Applications of Definite Integrals: Arc Length

Suppose that $y=f(x)$ is continuously differentiable on $[a,b]$. Figure 1

Figure 1 shows the curve $y=f(x)$ is approximated by a PL (Piecewise Linear) curve (in red). Suppose that the line segments of the PL curve are infinitesimal and denote by $dL$ a line element. Then it is given by \begin{align*}dL&=\sqrt{(dx)^2+(dy)^2}\\&=\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx\\&=\sqrt{1+[f'(x)]^2}dx\end{align*} Hence the length of the curve $y=f(x)$, $a\leq x\leq b$ is \begin{equation}\label{eq:arclength}L=\int_{x=a}^{x=b}dL=\int_a^b\sqrt{1+[f'(x)]^2}dx\end{equation}

Often a curve is represented by a parametric equation $r(t)=(x(t),y(t))$, $\alpha\leq t\leq\beta$ such that $x(t)$ and $y(t)$ are continuously differentiable. Such a curve is called smooth. A typical example is a motion in the plane (or in space). For a smooth curve $r(t)=(x(t),y(t))$, $\alpha\leq t\leq\beta$, $dL$ is \begin{align*}dL&=\sqrt{(dx)^2+(dy)^2}\\&=\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt\end{align*} and hence \begin{equation}\label{eq:arclength2}\int_\alpha^\beta\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt\end{equation} If $r(t)$ is the motion of a particle, $\left|\frac{dr(t)}{dt}\right|=\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}$ is the speed of the particle and so \eqref{eq:arclength2} represents the distance traveled by the particle over the time interval $\alpha\leq t\leq\beta$.

Example (The circumference of a circle). Find the length of the circle of radius $r$.

Solution.  The equation of the upper semicircle of radius r centered at the origin is $y=\sqrt{r^2-x^2}$, $-r\leq x\leq r$. $\frac{dy}{dx}=-\frac{x}{\sqrt{r^2-x^2}}$ and so the circumference is \begin{align*}2\int_{-r}^r\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx&=2\int_{-r}^r\frac{r}{\sqrt{r^2-x^2}}dx\\&=4\int_0^r\frac{r}{\sqrt{r^2-x^2}}dx\\&=4\int_0^r\frac{1}{\sqrt{1-\left(\frac{x}{r}\right)^2}}dx\\&=4r\int_0^1\frac{1}{\sqrt{1-u^2}}du\\&=4r[\sin^{-1}(u)]_0^1\\&=4r\left(\frac{\pi}{2}\right)=2\pi r\end{align*}

The circle $x^2+y^2=r^2$ of radius $r$ centered at the origin can be described by the parametric equations $$x=r\cos t,\ y=r\sin t,\ 0\leq t\leq 2\pi$$ $\frac{dx}{dt}=-r\sin t$ and $\frac{dy}{dt}=r\cos t$ so $\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}=r$. Therefore the circumference is $\int_0^{2\pi}rdt=2\pi r$. we find that using parametric equations of the circle comes with a much simpler calculation.

Example. Find the length of the asteroid $$x^{\frac{2}{3}}+y^{\frac{2}{3}}=1$$

Solution. The name asteroid is due to its graph as seen in Figure 2. Figure 2. Asteroid

The asteroid can be parametrized as $$x=\cos^3 t,\ y=\sin^3 t,\ 0\leq t\leq 2\pi$$ $\frac{dx}{dt}=-3\cos^2 t\sin t$ and $\frac{dy}{dt}=3\sin^2 t\cos t$ so $\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2=9\cos^2t\sin^2t$. For $0\leq t\leq\frac{\pi}{2}$, $\sin t\geq 0$ and $\cos t\geq 0$. Hence \begin{align*}L&=4\int_0^{\frac{\pi}{2}}\sqrt{9\cos^2t\sin^2t}dt\\&=12\int_0^{\frac{\pi}{2}}\cos t\sin tdt\\&=6\int_0^{\frac{\pi}{2}}\sin 2tdt\\&=3[-\cos 2t]_0^{\frac{\pi}{2}}=6\end{align*} The integral $\int_0^{\frac{\pi}{2}}\cos t\sin tdt$ also can be done using a substitution. Let $u=\sin t$. Then $du=\cos tdt$ so \begin{align*}\int_0^{\frac{\pi}{2}}\cos t\sin tdt&=\int_0^1udu\\&=\frac{1}{2}\end{align*}

Example. Find the length of the curve $$y=\frac{4\sqrt{2}}{3}x^{\frac{3}{2}}-1,\ 0\leq x\leq 1$$

Solution. $\frac{dy}{dx}=2\sqrt{2x}$ so $\sqrt{1+\left(\frac{dy}{dx}\right)^2}=\sqrt{1+8x}$ and \begin{align*}L&=\int_0^1\sqrt{1+8x}dx\\&=\frac{2}{3}\cdot\frac{1}{8}[(1+8x)^{\frac{3}{2}}]_0^1\\&=\frac{13}{6}\end{align*}

Example. Find the length of the curve $y=\left(\frac{x}{2}\right)^{\frac{2}{3}}$, $0\leq x\leq 2$.

Solution. $\frac{dy}{dx}=\frac{1}{3}\left(\frac{2}{x}\right)^{\frac{1}{3}}$. Note that this derivative does not exist at $x=0$. Rewrite the function as $x=2y^{\frac{3}{2}}$, $0\leq y\leq 1$. $\frac{dx}{dy}=3\sqrt{y}$ and \begin{align*}L&=\int_{y=0}^{y=1}dL\\&=\int_{y=0}^{y=1}\sqrt{(dx)^2+(dy)^2}\\&=\int_0^1\sqrt{\left(\frac{dx}{dy}\right)^2+1}dy\\&=\int_0^1\sqrt{9y+1}dy\\&=\frac{1}{9}\cdot\frac{2}{3}[(9y+1)^{\frac{3}{2}}]_0^1\\&=\frac{2}{27}(10\sqrt{10}-1)\end{align*}

Properties of Power Series

A power series is a series of the form \begin{equation}\label{eq:powerseries}\sum_{n=0}^\infty c_n(x-a)^n=c_0+c_1(x-a)+c_2(x-a)^2+\cdots\end{equation} The power series in \eqref{eq:powerseries} is particularly called a power series in $x-a$, a power series centered at $a$, or a power series about $a$. The number $a$ is called the center of the power series. The set of the values of $x$ for which the series converges is called the interval of convergence and the radius of the interval of convergence is called the radius of convergence. Since a power series in general does not converge everywhere, it is important to find the interval of convergence or the radius of convergence (if you find one, you consequently find the other). In the study of the convergence of power series, we mostly use the ratio test but the root test may come in handy sometimes.

Example. Find the interval and the radius of convergence for each of the following series.

1. $\sum_{n=0}^\infty\frac{x^n}{n!}$
2. $\sum_{n=0}^\infty n!x^n$
3. $\sum_{n=1}^\infty\frac{(x-3)^n}{n}$

Solution.

1.  \begin{align*}\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|&=\lim_{n\to\infty}\frac{\frac{|x|^{n+1}}{(n+1)!}}{\frac{|x|^n}{n!}}\\&=\lim_{n\to\infty}\frac{|x|}{n+1}\\&=0<1\end{align*} for all $x$. This means that the interval of convergence is $(-\infty,\infty)$ and the radius of convergence is $\infty$.
2. \begin{align*}\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|&=\lim_{n\to\infty}\frac{(n+1)!|x|^{n+1}}{n!|x|^n}\\&=\lim_{n\to\infty}(n+1)|x|\\&=\infty\end{align*} unless $x=0$. The interval of convergence is $\{0\}$ and the radius of convergence is 0.
3. \begin{align*}\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|&=\lim_{n\to\infty}\frac{\frac{|x-3|^{n+1}}{n+1}}{\frac{|x-3|^n}{n}}\\&=\lim_{n\to\infty}\frac{n}{n+1}|x-3|\\&=|x-3|\end{align*} In order for the series to converge, we require that $|x-3|<1$ or $-1<x-3<1$ i.e. $2<x<4$. For the last step, we must check if the series converges at the end points $x=2$ and $x=4$. If $x=2$, the series becomes $\sum_{n=1}^\infty\frac{(-1)^n}{n}$. This is an alternating harmonic series and we learned here that it converges. If $x=4$, the series becomes the harmonic series $\sum_{n=1}^\infty\frac{1}{n}$ which diverges. Therefore, the interval of convergence is $[2,4)$. Since the center of the power series is 3, the radius is 1.

Representing Functions as Power Series

As we will study later, functions that are infinitely many times differentiable (such functions are simply called smooth) in a open interval containing $a$ can be represented a power series centered at $a$ called a Taylor series. But even without knowing Taylor series expansion of a function, using the geometric series \begin{equation}\label{eq:geomseries}\frac{1}{1-x}=1+x+x^2+x^3+\cdots=\sum_{n=0}^\infty x^n\ \mbox{if}\ |x|<1\end{equation} we can represent many functions as power series.

Example. Express $\frac{1}{1+x^2}$ as a power series and find the interval of convergence.

Solution. \begin{align*}\frac{1}{1+x^2}&=\frac{1}{1-(-x^2)}\\&=\sum_{n=0}^\infty(-x^2)^n\\&=\sum_{n=0}^\infty(-1)^nx^{2n}\end{align*} provided $|x|^2<1$. Hence, the interval of convergence is $|x|<1$ or $(-1,1)$.

Example. Find a power series representation of $\frac{x^3}{x+2}$.

Solution. First we find a power series representation of $\frac{1}{x+2}$. \begin{align*}\frac{1}{x+2}&=\frac{1}{2}\frac{1}{1+\frac{x}{2}}\\&=\frac{1}{2}\frac{1}{1-\left(-\frac{x}{2}\right)}\\&=\frac{1}{2}\sum_{n=0}^\infty(-1)^n\frac{x^n}{2^n}\\&=\sum_{n=0}^\infty(-1)^n\frac{x^n}{2^{n+1}}\end{align*} provided $\left|\frac{x}{2}\right|<1$ or $|x|<2$. Therefore, $\frac{x^3}{x+2}=\sum_{n=0}^\infty(-1)^n\frac{x^{n+3}}{2^{n+1}}$ for $-2<x<2$.

Note: If one replaces $n$ by $n-3$, then the series can be rewritten equivalently as $\sum_{n=3}^\infty(-1)^{n-1}\frac{x^n}{2^{n-2}}$.

Differentiation and Integration of Power Series

It turns out that one can differentiate and integrate a power series term by term as he would’ve done with polynomials.

Theorem. If the power series $\sum_{n=0}^\infty c_n(x-a)^n$ has radius of convergence $R>0$ then the function $f$ defined by $$f(x)=\sum_{n=0}^\infty c_n(x-a)^n$$ is differentiable on the interval of convergence $(a-R,a+R)$ and $$f'(x)=\sum_{n=1}^\infty nc_n(x-a)^{n-1}$$ and $$\int f(x)dx=C+\sum_{n=0}^\infty\frac{c_n}{n+1}(x-a)^{n+1}$$ where $C$ is a constant. The two power series $f'(x)$ and $\int f(x)dx$ both have the same radius of convergence $R$.

This theorem along with \eqref{eq:geomseries} allows us to find the power series representations of a broader class of functions as we will see in the following examples.

Example. Express $\frac{1}{(1-x)^2}$ as a power series. What is the radius of convergence?

Solution. \begin{align*}\frac{1}{(1-x)^2}&=\frac{d}{dx}\frac{1}{1-x}\\&=\frac{d}{dx}\sum_{n=0}^\infty x^n\ (|x|<1)\\&=\sum_{n=1}^\infty nx^{n-1}\\&=\sum_{n=0}^\infty(n+1)x^n\end{align*} The radius of convergence is 1.

Example. Find a power series representation for $\ln (1-x)$ and its radius of convergence.

Solution. \begin{align*}\ln(1-x)&=-\int\frac{dx}{1-x}\\&=-\int\sum_{n=0}^\infty x^ndx\ (|x|<1)\\&=-\sum_{n=0}^\infty\frac{x^{n+1}}{n+1}+C\\&=-\sum_{n=1}^\infty\frac{x^n}{n}+C\end{align*} where $C$ is a constant. When $x=0$, $\ln(1-x)=0$ so we find $C=0$. Therefore, $$\ln(1-x)=-\sum_{n=1}^\infty\frac{x^n}{n}$$

Example. Find a power series representation for $f(x)=\tan^{-1}x$.

Solution. \begin{align*}\tan^{-1} x&=\int\frac{1}{1+x^2}dx\\&=\int\sum_{n=0}^\infty (-1)^nx^{2n}\ (|x|<1)\\&=\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{2n+1}+C\end{align*} where $C$ is a constant. Since $\tan^{-1}0=0$, we find $C=0$. Hence, $$\tan^{-1}x=\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{2n+1}=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots$$

Example. Find a power series representation of $\ln\left(\frac{1+x}{1-x}\right)$.

Solution.  First $\ln\left(\frac{1+x}{1-x}\right)=\ln(1+x)-\ln(1-x)$. \begin{align*}\ln(1+x)&=\int\frac{dx}{1+x}\\&=\int\sum_{n=0}^\infty (-1)^nx^ndx\ (|x|<1)\\&=\sum_{n=0}^\infty\frac{(-1)^nx^{n+1}}{n+1}+C\end{align*} Using $\ln(1+x)=0$ for $x=0$, we find $C=0$ and so $\ln(1+x)=\sum_{n=0}^\infty\frac{(-1)^nx^{n+1}}{n+1}$. Recall that we found a power series representation $\ln(1-x)=-\sum_{n=1}^\infty\frac{x^n}{n}$ in an example above. Hence, \begin{align*}\ln\left(\frac{1+x}{1-x}\right)&=\sum_{n=0}^\infty\frac{(-1)^nx^{n+1}}{n+1}+\sum_{n=0}^\infty\frac{x^{n+1}}{n+1}\\&=\sum_{n=1}^\infty\frac{(-1)^{n-1}x^n}{n}+\sum_{n=1}^\infty\frac{x^n}{n}\\&=\sum_{n=1}^\infty[(-1)^{n-1}+1]\frac{x^n}{n}\end{align*} Note $$(-1)^{n-1}+1=\left\{\begin{array}{ccc} 2 & \mbox{if} & n\ \mbox{is odd}\\0 & \mbox{if} & n\ \mbox{is even}\end{array}\right.$$ Therefore, $$\ln\left(\frac{1+x}{1-x}\right)=2\sum_{n=1}^\infty\frac{x^{2n-1}}{2n-1}=2\sum_{n=0}^\infty\frac{x^{2n+1}}{2n+1}$$

Remark. Since we already know that $\ln(1-x)=-\sum_{n=1}^\infty\frac{x^n}{n}$ and $\ln(1+x)=\ln(1-(-x))$, we can find a power series representation of $\ln(1+x)$ by replacing $x$ by $-x$ in the power series representation of $\ln(1-x)$: \begin{align*}\ln(1+x)&=-\sum_{n=0}^\infty\frac{(-x)^{n+1}}{n+1}\\&=-\sum_{n=0}^\infty\frac{(-1)^{n+1}x^{n+1}}{n+1}\\&=\sum_{n=0}^\infty\frac{(-1)^nx^{n+1}}{n+1}\end{align*}

Applications of Definite Integrals: Solids of Revolution (The Washer Method)

Let $y=f(x)$ and $y=g(x)$ be continuous functions on a closed interval $[a,b]$ such that $f(x)\geq g(x)$ on $[a,b]$ as shown in Figure 1. Figure 1

If the region bounded by $f(x)$ and $g(x)$ on $[a,b]$ is revolved about the $x$-axis, the volume of the resulting solid is obtained by \begin{align*}V&=\int_a^b\pi [f(x)]^2dx-\int_a^b\pi [g(x)]^2dx\\&=\int_a^b\pi ([f(x)]^2-[g(x)]^2)dx\end{align*} This is essentially the disk method but it is rather called the washer method because each cross section is a washer due to a hollow core within the solid.

Example. The region bounded by  the curve $y=x^2+1$ and the line $y=-x+3$ is revolved about the $x$-axis. Find the volume of the resulting solid.

Solution. Figure 2 shows the bounded region Figure 1

and Figure 2 shows the solid of revolution. Figure 2

The volume $V$ is given by $$V=\int_{-2}^1\pi[(-x+3)^2-(x^2+1)^2]dx=\frac{117\pi}{5}$$

Example. The region bounded by the parabola $y=x^2$ and the line $y=2x$ in the first quadrant is revolved about the $y$-axis to generate a solid. Find the volume of the solid.

Solution. Figure 3 shows the bounded region Figure 3

and Figure 4 shows the solid of revolution. Figure 4

The volume of the solid is given by $$V=\int_0^4\pi\left[(\sqrt{y})^2-\left(\frac{y}{2}\right)^2\right]dy=\frac{8\pi}{3}$$

Applications of Definite Integrals: Volumes by Cylindrical Shells

Suppose that the region bounded by $y=x^4+1$, $x$-axis, $x=1$ and $x=2$, as shown in Figure 1, is revolved about the $y$-axis and we want to find the volume of the resulting solid. Figure 1

You will find it difficult to apply the disk method or the washer method for this though it’s doable. Here we want to devise another method that can come in handy for this kind of problems. First let us take a look at Figure 2. Figure 2

Imagine that each rectangle is rotated about the $y$-axis. Then we would obtain layers of cylindrical shells and the sum of the volumes of these cylindrical shells would approximate the volume of the solid in question. If we use infinitesimally thin rectangles, the sum of the volumes of cylindrical shells would be exactly the volume of the solid. The radius of the cylindrical shell at $x$ is $x$ since the axis of rotation is the $y$-axis ($x=0$), its height is $x^4+1$, and its infinitesimal thickness is $dx$. So the volume of the cylindrical shell at $x$ is $2\pi x(x^4+1)dx$. Therefore the volume of the solid is $$V=\int_1^2 2\pi x(x^4+1)dx=24\pi$$ More generally we have

Cylindrical Shell Method

Suppose that the region under the curve $y=f(x)$, $a\leq x\leq b$ is revolved about the line $x=L$. Then the volume $V$ of the resulting solid is obtained by $$V=\left\{\begin{array}{ccc}\int_a^b 2\pi (x-L)f(x)dx & \mbox{if} & L\leq a\\\int_a^b 2\pi (L-x)f(x)dx & \mbox{if} & L\geq b\end{array}\right.$$

Example. The region enclosed by the $x$-axis and the parabola $y=3x-x^2$ is revolved about the vertical line $x=-1$. Find the volume of the resulting solid.

Solution. The region (in red) and the axis of rotation (in blue) are shown in Figure 3. Figure 3

By cylindrical shell method, the volume $V$ is given by $$V=\int_0^3 2\pi(x+1)(3x-x^2)dx=\frac{45\pi}{2}$$

Example. The region bounded by the curve $y=\sqrt{x}$, the $x$-axis and the line $x=4$ is revolved about the $y$-axis. Find the volume of the resulting solid.

Solution. Figure 4 shows the region. Figure 4

Since the axis of rotation is $x=0$, by cylindrical shell method the volume $V$ is given by $$V=\int_0^42\pi x\sqrt{x}dx=\frac{128\pi}{5}$$

Note that while cylindrical shell method is simpler, it is also easy to find the volume using washer method. $$V=\int_0^2\pi[4^2-(y^2)^2]dy=\int_0^2(16-y^4)dy=\frac{128\pi}{5}$$

Example. The same region in the previous example is revolved about the $x$-axis this time. Find the volume of the resulting solid using cylindrical shell method.

Solution. The solid and a cylindrical shell are shown in Figure 5. Figure 5

The shell radius is $y$ and the shell height $4-y^2$. Hence the volume is $$V=\int_0^2 2\pi y(4-y^2)dy=8\pi$$

Note that it would actually be easier to use disk method for this problem. Using disk method the volume $V$ is given by $$V=\int_0^4\pi(\sqrt{x})^2dx=\int_0^4\pi xdx=8\pi$$

Areas Between Curves

Let $y=f(x)$ and $y=g(x)$ be two continuous functions on a closed interval $[a,b]$ such that $f(x)\geq g(x)$ on $[a,b]$ as seen in Figure 1. Figure 1

The area $A$ of the region between the curves $f(x)$ and $g(x)$ is \begin{align*}A&=\int_a^b f(x)dx-\int_a^b g(x)dx\\&=\int_a^b[f(x)-g(x)]dx\end{align*}

Example. Find the area of the region enclosed by the parabola $y=2-x^2$ and the line $y=-x$.

Solution. First we need to find the $x$-coordinates of the points at which the two curves meet. Set $2-x^2=-x$. Then the quadratic equation $x^2-x-2=0$ has two solutions $x=-1,2$. From Figure 2 Figure 2

we see that the area is \begin{align*}A&=\int_{-1}^2[(2-x^2)-(-x)]dx\\&=\int_{-1}^2(2+x-x^2)dx\\&=\frac{9}{2}\end{align*}

Example. Find the area of the region in the first quadrant that is bounded above by $y=\sqrt{x}$ and below by the $x$-axis and the line $y=x-2$.

Solution. To find the $x$-coordinates of the points at which $y=\sqrt{x}$ and $y=x-2$ meet, set $\sqrt{x}=x-2$. Squaring this equation, we obtain the quadratic equation $x^2-5x+4=0$ whose solutions are $x=-1$ and $x=4$. Since we are considering only the first quadrant $x=-1$ is not our concern. The bounded region is shown in Figure 3. Figure 3

Note that on $[0,2]$ the part of the region is bounded by $y=\sqrt{x}$ and the $x$-axis and on $[2,4]$ the part of the reagion is bounded by $y=\sqrt{x}$ and $y=x-2$. Hence, the area is \begin{align*}A&=\int_0^2\sqrt{x}dx+\int_2^4[\sqrt{x}-(x-2)]dx\\&=\frac{10}{3}\end{align*}

If $x=f(y)$ and $x=g(y)$ are two continuous functions on a closed interval $[c,d]$ such that $f(y)\geq g(y)$ as seen in Figure 4, Figure 4

then the area $A$ bounded by $f(y)$ and $g(y)$ is given by $$A=\int_c^d[f(y)-g(y)]dy$$

Example. Redo the previous example by integrating with respect to $y$.

Solution. The same region in Figure 3 can be considered as the region bounded by $x=y^2$ ($y\geq 0$ and $x=y+2$. Hence, $$A=\int_0^2[y+2-y^2]dy=\frac{10}{3}$$

Often the fastest way to find an area may be to combine calculus with geometry. The same region in Figure 3 can be viewed as in Figure 5. Figure 5

The area of the right triangle in blue is $\frac{1}{2}\cdot 2\cdot 2=2$ hence, the area of the region is $$A=\int_0^4\sqrt{x}dx-2=\frac{10}{3}$$