# 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 $f\left(x\right)=y=\sqrt[3]{3x}.$ 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.

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 $x=1$

$y=\sqrt[3]{x}$ 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
Differentiate each from the first principle. y=x,y=1/x
I need help with calculus. Anyone help me.
yes
Hi
Usman
beautiful name usman
Fund
really
Usman
Hi guys
Macquitasha
Hello everyone here
abdulazeez
good day!
joel
hii
Shreya
You are welcome
abdulazeez
shreya
ashif
thanks
joel
hello Sar aapse Kuchh calculate ke sawal poochhne Hain
Sumit
integration seems interesting
it's like a multiple oparation in just one.
Efrain
Definitely integration
tangent line at a point/range on a function f(x) making f'(x)
Luis
Principles of definite integration?
ROHIT
For tangent they'll usually give an x='s value. In that case, solve for y, keep the ordered pair. then find f(x) prime. plug the given x value into the prime and the solution is the slope of the tangent line. Plug the ordered pair into the derived function in y=mx+b format as x and y to solve for B
Anastasia
parcing an area trough a function f(x)
Efrain
Find the length of the arc y = x^2 over 3 when x = 0 and x = 2.
integrate x ln dx from 1 to e
application of function
how i can need help
what ?
Bunyim
defination of math
azam
application of function
azam
azam
what is a circle
Ronnie
math is the science, logic, shape and arrangement
a circle is a hole shape
Jianna
a whole circumference have equal distance from one point
azam
please tell me books which write on function
azam
HE is a Nigerian, wrote the book INTEGRATED MATHEMATICS...CHECK IT OUT!!
Agboke
Woah this is working again
Bruce
show that the f^n f(x)=|x-1| is not differentiable at x=1.
is there any solution manual to calculuse 1 for Gilbert Strang ?
I am beginner
Abdul
I am a beginner
ephraim
l am also beginner
just began, bois!!
Luis
Hello
abdulazeez
abdulazeez
Hey
Bonface
Hi
Jianna
what is mathematics
logical usage of numbers
Leo
thanks
Henry
you welcome
Leo
what's career can one specialize in by doing pure maths
Lucy
Lucy Omollo...... The World is Yours by specializing in pure math. Analytics, Financial engineering ,programming, education, combinatorial mathematics, Game Theory. your skill-set will be like water a necessary element of survival.
David
***onetonline.org/find/descriptor/result/1.A.1.c.1
Bruce
mathematics seems to be anthropocentric deductive reasoning and a little high order logic. I only say this because I can only find two things going on which is infinitely smaller than 0 and anything over 1
David
More comprehensive list here: ***onetonline.org/find/descriptor/result/1.A.1.c.1?a=1
Bruce
so how can we differentiate inductive reasoning and deductive reasoning
Henry
thanks very much Mr David
Henry
hi everyone
Sabir
is there anyone who can guide me in learning the mathematics easily
Sabir
Hi Sabir first step of learning mathematics is by falling in love with it and secondly, watch videos on simple algebra then read and solved problems on it
Leo
yes sabir just do every time practice that is the solution
Henry
it will be work over to you ,u know how mind work ,it prossed the information easily when u are practising regularly
Henry
in calculas,does a self inverse function exist
Lucy
I'm lost in all functions need help
Jonathan
hello i need help in rate of change
Moises
***questioncove.com/invite/QzOQGp
Bruce
Hello
hassan
hi
MJ
hi
Masaniel
so difficult
Masaniel
hello my name is Charles Christian
Hello Charles
Jianna
Hi! I am Dante
Dante
Hi! I'm ashwini
ashwini
halĺo
Roben
Hi
Leo
hello leo
Agboke
can anyone prove why AU(BnC)=(AUB)n(AUC)
Agboke
this one it can't be proven these are assumption
Henry
hello agboke there is no proof for such
Leo
Hi
hi this is wasim
wasim
can anybody put me through flowchart and algorithm here
Agboke
Leo
Luis
music while you math
Luis
dy/dx= 1-cos4x/sin4x
what is the derivatives of 1-cos4x/sin4x
Alma
what is the derivate of Sec2x
Johar
d/dx(sec(2 x)) = 2 tan(2 x) sec(2 x)
AYAN
who knows more about mathematical induction?
Agboke
who know anything about the whole calculus thing 🤔 its killing me 😶
matbakh
Yes
hii
Gagan