Topic 9 Further Applications of Integration

9.1 Arc length

Suppose $f$ is a function such that $f^{\prime }$ is continuous. Locally, that is, over a small interval of $x$, the curve defined by $y=f\left(x\right)$ is approximately a straight line segment. The length of the small line segment is $$\sqrt{(dx{)}^2+(dy{)}^2}=\sqrt{1+\left(f^{\prime }\right(x){)}^2}dx.$$

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 $${\int }_{x_a}^{x_b}\sqrt{1+\left(f^{\prime }\right(x){)}^2}dx.$$

Remark. If the curve is define by $x=g\left(y\right)$, we may switch $x$ and $y$. The length of the curve is $${\int }_{y_a}^{y_b}\sqrt{1+\left(g\prime \right(y){)}^2}dy.$$

If the curve is parametrized by $t$, that is a point on the curve is given by $\left(x\right(t),y(t\left)\right)$, where $x\left(t\right)$ and $y\left(t\right)$ are functions of $t$, then $$\sqrt{(dx{)}^2+(dy{)}^2}=\sqrt{(\frac{dx}{dt}{)}^2+(\frac{dy}{dt}{)}^2}dt=\sqrt{\left(x^{\prime }\right(t){)}^2+(y^{\prime }\left(t\right){)}^2}dt$$ and the length of the curve parametrized by $t$ in $[a,b]$ is $${\int }_a^b\sqrt{\left(x^{\prime }\right(t){)}^2+(y^{\prime }\left(t\right){)}^2}dt.$$

If we fixed an initial point $(a,b)$ on the curve $y=f\left(x\right)$, then the arc length of the curve from $(a,b)$ to $(x,y)$ defines a function $$s\left(x\right)={\int }_a^x\sqrt{1+\left(f^{\prime }\right(t){)}^2}dt$$ which is called the arc length function.

The differential $ds=\sqrt{1+\left(f^{\prime }\right(x){)}^2}dx$ will be used to calculate the surface area of the revolution of a curve.

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

Solution.

Since $y>0$, solve for $y$, we get $y=2(x+4{)}^{\frac{3}{2}}$.

Then $$\frac{dy}{dx}=3(x+4{)}^{\frac{1}{2}}$$

The arc length of the curve is given by the integral $$\begin{array}{cc}& {\int }_0^2\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}dx\\ =& {\int }_0^2\sqrt{1+9(x+4)}dx\\ =& {\int }_0^2\sqrt{9x+37}dx\\ =& \frac{2}{27}(9x+37{)}^{\frac{3}{2}}{|}_0^2\\ =& \frac{2}{27}(55\sqrt{55}-37\sqrt{37}).\end{array}$$

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

Solution.

It is easier to find $\frac{dx}{dy}$ which is $$\frac{dx}{dy}=\frac{y^3}{2}-\frac{y^{-3}}{2}.$$

The arc length of the curve is given by the integral $$\begin{array}{cc}& {\int }_1^2\sqrt{1+{\left(\frac{dx}{dy}\right)}^2}dy\\ =& {\int }_1^2\sqrt{1+{(\frac{y^3}{2}-\frac{y^{-3}}{2})}^2}dx\\ =& {\int }_1^2\sqrt{{(\frac{y^3}{2}+\frac{y^{-3}}{2})}^2}dx\\ =& {\int }_1^2(\frac{y^3}{2}+\frac{y^{-3}}{2})dx\\ =& (\frac{y^4}{8}-\frac{y^{-2}}{4}){|}_1^2\\ =& \frac{33}{16}.\end{array}$$

Exercise 9.3 Find the exact length of the curve $$y=\ln \left(\cos x\right),0\le x\le \pi /3$$

Solution.

We first find the derivative $$\frac{dy}{dx}=-\frac{\sin x}{\cos x}=-\tan x.$$

The arc length of the curve is given by the integral $$\begin{array}{cc}& {\int }_0^{\pi /3}\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}dx\\ =& {\int }_0^{\pi /3}\sqrt{1+{(-\tan x)}^2}dx\\ =& {\int }_0^{\pi /3}\sec xdx\\ =& \ln (\sec x+\tan x){|}_0^{\pi /3}\\ =& \ln (2+\sqrt{3})\end{array}$$

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

Solution.

First find the derivative $$\frac{dy}{dx}=\frac{x}{2}-\frac{x^{-1}}{2}.$$

The arc length of the curve is given by the integral $$\begin{array}{cc}& {\int }_1^2\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}dx\\ =& {\int }_1^2\sqrt{1+{(\frac{x}{2}-\frac{1}{2x})}^2}dx\\ =& {\int }_1^2\frac{x}{2}+\frac{1}{2x}dx\\ =& (\frac{x^2}{4}+\frac{1}{2}\ln x){|}_1^2\\ =& \frac{3}{4}+\frac{1}{2}\ln 2\end{array}$$

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 $x={\sin }^{3}t$ and $y={\cos }^{3}t$ 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 $$\begin{array}{cc}& \sqrt{(dx{)}^2+(dy{)}^2}\\ =& \sqrt{(3{\sin }^{2}t\cos t{)}^2+(-3{\cos }^{2}t\sin t{)}^2}dt\\ =& 3\sin t\cos tdt\\ =& \frac{3}{2}\sin 2tdt.\end{array}$$

The arc length of the curve for $0\le t\le \pi /2$ is given by the integral $$\begin{array}{cc}& {\int }_0^{\pi /2}\frac{3}{2}\sin 2tdt\\ =& -\frac{3}{4}\cos 2t{|}_0^{\pi /2}\\ =& \frac{3}{2}.\end{array}$$

So the total length of the curve is $4\cdot \frac{3}{2}=6$.

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 $$2\pi rds$$ where $s$ is the arc length function.

Using integral, the area of the surface of revolution can be calculated by $$\int 2\pi rds.$$

The explicit form of $r$ and $ds$ 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. $$9x=y^2+18,2\le x\le 6$$

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 $$\begin{array}{cc}& \sqrt{1+{\left(\frac{dx}{dy}\right)}^2}\\ =& \sqrt{1+{\left(\frac{2}{9}y\right)}^2}dy\\ =& \frac{1}{9}\sqrt{81+4y^2}dy\end{array}$$

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 $$\begin{array}{cc}& 2\pi {\int }_0^6\frac{1}{9}y\sqrt{81+4y^2}dy\\ =& \frac{2\pi }{9}{\int }_0^{6}y\sqrt{81+4y^2}dy\\ =& \frac{2\pi }{72}{\int }_0^6\sqrt{81+4y^2}d(81+4y^2)\\ =& \frac{\pi }{54}(81+4y^2{)}^{\frac{3}{2}}{|}_0^6\\ =& 49\pi .\end{array}$$

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

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 $$\begin{array}{cc}& \sqrt{1+{\left(\frac{dy}{dx}\right)}^2}\\ =& \sqrt{1+{\left(\pi \cos \right(\pi x\left)\right)}^2}dx\end{array}$$

Then, the area of the surface of revolution is given by $$2\pi {\int }_0^1\sin \left(\pi x\right)\sqrt{1+{\left(\pi \cos \right(\pi x\left)\right)}^2}dx.$$

We first find the indefinite integral $$\begin{array}{cc}& \int \sin \left(\pi x\right)\sqrt{1+{\left(\pi \cos \right(\pi x\left)\right)}^2}dx\\ =& \int \sqrt{1+u^2}(-\frac{du}{{\pi }^2})u=\pi \cos \left(\pi x\right)\\ =& -\frac{1}{{\pi }^2}\int {\sec }^{3}tdtu=\tan t\\ =& -\frac{1}{{\pi }^2}(\frac{\sec t\tan t}{2}+\frac{1}{2}\ln |\tan t+\sec t|)+C\text{by integration by parts}\\ =& -\frac{1}{2{\pi }^2}(u\sqrt{1+u^2}+\ln (u+\sqrt{1+u^2}\left)\right)+C\text{by integration by parts}\\ =& -\frac{1}{2{\pi }^2}\left(\pi \cos \right(\pi x)\sqrt{1+{\pi }^2{\cos }^2\left(\pi x\right)}+\ln \left(\pi \cos \right(\pi x)+\sqrt{1+{\pi }^2{\cos }^2\left(\pi x\right)}))+C\end{array}$$

By FTC, $$\begin{array}{cc}& 2\pi {\int }_0^1\sin \left(\pi x\right)\sqrt{1+{\left(\pi \cos \right(\pi x\left)\right)}^2}dx\\ =& -\frac{1}{\pi }\left(\pi \cos \right(\pi x)\sqrt{1+{\pi }^2{\cos }^2\left(\pi x\right)}+\ln \left(\pi \cos \right(\pi x)+\sqrt{1+{\pi }^2{\cos }^2\left(\pi x\right)})){|}_0^1\\ =& 2\sqrt{1+{\pi }^2}+\frac{\ln (2{\pi }^2+2\pi \sqrt{1+{\pi }^2}+1)}{\pi }\end{array}$$

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

Solution.

The radius is given by $x=y^3$. The derivative is $\frac{dx}{dy}=3y^2$.

Then the area of the surface of revolution is given by $$\begin{array}{cc}& 2\pi {\int }_1^2{y}^3\sqrt{1+{\left(3y^2\right)}^2}dy\\ =& 2\pi {\int }_{10}^{145}\sqrt{u}d\left(\frac{u}{36}\right)u=1+9y^4\\ =& \frac{\pi }{18}\left(\frac{2}{3}{u}^{\frac{3}{2}}\right){|}_{10}^{145}\\ =& \frac{\pi }{27}(145\sqrt{145}-10\sqrt{10}).\end{array}$$

Exercise 9.9 Find the exact area of the surface obtained by rotating the curve about the $y$-axis.

$$y=\frac{1}{4}{x}^2-\frac{1}{2}\ln x,1\le x\le 2$$

Solution.

The radius is $x=y^3$. The derivative is $$\frac{dy}{dx}=\frac{x}{2}-\frac{1}{2x}.$$

Then the area of the surface of revolution is given by $$\begin{array}{cc}& 2\pi {\int }_1^{2}x\sqrt{1+{(\frac{x}{2}-\frac{1}{2x})}^2}dy\\ =& 2\pi {\int }_1^{2}x(\frac{x}{2}+\frac{1}{2x})dx\\ =& 2\pi (\frac{x^3}{6}+\frac{x}{2}){|}_1^2\\ =& \frac{10\pi }{3}.\end{array}$$

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 $$\sqrt{(dx{)}^2+(dy{)}^2}=rdt.$$

Then the area of the surface is $$\frac{dy}{dx}=\frac{x}{2}-\frac{1}{2x}.$$

Then the area of the surface of revolution is given by $$\begin{array}{cc}& 2\pi {\int }_0^{2\pi }(r-r\cos t)rdt\\ =& 2\pi r^2(t-\sin t){|}_0^{2\pi }\\ =& 4{\pi }^2{r}^2.\end{array}$$