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.

A series of half circles drawn above the x-axis with x intercepts being multiples of 2π. The half circle between 0 and 2π is highlighted. On the graph there are also written two equations: x(t) = t – sin(t) and y(t) = 1 – cos(t).

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

x = x (t), y = y (t), a \leq t \leq b

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

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.

A curved line is drawn in the first quadrant. Below it are a series of rectangles marked that begin at the x-axis and reach up to the curved line; the rectangle’s height is determined by the location of the curved line at the leftmost point of the rectangle. These lines are noted as x(t0), x(t1), …, x(tn).

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 (x ({\bar{t}}_{i})) (x (t_{i}) - x (t_{i - 1}))

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

Then a Riemann sum for the area is

A_{n} = \sum_{i = 1}^{n} y (x ({\bar{t}}_{i})) (x (t_{i}) - x (t_{i - 1}))

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

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

A_{n} = \sum_{i = 1}^{n} y (x ({\bar{t}}_{i})) (\frac{x (t_{i}) - x (t_{i - 1})}{t_{i} - t_{i - 1}}) (t_{i} - t_{i - 1}) = \sum_{i = 1}^{n} y (x ({\bar{t}}_{i})) (\frac{x (t_{i}) - x (t_{i - 1})}{Δ t}) Δ t

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

Taking the limit as $n$ approaches infinity gives

A = lim_{n \to \infty} A_{n} = \int_{a}^{b} y (t) x^{'} (t) d t

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

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^{'} (c) = \frac{f (b) - f (a)}{b - a}

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

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 (t), y = y (t), a \leq t \leq b

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

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

A = \int_{a}^{b} y (t) x^{'} (t) d t

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

Example: Finding the Area under a Parametric Curve

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

x (t) = t - \sin t, y (t) = 1 - \cos t, 0 \leq t \leq 2 π

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

Show Solution

Using the theorem, we have

\begin{array}{cc} A & = \int_{a}^{b} y (t) x^{'} (t) d t \\ = \int_{0}^{2 π} (1 - \cos t) (1 - \cos t) d t \\ = \int_{0}^{2 π} (1 - 2 \cos t + \cos^{2} t) d t \\ = \int_{0}^{2 π} (1 - 2 \cos t + \frac{1 + \cos 2 t}{2}) d t \\ = \int_{0}^{2 π} (\frac{3}{2} - 2 \cos t + \frac{\cos 2 t}{2}) d t \\ = {\frac{3 t}{2} - 2 \sin t + \frac{\sin 2 t}{4} |}_{0}^{2 π} \\ = 3 π . \end{array}

$\begin{array}{cc}\hfill A& ={\displaystyle\int }_{a}^{b}y\left(t\right){x}^{\prime }\left(t\right)dt\hfill \\ & ={\displaystyle\int }_{0}^{2\pi }\left(1-\cos{t}\right)\left(1-\cos{t}\right)dt\hfill \\ & ={\displaystyle\int }_{0}^{2\pi }\left(1 - 2\cos{t}+{\cos}^{2}t\right)dt\hfill \\ & ={\displaystyle\int }_{0}^{2\pi }\left(1 - 2\cos{t}+\frac{1+\cos2t}{2}\right)dt\hfill \\ & ={\displaystyle\int }_{0}^{2\pi }\left(\frac{3}{2}-2\cos{t}+\frac{\cos2t}{2}\right)dt\hfill \\ & ={\frac{3t}{2}-2\sin{t}+\frac{\sin2t}{4}|}_{0}^{2\pi }\hfill \\ & =3\pi .\hfill \end{array}$

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.

You can view the transcript for this segmented clip of “7.2 Calculus of Parametric Curves” here (opens in new window).

try it

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

x (t) = 3 \cos t + \cos 3 t, y (t) = 3 \sin t - \sin 3 t, 0 \leq t \leq π

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

Hint

Use the theorem, along with the identities $\sin\alpha \sin\beta =\frac{1}{2}\left[\cos\left(\alpha -\beta \right)-\cos\left(\alpha +\beta \right)\right]$ and ${\sin}^{2}t=\frac{1-\cos2t}{2}$ .

Show Solution

$A=3\pi$ (Note that the integral formula actually yields a negative answer. This is due to the fact that $x\left(t\right)$ is a decreasing function over the interval $\left[0,2\pi \right]$ ; that is, the curve is traced from right to left.)

Try It

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.

A curved line in the first quadrant with points marked for x = 1, 2, 3, 4, and 5. These points have values roughly 2.1, 2.7, 3, 2.7, and 2.1, respectively. The points for x = 1 and 5 are marked A and B, respectively.

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{matrix} d_{1} = \sqrt{{(x (t_{1}) - x (t_{0}))}^{2} + {(y (t_{1}) - y (t_{0}))}^{2}} \\ d_{2} = \sqrt{{(x (t_{2}) - x (t_{1}))}^{2} + {(y (t_{2}) - y (t_{1}))}^{2}} etc . \end{matrix}

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

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

s \approx \sum_{k = 1}^{n} s_{k} = \sum_{k = 1}^{n} \sqrt{{(x (t_{k}) - x (t_{k - 1}))}^{2} + {(y (t_{k}) - y (t_{k - 1}))}^{2}}

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

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{matrix} x (t_{k}) - x (t_{k - 1}) = x^{'} ({\hat{t}}_{k}) (t_{k} - t_{k - 1}) = x^{'} ({\hat{t}}_{k}) Δ t \\ y (t_{k}) - y (t_{k - 1}) = y^{'} ({\tilde{t}}_{k}) (t_{k} - t_{k - 1}) = y^{'} ({\tilde{t}}_{k}) Δ t . \end{matrix}

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

Therefore $s$ becomes

\begin{array}{cc} s & \approx \sum_{k = 1}^{n} s_{k} \\ = \sum_{k = 1}^{n} \sqrt{{(x^{'} ({\hat{t}}_{k}) Δ t)}^{2} + {(y^{'} ({\tilde{t}}_{k}) Δ t)}^{2}} \\ = \sum_{k = 1}^{n} \sqrt{{(x^{'} ({\hat{t}}_{k}))}^{2} {(Δ t)}^{2} + {(y^{'} ({\tilde{t}}_{k}))}^{2} {(Δ t)}^{2}} \\ = (\sum_{k = 1}^{n} \sqrt{{(x^{'} ({\hat{t}}_{k}))}^{2} + {(y^{'} ({\tilde{t}}_{k}))}^{2}}) Δ t . \end{array}

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

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} s & = lim_{n \to \infty} \sum_{k = 1}^{n} s_{k} \\ = lim_{n \to \infty} (\sum_{k = 1}^{n} \sqrt{{(x^{'} ({\hat{t}}_{k}))}^{2} + {(y^{'} ({\tilde{t}}_{k}))}^{2}}) Δ t \\ = \int_{a}^{b} \sqrt{{(x^{'} (t))}^{2} + {(y^{'} (t))}^{2}} d t . \end{array}

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

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 (t), y = y (t), t_{1} \leq t \leq t_{2}

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

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 = \int_{t_{1}}^{t_{2}} \sqrt{{(\frac{d x}{d t})}^{2} + {(\frac{d y}{d t})}^{2}} d t

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

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} s & = \int_{t_{1}}^{t_{2}} \sqrt{{(\frac{d x}{d t})}^{2} + {(\frac{d y}{d t})}^{2}} d t \\ = \int_{t_{1}}^{t_{2}} \sqrt{{(\frac{d x}{d t})}^{2} + {(F^{'} (x) \frac{d x}{d t})}^{2}} d t \\ = \int_{t_{1}}^{t_{2}} \sqrt{{(\frac{d x}{d t})}^{2} (1 + {(F^{'} (x))}^{2})} d t \\ = \int_{t_{1}}^{t_{2}} x^{'} (t) \sqrt{1 + {(\frac{d y}{d x})}^{2}} d t . \end{array}

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

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 = \int_{a}^{b} \sqrt{1 + {(\frac{d y}{d x})}^{2}} d x

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

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 (t) = 3 \cos t, y (t) = 3 \sin t, 0 \leq t \leq π

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

Show Solution

The values $t=0$ to $t=\pi$ trace out the red curve in Figure 9. To determine its length, use the theorem:

\begin{array}{cc} s & = \int_{t_{1}}^{t_{2}} \sqrt{{(\frac{d x}{d t})}^{2} + {(\frac{d y}{d t})}^{2}} d t \\ = \int_{0}^{π} \sqrt{{(- 3 \sin t)}^{2} + {(3 \cos t)}^{2}} d t \\ = \int_{0}^{π} \sqrt{9 \sin^{2} t + 9 \cos^{2} t} d t \\ = \int_{0}^{π} \sqrt{9 (\sin^{2} t + \cos^{2} t)} d t \\ = \int_{0}^{π} 3 d t = {3 t |}_{0}^{π} = 3 π . \end{array}

$\begin{array}{cc}\hfill s& ={\displaystyle\int }_{{t}_{1}}^{{t}_{2}}\sqrt{{\left(\frac{dx}{dt}\right)}^{2}+{\left(\frac{dy}{dt}\right)}^{2}}dt\hfill \\ & ={\displaystyle\int }_{0}^{\pi }\sqrt{{\left(-3\sin{t}\right)}^{2}+{\left(3\cos{t}\right)}^{2}}dt\hfill \\ & ={\displaystyle\int }_{0}^{\pi }\sqrt{9{\sin}^{2}t+9{\cos}^{2}t}dt\hfill \\ & ={\displaystyle\int }_{0}^{\pi }\sqrt{9\left({\sin}^{2}t+{\cos}^{2}t\right)}dt\hfill \\ & ={\displaystyle\int }_{0}^{\pi }3dt={3t|}_{0}^{\pi }=3\pi .\hfill \end{array}$

Note that the formula for the arc length of a semicircle is $\pi r$ and the radius of this circle is 3. This is a great example of using calculus to derive a known formula of a geometric quantity.

A semicircle is drawn with radius 3. There is an arrow pointing counterclockwise. On the graph there are also written three equations: x(t) = 3 cos(t), y(t) = 3 sin(t), and 0 ≤ t ≤ π.

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

You can view the transcript for this segmented clip of “7.2 Calculus of Parametric Curves” here (opens in new window).

try it

Find the arc length of the curve defined by the equations

x (t) = 3 t^{2}, y (t) = 2 t^{3}, 1 \leq t \leq 3

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

Hint

Show Solution

$s=2\left({10}^{\frac{3}{2}}-{2}^{\frac{3}{2}}\right)\approx 57.589$

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 (t) = 140 t, y (t) = - 16 t^{2} + 2 t

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

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 (v) = 140 v, y (v) = - 16 v^{2} + 2 v

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

Then we write the arc length formula as follows:

\begin{array}{cc} s (t) & = \int_{0}^{t} \sqrt{{(\frac{d x}{d v})}^{2} + {(\frac{d y}{d v})}^{2}} d v \\ = \int_{0}^{t} \sqrt{140^{2} + {(- 32 v + 2)}^{2}} d v . \end{array}

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

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,

\int \sqrt{a^{2} + u^{2}} d u = \frac{u}{2} \sqrt{a^{2} + u^{2}} + \frac{a^{2}}{2} ln | u + \sqrt{a^{2} + u^{2}} | + C

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

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

\begin{array}{cc} \int \sqrt{140^{2} + {(- 32 v + 2)}^{2}} d v & = - \frac{1}{32} \int \sqrt{a^{2} + u^{2}} d u \\ = - \frac{1}{32} [\begin{matrix} \frac{(- 32 v + 2)}{2} \sqrt{140^{2} + {(- 32 v + 2)}^{2}} \\ + \frac{140^{2}}{2} ln | (- 32 v + 2) + \sqrt{140^{2} + {(- 32 v + 2)}^{2}} | \end{matrix}] + C \end{array}

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

and

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

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

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}{d x} \int_{a}^{x} f (u) d u = f (x)

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

Therefore

\begin{array}{cc} s^{'} (t) & = \frac{d}{d t} [s (t)] \\ = \frac{d}{d t} [\int_{0}^{t} \sqrt{140^{2} + {(- 32 v + 2)}^{2}} d v] \\ = \sqrt{140^{2} + {(- 32 t + 2)}^{2}} \\ = \sqrt{1024 t^{2} - 128 t + 19604} \\ = 2 \sqrt{256 t^{2} - 32 t + 4901} . \end{array}

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

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

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

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

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

s^{'} (\frac{1}{3}) = 2 \sqrt{256 {(\frac{1}{3})}^{2} - 16 (\frac{1}{3}) + 4901} \approx 140.34 ft/s

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

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 π \int_{a}^{b} f (x) \sqrt{1 + {(f^{'} (x))}^{2}} d x

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

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.

A curve is drawn in the first quadrant with endpoints marked t = a and t = b. On this curve, there is a point marked (x(t), y(t)). There is a circle with an arrow drawn around the x-axis that seems to indicate a rotation about the x axis, and there is a shape that accompanies that curve that seems to be what you would obtain if you rotated the curve about the x axis.

The analogous formula for a parametrically defined curve is

S = 2 π \int_{a}^{b} y (t) \sqrt{{(x^{'} (t))}^{2} + {(y^{'} (t))}^{2}} d t

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

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.

Show Solution

We start with the curve defined by the equations

x (t) = r \cos t, y (t) = r \sin t, 0 \leq t \leq π

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

This generates an upper semicircle of radius r centered at the origin as shown in the following graph.

A semicircle is drawn with radius r. On the graph there are also written three equations: x(t) = r cos(t), y(t) = r sin(t), and 0 ≤ t ≤ π.

When this curve is revolved around the $x$ -axis, it generates a sphere of radius $r$ . To calculate the surface area of the sphere, we use the above equation:

\begin{array}{cc} S & = 2 π \int_{a}^{b} y (t) \sqrt{{(x^{'} (t))}^{2} + {(y^{'} (t))}^{2}} d t \\ = 2 π \int_{0}^{π} r \sin t \sqrt{{(- r \sin t)}^{2} + {(r \cos t)}^{2}} d t \\ = 2 π \int_{0}^{π} r \sin t \sqrt{r^{2} \sin^{2} t + r^{2} \cos^{2} t} d t \\ = 2 π \int_{0}^{π} r \sin t \sqrt{r^{2} (\sin^{2} t + \cos^{2} t)} d t \\ = 2 π \int_{0}^{π} r^{2} \sin t d t \\ = 2 π r^{2} (- \cos t |_{0}^{π}) \\ = 2 π r^{2} (- \cos π + \cos 0) \\ = 4 π r^{2} . \end{array}

$\begin{array}{cc}\hfill S&=2\pi {\displaystyle\int_{a}^{b}} y\left(t\right)\sqrt{{\left({x}^{\prime }\left(t\right)\right)}^{2}+{\left({y}^{\prime}\left(t\right)\right)}^{2}}dt\hfill \\ &=2\pi {\displaystyle\int_{0}^{\pi}} r\sin{t}\sqrt{{\left(-r\sin{t}\right)}^{2}+{\left(r\cos{t}\right)}^{2}}dt\hfill \\ &=2\pi {\displaystyle\int }_{0}^{\pi }r\sin{t}\sqrt{{r}^{2}{\sin}^{2}t+{r}^{2}{\cos}^{2}t}dt\hfill \\ &=2\pi {\displaystyle\int }_{0}^{\pi }r\sin{t}\sqrt{{r}^{2}\left({\sin}^{2}t+{\cos}^{2}t\right)}dt\hfill \\ &=2\pi {\displaystyle\int_{0}^{\pi}{r}^{2}\sin{t}dt}\hfill \\ &=2\pi{r}^{2}\left(-\cos{t}|_{0}^{\pi}\right)\hfill \\ & =2\pi {r}^{2}\left(-\cos\pi +\cos0\right)\hfill \\ & =4\pi {r}^{2}.\hfill \end{array}$

This is, in fact, the formula for the surface area of a sphere.

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

You can view the transcript for this segmented clip of “7.2 Calculus of Parametric Curves” here (opens in new window).

try it

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

x (t) = t^{3}, y (t) = t^{2}, 0 \leq t \leq 1

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

is revolved around the x-axis.

Hint

Show Solution

$A=\frac{\pi \left(494\sqrt{13}+128\right)}{1215}$

Module 7: Parametric Equations and Polar Coordinates