\)-axis, the radius of a band on the surface is \(y\).

An arc above an interval near \((x, y)\) is about \[ \begin{aligned} &\sqrt{1+\left(\frac{\mathrm{d} y}{\mathrm{d} x}\right)^2}\\ =&\sqrt{1+\left(\pi\cos(\pi x) \right)^2} \mathrm{d} x \end{aligned} \]

Then, the area of the surface of revolution is given by \[ 2\pi \int_0^1 \sin(\pi x)\sqrt{1+\left(\pi\cos(\pi x) \right)^2} \mathrm{d} x. \]

We first find the indefinite integral \[ \begin{aligned} &\int \sin(\pi x)\sqrt{1+\left(\pi\cos(\pi x) \right)^2} \mathrm{d} x\\ =&\int \sqrt{1+u^2} ~ \left(-\frac{\mathrm{d} u}{\pi^2}\right) \qquad u=\pi\cos(\pi x)\\ =&-\frac{1}{\pi^2}\int \sec^3t \mathrm{d}t \qquad u=\tan t\\ =&-\frac{1}{\pi^2}\left(\frac{\sec t \tan t}{2}+\frac{1}{2} \ln |\tan t+\sec t|\right)+C \qquad \text{by integration by parts}\\ =&-\frac{1}{2\pi^2}\left(u\sqrt{1+u^2} + \ln(u+\sqrt{1+u^2})\right)+C \qquad \text{by integration by parts}\\ =&-\frac{1}{2\pi^2}\left(\pi\cos(\pi x)\sqrt{1+\pi^2\cos^2(\pi x)} + \ln\left(\pi\cos(\pi x)+\sqrt{1+\pi^2\cos^2(\pi x)}\right)\right)+C\\ \end{aligned} \]

By FTC, \[ \begin{aligned} & 2\pi \int_0^1 \sin(\pi x)\sqrt{1+\left(\pi\cos(\pi x) \right)^2} \mathrm{d} x\\ =& -\frac{1}{\pi}\left(\pi\cos(\pi x)\sqrt{1+\pi^2\cos^2(\pi x)} + \ln\left(\pi\cos(\pi x)+\sqrt{1+\pi^2\cos^2(\pi x)}\right)\right)\bigg|_0^1\\ =&2\sqrt{1+\pi^{2}}+\frac{\ln \left(2 \pi^{2}+2 \pi \sqrt{1+\pi^{2}}+1\right)}{\pi} \end{aligned} \]

**Exercise 9.8 **Find the exact area of the surface obtained by rotating the curve about the \(y\)-axis.
\[
y=\sqrt[3]{x}, \quad 1 \le y \le 2
\]

*Solution*.

The radius is given by \(x=y^3\). The derivative is \(\frac{\mathrm{d} x}{\mathrm{d} y}=3y^2\).

Then the area of the surface of revolution is given by \[ \begin{aligned} &2\pi \int_1^2 y^3\sqrt{1+\left(3y^2\right)^2} \mathrm{d} y\\ =&2\pi \int_{10}^{145} \sqrt{u}~\mathrm{d}\left(\frac{u}{36}\right) \qquad u=1+9y^4\\ =&\frac{\pi}{18}\left(\frac23u^{\frac32}\right)\big|_{10}^{145}\\ =&\frac{\pi}{27}(145\sqrt{145}-10\sqrt{10}). \end{aligned} \]

**Exercise 9.9 **Find the exact area of the surface obtained by rotating the curve about the \(y\)-axis.

\[ y=\frac{1}{4} x^{2}-\frac{1}{2} \ln x, \quad 1 \le x \le 2 \]

*Solution*.

The radius is \(x=y^3\). The derivative is \[ \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{x}{2}-\frac{1}{2x}. \]

Then the area of the surface of revolution is given by \[ \begin{aligned} & 2\pi \int_1^2 x\sqrt{1+\left(\frac{x}{2}-\frac{1}{2x}\right)^2} \mathrm{d} y\\ =& 2\pi \int_1^2 x\left(\frac{x}{2}+\frac{1}{2x}\right)\mathrm{d} x\\ =& 2\pi\left(\frac{x^3}{6}+\frac{x}{2}\right)\bigg|_1^2\\ =& \frac{10\pi}{3}. \end{aligned} \]

**Exercise 9.10 **Find the area of the surface obtained by rotating the circle \(x^{2}+y^{2}=r^{2}\) about the line \(x=r\).

*Solution*.

We parametrize the curve by \(x=r\cos t\) and \(y=r\sin t\) with \(0\le t\le 2\pi\).

Since the rotation axis is \(x=r\), the radius is \[ |x-r|=r-x=r-r\cos t. \]

The length of a small arc is \[ \sqrt{(\mathrm{d} x)^2+(\mathrm{d} y)^2}=r\mathrm{d} t. \]

Then the area of the surface is \[ \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{x}{2}-\frac{1}{2x}. \]

Then the area of the surface of revolution is given by \[ \begin{aligned} &2\pi \int_0^{2\pi}(r-r\cos t) r\mathrm{d} t\\ =& 2\pi r^2(t-\sin t)|_0^{2\pi}\\ =&4\pi^2 r^2. \end{aligned} \]