Question

\[y=\frac 1 x\]
If the region R{(x,y)|x>=1,0<=y<= (1/x)} is rotated about the x-axis, the volume of the resulting solid is infinite. Show that the surface area is infinite. (aka Gabriel's horn)

Answer 1

\[\int_{1}^{\infty} 2\pi\frac 1 x \sqrt(1+(\frac{-1}{x^2})^2)dx\]

Answer 2

alright continue

Answer 3

Lemme clean that up.
\[2\pi \int_1^\infty \frac{1}{x}\sqrt{1+\frac{1}{x^4}}\ dx\]

Answer 4

alright kep going

Answer 5

hint do someting with the radical

Answer 6

substitution?

Answer 7

not quite , create a common denominator

Answer 8

\[2\pi\int_{1}^{\infty} \frac 1 x \sqrt(1+\frac{1}{x^4})dx\]
\[\sqrt(1+\frac{1}{x^4})\]
\[\sqrt(\frac{1+x^4}{x^4})\]
\[ \frac{\sqrt(1+x^4)}{x^2}\]

Answer 9

correct

Answer 10

continue

Answer 11

\[2\pi\int_{1}^{\infty} \frac 1 x \frac{\sqrt(1+x^4)}{x^2}dx\]
\[2\pi\int_{1}^{\infty} \frac{\sqrt(1+x^4)}{x^3}dx\]

Answer 12

OH MY GOD!
\[2\pi\int_{1}^{\infty} \frac{\sqrt(1+x^4)}{x^3}dx\]
u=1+x^4
du=4x^3
(1/4)du=x^3

Answer 13

lol

Answer 14

?

Answer 15

I realized that I can finally make the substitution to solve the integral

Answer 16

oh hah

Answer 17

that's not right though because the x^3 is in the denominator

Answer 18

\[\frac\pi 2\int_{1}^{\infty} \frac{\sqrt(u)}{x^3}du\]
what about the \[x^3\] that's in the denominator?

Answer 19

trig sub =P

Answer 20

do you know this?

Answer 21

I thinks so ...lets see
\[\]

Answer 22

\[2\pi\int_{1}^{\infty} \frac{sec^2}{x^3}dx\]
\[sec^2\] because 1+tan^4 =sec^4
x=tan^4 x
dx=4 tan^3 x sec^2 x

Answer 23

one sec

Answer 24

\[2\pi\int_{1}^{\infty} \frac{sec^2\theta}{(tan^4 \theta)^3} 4tan^3 \theta sec^2\theta d\theta\]

Answer 25

\[8\pi\int_{1}^{\infty} \frac{sec^4\theta}{tan^4 \theta} d\theta\]

Answer 26

let x=atan(x)
dx=asec^2x

Answer 27

so what I have written is wrong?

Answer 28

\[2\pi\int_{1}^{\infty} \frac{\sqrt(1+tanx^4)}{tanx^3}sec^2xdx\]

Answer 29

i'm just looking to make sure

x=tan(x)
dx=sec^2x
\[\sqrt{a^2+x^2}=\sqrt{1+tan^2x}=\sqrt{sec^2x}=sec(x)\]

Answer 30

the numerator seems to be \[sec^4x\]

Answer 31

is the denominator ok?

Answer 32

wait it should be

\[x^2=tan(x)\]

Answer 33

oh

Answer 34

\[\pi\int_{1}^{\infty} \frac{\sqrt(1+tan^2 x)}{xtanx}sec^2xdx\]
dx=.5sec^2 x?

Answer 35

i'm not so sure now lol

Answer 36

\[8\pi\int_{1}^{\infty} \frac{sec^4\theta}{tan^4 \theta} d\theta\]
this look pretty...

Answer 37

\[x = \sqrt{\tan u}\]
\[dx = \frac{\sec^{2} u}{2\sqrt{\tan u}}\]
\[\rightarrow 2\pi \int\limits_{1}^{\infty} \frac{\sec u}{\tan u \sqrt{\tan u}}*\frac{\sec^{2} u}{2\sqrt{\tan u}} du\]
\[= \pi \int\limits_{1}^{\infty}\frac{\sec^{3} u}{\tan^{2} u} du\]

Answer 38

ahh so i was right -.-

Answer 39

sorry sofia i'm just really busy atm

Answer 40

lol no prob.

Answer 41

thanks smartcow