## Geometric Calculations of Parametric Curves

### Learning Outcomes

• Find the area under a parametric curve
• Use the equation for arc length of a parametric curve
• Apply the formula for surface area to a volume generated by a parametric curve

## Integrals Involving Parametric Equations

Now that we have seen how to calculate the derivative of a plane curve, the next question is this: How do we find the area under a curve defined parametrically? Recall the cycloid defined by the equations $x\left(t\right)=t-\sin{t},y\left(t\right)=1-\cos{t}$. Suppose we want to find the area of the shaded region in the following graph.

To derive a formula for the area under the curve defined by the functions

$x=x\left(t\right),y=y\left(t\right),a\le t\le b$,

we assume that $x\left(t\right)$ is differentiable and start with an equal partition of the interval $a\le t\le b$. Suppose ${t}_{0}=a<{t}_{1}<{t}_{2}<\cdots <{t}_{n}=b$ and consider the following graph.

We use rectangles to approximate the area under the curve. The height of a typical rectangle in this parametrization is $y\left(x\left({\overline{t}}_{i}\right)\right)$ for some value ${\overline{t}}_{i}$ in the ith subinterval, and the width can be calculated as $x\left({t}_{i}\right)-x\left({t}_{i - 1}\right)$. Thus the area of the ith rectangle is given by

${A}_{i}=y\left(x\left({\overline{t}}_{i}\right)\right)\left(x\left({t}_{i}\right)-x\left({t}_{i - 1}\right)\right)$.

Then a Riemann sum for the area is

${A}_{n}=\displaystyle\sum _{i=1}^{n}y\left(x\left({\overline{t}}_{i}\right)\right)\left(x\left({t}_{i}\right)-x\left({t}_{i - 1}\right)\right)$.

Multiplying and dividing each area by ${t}_{i}-{t}_{i - 1}$ gives

${A}_{n}=\displaystyle\sum _{i=1}^{n}y\left(x\left({\overline{t}}_{i}\right)\right)\left(\frac{x\left({t}_{i}\right)-x\left({t}_{i - 1}\right)}{{t}_{i}-{t}_{i - 1}}\right)\left({t}_{i}-{t}_{i - 1}\right)=\displaystyle\sum _{i=1}^{n}y\left(x\left({\overline{t}}_{i}\right)\right)\left(\frac{x\left({t}_{i}\right)-x\left({t}_{i - 1}\right)}{\Delta t}\right)\Delta t$.

Taking the limit as $n$ approaches infinity gives

$A=\underset{n\to \infty }{\text{lim}}{A}_{n}={\displaystyle\int }_{a}^{b}y\left(t\right){x}^{\prime }\left(t\right)dt$.

Note that as the value of $n$ approaches $\infty$, the change in $x$ over smaller and smaller time intervals can be rewritten as the instantaneous rate of change of $x$ with respect to $t$ at some value within the sub-interval of $t$. Recall that this fact is known as the Mean Value Theorem, and we briefly review it to clarify the proofs within this section.

### Recall: Mean Value Theorem

Let $f$ be continuous over the closed interval $[a,b]$ and differentiable over the open interval $(a,b)$. Then, there exists at least one point $c \in (a,b)$ such that

$f^{\prime}(c)=\dfrac{f(b)-f(a)}{b-a}$

The preceding result leads to the following theorem.

### theorem: Area under a Parametric Curve

Consider the non-self-intersecting plane curve defined by the parametric equations

$x=x\left(t\right),y=y\left(t\right),a\le t\le b$

and assume that $x\left(t\right)$ is differentiable. The area under this curve is given by

$A={\displaystyle\int }_{a}^{b}y\left(t\right){x}^{\prime }\left(t\right)dt$.

### Example: Finding the Area under a Parametric Curve

Find the area under the curve of the cycloid defined by the equations

$x\left(t\right)=t-\sin{t},y\left(t\right)=1-\cos{t},0\le t\le 2\pi$.

Watch the following video to see the worked solution to Example: Finding the Area under a Parametric Curve.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

### try it

Find the area under the curve of the hypocycloid defined by the equations

$x\left(t\right)=3\cos{t}+\cos3t,y\left(t\right)=3\sin{t}-\sin3t,0\le t\le \pi$.

## Arc Length of a Parametric Curve

In addition to finding the area under a parametric curve, we sometimes need to find the arc length of a parametric curve. In the case of a line segment, arc length is the same as the distance between the endpoints. If a particle travels from point A to point B along a curve, then the distance that particle travels is the arc length. To develop a formula for arc length, we start with an approximation by line segments as shown in the following graph.

Given a plane curve defined by the functions $x=x\left(t\right),y=y\left(t\right),a\le t\le b$, we start by partitioning the interval $\left[a,b\right]$ into n equal subintervals: ${t}_{0}=a<{t}_{1}<{t}_{2}<\cdots <{t}_{n}=b$. The width of each subinterval is given by $\Delta t=\frac{\left(b-a\right)}{n}$. We can calculate the length of each line segment:

$\begin{array}{}\\ {d}_{1}=\sqrt{{\left(x\left({t}_{1}\right)-x\left({t}_{0}\right)\right)}^{2}+{\left(y\left({t}_{1}\right)-y\left({t}_{0}\right)\right)}^{2}}\hfill \\ {d}_{2}=\sqrt{{\left(x\left({t}_{2}\right)-x\left({t}_{1}\right)\right)}^{2}+{\left(y\left({t}_{2}\right)-y\left({t}_{1}\right)\right)}^{2}}\text{etc}.\hfill \end{array}$

Then add these up. We let s denote the exact arc length and ${s}_{n}$ denote the approximation by n line segments:

$s\approx \displaystyle\sum _{k=1}^{n}{s}_{k}=\displaystyle\sum _{k=1}^{n}\sqrt{{\left(x\left({t}_{k}\right)-x\left({t}_{k - 1}\right)\right)}^{2}+{\left(y\left({t}_{k}\right)-y\left({t}_{k - 1}\right)\right)}^{2}}$.

If we assume that $x\left(t\right)$ and $y\left(t\right)$ are differentiable functions of t, then the Mean Value Theorem (Introduction to the Applications of Derivatives) applies, so in each subinterval $\left[{t}_{k - 1},{t}_{k}\right]$ there exist $\hat{t}_{k}$ and $\tilde{t}_{k}$ such that

$\begin{array}{}\\ x\left({t}_{k}\right)-x\left({t}_{k - 1}\right)={x}^{\prime }\left(\hat{t}_{k}\right)\left({t}_{k}-{t}_{k - 1}\right)={x}^{\prime }\left(\hat{t}_{k}\right)\Delta t\hfill \\ y\left({t}_{k}\right)-y\left({t}_{k - 1}\right)={y}^{\prime }\left(\tilde{t}_{k}\right)\left({t}_{k}-{t}_{k - 1}\right)={y}^{\prime }\left(\tilde{t}_{k}\right)\Delta t.\hfill \end{array}$

Therefore $s$ becomes

$\begin{array}{cc}\hfill s& \approx {\displaystyle\sum _{k=1}^{n}}{s}_{k}\hfill \\ & ={\displaystyle\sum _{k=1}^{n}}\sqrt{{\left({x}^{\prime}\left(\hat{t}_{k}\right)\Delta t\right)}^{2}+{\left({y}^{\prime}\left(\tilde{t}_{k}\right)\Delta t\right)}^{2}}\hfill \\ & ={\displaystyle\sum _{k=1}^{n}}\sqrt{{\left({x}^{\prime}\left(\hat{t}_{k}\right)\right)}^{2}{\left(\Delta t\right)}^{2}+{\left({y}^{\prime}\left(\tilde{t}_{k}\right)\right)}^{2}{\left(\Delta t\right)}^{2}}\hfill \\ & =\left({\displaystyle\sum _{k=1}^{n}}\sqrt{{\left({x}^{\prime}\left(\hat{t}_{k}\right)\right)}^{2}+{\left({y}^{\prime}\left(\tilde{t}_{k}\right)\right)}^{2}}\right)\Delta t.\hfill \end{array}$

This is a Riemann sum that approximates the arc length over a partition of the interval $\left[a,b\right]$. If we further assume that the derivatives are continuous and let the number of points in the partition increase without bound, the approximation approaches the exact arc length. This gives

$\begin{array}{cc}\hfill s& ={\underset{n\to\infty}\lim} {\displaystyle\sum _{k=1}^{n}} {s}_{k}\hfill \\ & = {\underset{n\to\infty}\lim}\left({\displaystyle\sum _{k=1}^{n}}\sqrt{{\left({x}^{\prime}\left(\hat{t}_{k}\right)\right)}^{2}+{\left({y}^{\prime}\left(\tilde{t}_{k}\right)\right)}^{2}}\right)\Delta t\hfill \\ & ={\displaystyle\int_{a}^{b}}\sqrt{{\left({x}^{\prime }\left(t\right)\right)}^{2}+{\left({y}^{\prime }\left(t\right)\right)}^{2}}dt.\hfill \end{array}$

When taking the limit, the values of $\hat{t}_{k}$ and $\tilde{t}_{k}$ are both contained within the same ever-shrinking interval of width $\Delta t$, so they must converge to the same value.

We can summarize this method in the following theorem.

### theorem: Arc Length of a Parametric Curve

Consider the plane curve defined by the parametric equations

$x=x\left(t\right),y=y\left(t\right),{t}_{1}\le t\le {t}_{2}$

and assume that $x\left(t\right)$ and $y\left(t\right)$ are differentiable functions of t. Then the arc length of this curve is given by

$s={\displaystyle\int }_{{t}_{1}}^{{t}_{2}}\sqrt{{\left(\frac{dx}{dt}\right)}^{2}+{\left(\frac{dy}{dt}\right)}^{2}}dt$.

At this point a side derivation leads to a previous formula for arc length. In particular, suppose the parameter can be eliminated, leading to a function $y=F\left(x\right)$. Then $y\left(t\right)=F\left(x\left(t\right)\right)$ and the Chain Rule gives ${y}^{\prime }\left(t\right)={F}^{\prime }\left(x\left(t\right)\right){x}^{\prime }\left(t\right)$. Substituting this into the theorem gives

$\begin{array}{cc}\hfill s& ={\displaystyle\int }_{{t}_{1}}^{{t}_{2}}\sqrt{{\left(\frac{dx}{dt}\right)}^{2}+{\left(\frac{dy}{dt}\right)}^{2}}dt\hfill \\ & ={\displaystyle\int }_{{t}_{1}}^{{t}_{2}}\sqrt{{\left(\frac{dx}{dt}\right)}^{2}+{\left({F}^{\prime }\left(x\right)\frac{dx}{dt}\right)}^{2}}dt\hfill \\ & ={\displaystyle\int }_{{t}_{1}}^{{t}_{2}}\sqrt{{\left(\frac{dx}{dt}\right)}^{2}\left(1+{\left({F}^{\prime }\left(x\right)\right)}^{2}\right)}dt\hfill \\ & ={\displaystyle\int }_{{t}_{1}}^{{t}_{2}}{x}^{\prime }\left(t\right)\sqrt{1+{\left(\frac{dy}{dx}\right)}^{2}}dt.\hfill \end{array}$

Here we have assumed that ${x}^{\prime }\left(t\right)>0$, which is a reasonable assumption. The Chain Rule gives $dx={x}^{\prime }\left(t\right)dt$, and letting $a=x\left({t}_{1}\right)$ and $b=x\left({t}_{2}\right)$ we obtain the formula

$s={\displaystyle\int }_{a}^{b}\sqrt{1+{\left(\frac{dy}{dx}\right)}^{2}}dx$,

which is the formula for arc length obtained in the Introduction to the Applications of Integration.

### Example: Finding the Arc Length of a Parametric Curve

Find the arc length of the semicircle defined by the equations

$x\left(t\right)=3\cos{t},y\left(t\right)=3\sin{t},0\le t\le \pi$.

Watch the following video to see the worked solution to Example: Finding the Arc Length of a Parametric Curve.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

### try it

Find the arc length of the curve defined by the equations

$x\left(t\right)=3{t}^{2},y\left(t\right)=2{t}^{3},1\le t\le 3$.

### Try It

We now return to the problem posed at the beginning of the section about a baseball leaving a pitcher’s hand. Ignoring the effect of air resistance (unless it is a curve ball!), the ball travels a parabolic path. Assuming the pitcher’s hand is at the origin and the ball travels left to right in the direction of the positive x-axis, the parametric equations for this curve can be written as

$x\left(t\right)=140t,y\left(t\right)=-16{t}^{2}+2t$

where t represents time. We first calculate the distance the ball travels as a function of time. This distance is represented by the arc length. We can modify the arc length formula slightly. First rewrite the functions $x\left(t\right)$ and $y\left(t\right)$ using v as an independent variable, so as to eliminate any confusion with the parameter t:

$x\left(v\right)=140v,y\left(v\right)=-16{v}^{2}+2v$.

Then we write the arc length formula as follows:

$\begin{array}{cc}\hfill s\left(t\right)& ={\displaystyle\int }_{0}^{t}\sqrt{{\left(\frac{dx}{dv}\right)}^{2}+{\left(\frac{dy}{dv}\right)}^{2}}dv\hfill \\ & ={\displaystyle\int }_{0}^{t}\sqrt{{140}^{2}+{\left(-32v+2\right)}^{2}}dv.\hfill \end{array}$

The variable v acts as a dummy variable that disappears after integration, leaving the arc length as a function of time t. To integrate this expression we can use a formula from Appendix A,

$\displaystyle\int \sqrt{{a}^{2}+{u}^{2}}du=\frac{u}{2}\sqrt{{a}^{2}+{u}^{2}}+\frac{{a}^{2}}{2}\text{ln}|u+\sqrt{{a}^{2}+{u}^{2}}|+C$.

We set $a=140$ and $u=-32v+2$. This gives $du=-32dv$, so $dv=-\frac{1}{32}du$. Therefore

$\begin{array}{cc}\hfill {\displaystyle\int \sqrt{{140}^{2}+{\left(-32v+2\right)}^{2}}dv}& =-\frac{1}{32}{\displaystyle\int \sqrt{{a}^{2}+{u}^{2}}du}\hfill \\ & =-\frac{1}{32}\left[\begin{array}{c}\frac{\left(-32v+2\right)}{2}\sqrt{{140}^{2}+{\left(-32v+2\right)}^{2}}\hfill \\ +\frac{{140}^{2}}{2}\text{ln}|\left(-32v+2\right)+\sqrt{{140}^{2}+{\left(-32v+2\right)}^{2}}|\hfill \end{array}\right]+C\hfill \end{array}$

and

$\begin{array}{cc}\hfill s\left(t\right)& =-\frac{1}{32}\left[\frac{\left(-32t+2\right)}{2}\sqrt{{140}^{2}+{\left(-32t+2\right)}^{2}}+\frac{{140}^{2}}{2}\text{ln}|\left(-32t+2\right)+\sqrt{{140}^{2}+{\left(-32t+2\right)}^{2}}|\right]\hfill \\ & +\frac{1}{32}\left[\sqrt{{140}^{2}+{2}^{2}}+\frac{{140}^{2}}{2}\text{ln}|2+\sqrt{{140}^{2}+{2}^{2}}|\right]\hfill \\ & =\left(\frac{t}{2}-\frac{1}{32}\right)\sqrt{1024{t}^{2}-128t+19604}-\frac{1225}{4}\text{ln}|\left(-32t+2\right)+\sqrt{1024{t}^{2}-128t+19604}|\hfill \\ & +\frac{\sqrt{19604}}{32}+\frac{1225}{4}\text{ln}\left(2+\sqrt{19604}\right).\hfill \end{array}$

This function represents the distance traveled by the ball as a function of time. To calculate the speed, take the derivative of this function with respect to t. While this may seem like a daunting task, it is possible to obtain the answer directly from the Fundamental Theorem of Calculus:

$\frac{d}{dx}{\displaystyle\int }_{a}^{x}f\left(u\right)du=f\left(x\right)$.

Therefore

$\begin{array}{cc}\hfill {s}^{\prime }\left(t\right)& =\frac{d}{dt}\left[s\left(t\right)\right]\hfill \\ & =\frac{d}{dt}\left[{\displaystyle\int }_{0}^{t}\sqrt{{140}^{2}+{\left(-32v+2\right)}^{2}}dv\right]\hfill \\ & =\sqrt{{140}^{2}+{\left(-32t+2\right)}^{2}}\hfill \\ & =\sqrt{1024{t}^{2}-128t+19604}\hfill \\ & =2\sqrt{256{t}^{2}-32t+4901}.\hfill \end{array}$

One third of a second after the ball leaves the pitcher’s hand, the distance it travels is equal to

$\begin{array}{cc}\hfill s\left(\frac{1}{3}\right)& =\left(\frac{\frac{1}{3}}{2}-\frac{1}{32}\right)\sqrt{1024{\left(\frac{1}{3}\right)}^{2}-128\left(\frac{1}{3}\right)+19604}\hfill \\ & -\frac{1225}{4}\text{ln}|\left(-32\left(\frac{1}{3}\right)+2\right)+\sqrt{1024{\left(\frac{1}{3}\right)}^{2}-128\left(\frac{1}{3}\right)+19604}|\hfill \\ & +\frac{\sqrt{19604}}{32}+\frac{1225}{4}\text{ln}\left(2+\sqrt{19604}\right)\hfill \\ & \approx 46.69\text{feet}.\hfill \end{array}$

This value is just over three quarters of the way to home plate. The speed of the ball is

${s}^{\prime }\left(\frac{1}{3}\right)=2\sqrt{256{\left(\frac{1}{3}\right)}^{2}-16\left(\frac{1}{3}\right)+4901}\approx 140.34\text{ft/s}$.

This speed translates to approximately 95 mph—a major-league fastball.

## Surface Area Generated by a Parametric Curve

Recall the problem of finding the surface area of a volume of revolution. In Curve Length and Surface Area, we derived a formula for finding the surface area of a volume generated by a function $y=f\left(x\right)$ from $x=a$ to $x=b$, revolved around the x-axis:

$S=2\pi {\displaystyle\int }_{a}^{b}f\left(x\right)\sqrt{1+{\left({f}^{\prime }\left(x\right)\right)}^{2}}dx$.

We now consider a volume of revolution generated by revolving a parametrically defined curve $x=x\left(t\right),y=y\left(t\right),a\le t\le b$ around the x-axis as shown in the following figure.

The analogous formula for a parametrically defined curve is

$S=2\pi {\displaystyle\int }_{a}^{b}y\left(t\right)\sqrt{{\left({x}^{\prime }\left(t\right)\right)}^{2}+{\left({y}^{\prime }\left(t\right)\right)}^{2}}dt$

provided that $y\left(t\right)$ is not negative on $\left[a,b\right]$.

### Example: Finding Surface Area

Find the surface area of a sphere of radius $r$ centered at the origin.

Watch the following video to see the worked solution to Example: Finding Surface Area.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

### try it

Find the surface area generated when the plane curve defined by the equations

$x\left(t\right)={t}^{3},y\left(t\right)={t}^{2},0\le t\le 1$

is revolved around the x-axis.