# 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. Graph of the curve $$x^{2/3}+y^{2/3}=1$$

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. $y=\sqrt{x}, \quad 1 \le y \le 2$

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}