# 6.4 Arc length of a curve and surface area  (Page 5/8)

 Page 5 / 8

## Calculating the surface area of a surface of revolution 2

Let $f\left(x\right)=y=\sqrt{3x}.$ Consider the portion of the curve where $0\le y\le 2.$ Find the surface area of the surface generated by revolving the graph of $f\left(x\right)$ around the $y\text{-axis}.$

Notice that we are revolving the curve around the $y\text{-axis},$ and the interval is in terms of $y,$ so we want to rewrite the function as a function of y . We get $x=g\left(y\right)=\left(1\text{/}3\right){y}^{3}.$ The graph of $g\left(y\right)$ and the surface of rotation are shown in the following figure. (a) The graph of g ( y ) . (b) The surface of revolution.

We have $g\left(y\right)=\left(1\text{/}3\right){y}^{3},$ so ${g}^{\prime }\left(y\right)={y}^{2}$ and ${\left({g}^{\prime }\left(y\right)\right)}^{2}={y}^{4}.$ Then

$\begin{array}{cc}\hfill \text{Surface Area}& ={\int }_{c}^{d}\left(2\pi g\left(y\right)\sqrt{1+{\left({g}^{\prime }\left(y\right)\right)}^{2}}\right)dy\hfill \\ & ={\int }_{0}^{2}\left(2\pi \left(\frac{1}{3}{y}^{3}\right)\sqrt{1+{y}^{4}}\right)dy\hfill \\ & =\frac{2\pi }{3}{\int }_{0}^{2}\left({y}^{3}\sqrt{1+{y}^{4}}\right)dy.\hfill \end{array}$

Let $u={y}^{4}+1.$ Then $du=4{y}^{3}dy.$ When $y=0,$ $u=1,$ and when $y=2,$ $u=17.$ Then

$\begin{array}{cc}\hfill \frac{2\pi }{3}{\int }_{0}^{2}\left({y}^{3}\sqrt{1+{y}^{4}}\right)dy& =\frac{2\pi }{3}{\int }_{1}^{17}\frac{1}{4}\sqrt{u}du\hfill \\ & =\frac{\pi }{6}{\left[\frac{2}{3}{u}^{3\text{/}2}\right]\phantom{\rule{0.2em}{0ex}}|}_{1}^{17}=\frac{\pi }{9}\left[{\left(17\right)}^{3\text{/}2}-1\right]\approx 24.118.\hfill \end{array}$

Let $g\left(y\right)=\sqrt{9-{y}^{2}}$ over the interval $y\in \left[0,2\right].$ Find the surface area of the surface generated by revolving the graph of $g\left(y\right)$ around the $y\text{-axis}.$

$12\pi$

## Key concepts

• The arc length of a curve can be calculated using a definite integral.
• The arc length is first approximated using line segments, which generates a Riemann sum. Taking a limit then gives us the definite integral formula. The same process can be applied to functions of $y.$
• The concepts used to calculate the arc length can be generalized to find the surface area of a surface of revolution.
• The integrals generated by both the arc length and surface area formulas are often difficult to evaluate. It may be necessary to use a computer or calculator to approximate the values of the integrals.

## Key equations

• Arc Length of a Function of x
$\text{Arc Length}={\int }_{a}^{b}\sqrt{1+{\left[{f}^{\prime }\left(x\right)\right]}^{2}}\phantom{\rule{0.2em}{0ex}}dx$
• Arc Length of a Function of y
$\text{Arc Length}={\int }_{c}^{d}\sqrt{1+{\left[{g}^{\prime }\left(y\right)\right]}^{2}}\phantom{\rule{0.2em}{0ex}}dy$
• Surface Area of a Function of x
$\text{Surface Area}={\int }_{a}^{b}\left(2\pi f\left(x\right)\sqrt{1+{\left({f}^{\prime }\left(x\right)\right)}^{2}}\right)dx$

For the following exercises, find the length of the functions over the given interval.

$y=5x\phantom{\rule{0.2em}{0ex}}\text{from}\phantom{\rule{0.2em}{0ex}}x=0\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}x=2$

$2\sqrt{26}$

$y=-\frac{1}{2}x+25\phantom{\rule{0.2em}{0ex}}\text{from}\phantom{\rule{0.2em}{0ex}}x=1\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}x=4$

$x=4y\phantom{\rule{0.2em}{0ex}}\text{from}\phantom{\rule{0.2em}{0ex}}y=-1\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}y=1$

$2\sqrt{17}$

Pick an arbitrary linear function $x=g\left(y\right)$ over any interval of your choice $\left({y}_{1},{y}_{2}\right).$ Determine the length of the function and then prove the length is correct by using geometry.

Find the surface area of the volume generated when the curve $y=\sqrt{x}$ revolves around the $x\text{-axis}$ from $\left(1,1\right)$ to $\left(4,2\right),$ as seen here. $\frac{\pi }{6}\left(17\sqrt{17}-5\sqrt{5}\right)$

Find the surface area of the volume generated when the curve $y={x}^{2}$ revolves around the $y\text{-axis}$ from $\left(1,\phantom{\rule{0.2em}{0ex}}1\right)$ to $\left(3,9\right).$ For the following exercises, find the lengths of the functions of $x$ over the given interval. If you cannot evaluate the integral exactly, use technology to approximate it.

$y={x}^{3\text{/}2}$ from $\left(0,0\right)\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}\left(1,1\right)$

$\frac{13\sqrt{13}-8}{27}$

$y={x}^{2\text{/}3}$ from $\left(1,1\right)\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}\left(8,4\right)$

$y=\frac{1}{3}{\left({x}^{2}+2\right)}^{3\text{/}2}$ from $x=0\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}x=1$

$\frac{4}{3}$

$y=\frac{1}{3}{\left({x}^{2}-2\right)}^{3\text{/}2}$ from $x=2$ to $x=4$

[T] $y={e}^{x}$ on $x=0$ to $x=1$

$2.0035$

$y=\frac{{x}^{3}}{3}+\frac{1}{4x}$ from $x=1\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}x=3$

$y=\frac{{x}^{4}}{4}+\frac{1}{8{x}^{2}}$ from $x=1\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}x=2$

$\frac{123}{32}$

$y=\frac{2{x}^{3\text{/}2}}{3}-\frac{{x}^{1\text{/}2}}{2}$ from $x=1\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}x=4$

$y=\frac{1}{27}{\left(9{x}^{2}+6\right)}^{3\text{/}2}$ from $x=0\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}x=2$

$10$

[T] $y=\text{sin}\phantom{\rule{0.2em}{0ex}}x$ on $x=0\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}x=\pi$

For the following exercises, find the lengths of the functions of $y$ over the given interval. If you cannot evaluate the integral exactly, use technology to approximate it.

$y=\frac{5-3x}{4}$ from $y=0$ to $y=4$

$\frac{20}{3}$

$x=\frac{1}{2}\left({e}^{y}+{e}^{\text{−}y}\right)$ from $y=-1\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}y=1$

$x=5{y}^{3\text{/}2}$ from $y=0$ to $y=1$

$\frac{1}{675}\left(229\sqrt{229}-8\right)$

[T] $x={y}^{2}$ from $y=0$ to $y=1$

$x=\sqrt{y}$ from $y=0\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}y=1$

$\frac{1}{8}\left(4\sqrt{5}+\text{ln}\left(9+4\sqrt{5}\right)\right)$

$x=\frac{2}{3}{\left({y}^{2}+1\right)}^{3\text{/}2}$ from $y=1$ to $y=3$

[T] $x=\text{tan}\phantom{\rule{0.2em}{0ex}}y$ from $y=0$ to $y=\frac{3}{4}$

$1.201$

[T] $x={\text{cos}}^{2}y$ from $y=-\frac{\pi }{2}$ to $y=\frac{\pi }{2}$

[T] $x={4}^{y}$ from $y=0\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}y=2$

$15.2341$

[T] $x=\text{ln}\left(y\right)$ on $y=\frac{1}{e}$ to $y=e$

For the following exercises, find the surface area of the volume generated when the following curves revolve around the $x\text{-axis}.$ If you cannot evaluate the integral exactly, use your calculator to approximate it.

$y=\sqrt{x}$ from $x=2$ to $x=6$

$\frac{49\pi }{3}$

$y={x}^{3}$ from $x=0$ to $x=1$

$y=7x$ from $x=-1\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}x=1$

$70\pi \sqrt{2}$

[T] $y=\frac{1}{{x}^{2}}$ from $x=1\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}x=3$

$y=\sqrt{4-{x}^{2}}$ from $x=0\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}x=2$

$8\pi$

$y=\sqrt{4-{x}^{2}}$ from $x=-1\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}x=1$

$y=5x$ from $x=1\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}x=5$

$120\pi \sqrt{26}$

[T] $y=\text{tan}\phantom{\rule{0.2em}{0ex}}x$ from $x=-\frac{\pi }{4}\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}x=\frac{\pi }{4}$

For the following exercises, find the surface area of the volume generated when the following curves revolve around the $y\text{-axis}\text{.}$ If you cannot evaluate the integral exactly, use your calculator to approximate it.

$y={x}^{2}$ from $x=0\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}x=2$

$\frac{\pi }{6}\left(17\sqrt{17}-1\right)$

$y=\frac{1}{2}{x}^{2}+\frac{1}{2}$ from $x=0\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}x=1$

$y=x+1$ from $x=0\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}x=3$

$9\sqrt{2}\pi$

[T] $y=\frac{1}{x}$ from $x=\frac{1}{2}$ to $x=1$

$y=\sqrt{x}$ from $x=1\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}x=27$

$\frac{10\sqrt{10}\pi }{27}\left(73\sqrt{73}-1\right)$

[T] $y=3{x}^{4}$ from $x=0$ to $x=1$

[T] $y=\frac{1}{\sqrt{x}}$ from $x=1$ to $x=3$

$25.645$

[T] $y=\text{cos}\phantom{\rule{0.2em}{0ex}}x$ from $x=0$ to $x=\frac{\pi }{2}$

The base of a lamp is constructed by revolving a quarter circle $y=\sqrt{2x-{x}^{2}}$ around the $y\text{-axis}$ from $x=1$ to $x=2,$ as seen here. Create an integral for the surface area of this curve and compute it. $2\pi$

A light bulb is a sphere with radius $1\text{/}2$ in. with the bottom sliced off to fit exactly onto a cylinder of radius $1\text{/}4$ in. and length $1\text{/}3$ in., as seen here. The sphere is cut off at the bottom to fit exactly onto the cylinder, so the radius of the cut is $1\text{/}4$ in. Find the surface area (not including the top or bottom of the cylinder). [T] A lampshade is constructed by rotating $y=1\text{/}x$ around the $x\text{-axis}$ from $y=1$ to $y=2,$ as seen here. Determine how much material you would need to construct this lampshade—that is, the surface area—accurate to four decimal places. $10.5017$

[T] An anchor drags behind a boat according to the function $y=24{e}^{\text{−}x\text{/}2}-24,$ where $y$ represents the depth beneath the boat and $x$ is the horizontal distance of the anchor from the back of the boat. If the anchor is $23$ ft below the boat, how much rope do you have to pull to reach the anchor? Round your answer to three decimal places.

[T] You are building a bridge that will span $10$ ft. You intend to add decorative rope in the shape of $y=5|\text{sin}\left(\left(x\pi \right)\text{/}5\right)|,$ where $x$ is the distance in feet from one end of the bridge. Find out how much rope you need to buy, rounded to the nearest foot.

$23$ ft

For the following exercises, find the exact arc length for the following problems over the given interval.

$y=\text{ln}\left(\text{sin}\phantom{\rule{0.2em}{0ex}}x\right)$ from $x=\pi \text{/}4$ to $x=\left(3\pi \right)\text{/}4.$ ( Hint: Recall trigonometric identities.)

Draw graphs of $y={x}^{2},$ $y={x}^{6},$ and $y={x}^{10}.$ For $y={x}^{n},$ as $n$ increases, formulate a prediction on the arc length from $\left(0,0\right)$ to $\left(1,1\right).$ Now, compute the lengths of these three functions and determine whether your prediction is correct.

$2$

Compare the lengths of the parabola $x={y}^{2}$ and the line $x=by$ from $\left(0,0\right)\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}\left({b}^{2},b\right)$ as $b$ increases. What do you notice?

Solve for the length of $x={y}^{2}$ from $\left(0,0\right)\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}\left(1,1\right).$ Show that $x=\left(1\text{/}2\right){y}^{2}$ from $\left(0,0\right)$ to $\left(2,\phantom{\rule{0.2em}{0ex}}2\right)$ is twice as long. Graph both functions and explain why this is so.

[T] Which is longer between $\left(1,\phantom{\rule{0.2em}{0ex}}1\right)$ and $\left(2,1\text{/}2\right)\text{:}$ the hyperbola $y=1\text{/}x$ or the graph of $x+2y=3?$

Explain why the surface area is infinite when $y=1\text{/}x$ is rotated around the $x\text{-axis}$ for $1\le x<\infty ,$ but the volume is finite.

find the integral of tan
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