Question

surface area of revolved surface between x=sqrt(x^2+1)
x=0
x=3
about the y-axis

Answer 1

im sorry its the x-axis not the y-axis

Answer 2

have you found dx/dy yet?

Answer 3

wait

Answer 4

I'm confused... I though you had x=sqrt(y^2+1)

Answer 5

but you have x=sqrt(x^2+1)

Answer 6

this is an equation
not a relation between x and y

Answer 7

oh sorry its y= sqrt( x^2 +1)
typo on my part

Answer 8

ok so have you tried to find dy/dx?

Answer 9

i got
x/sqrt(x^2+1)

Answer 10

and also if it is about the x-axis then we have:
\[\int\limits 2 \pi y ds \\ \text{ where you can make either choice in } ds \text{ :} \\ ds=\sqrt{1+(y')^2} dx \text{ if } y=f(x), a \le x \le b \\ \text{ or } \\ ds=\sqrt{1+(x')^2} dy \text{ if } x=h(y), c \le y \le d\]

checking your y'...
\[y=\sqrt{x^2+1} \\ y^2=x^2+1 \\ 2y dy=2x dx \\ y dy =x dx \\ y \frac{dy}{dx}=x \\ \frac{dy}{dx}=\frac{x}{y} \\ \frac{dy}{dx}=\frac{x}{\sqrt{x^2+1}} \]
that is right

Answer 11

\[\int\limits_0^3 2 \pi y \sqrt{1+(\frac{x}{\sqrt{x^2+1}})^2} dx \\ \\ \int\limits_0^3 2 \pi \sqrt{x^2+1} \sqrt{1+(\frac{x}{\sqrt{x^2+1}})^2} dx \]

Answer 12

are you able to complete the problem
or are you getting stuck somewhere

Answer 13

im lagging bad

Answer 14

yeah im stuck on what to do next

Answer 15

can you show me what you have so far?

Answer 16

or were you not able to do anything from what I last wrote?

Answer 17

yeah i couldn't get past that last part

Answer 18

\[\int\limits_0^3 2\pi \sqrt{x^2+1} \sqrt{1+\frac{x^2}{x^2+1}} dx \\ \int\limits_0^3 2\pi \sqrt{x^2+1} \sqrt{\frac{x^2+1}{x^2+1}+\frac{x^2}{x^2+1}} dx \\ \int\limits_0^3 2 \pi \sqrt{x^2+1} \sqrt{\frac{x^2+1+x^2}{x^2+1}} dx \\ \int\limits_0^3 2 \pi \sqrt{x^2+1 } \sqrt{\frac{2x^2+1}{x^2+1} } dx \\ \int\limits _0^3 2 \pi \frac{\sqrt{x^2+1}}{\sqrt{x^2+1}} \sqrt{2x^2+1} dx \\ \int\limits_0^3 2 \pi \sqrt{2x^2+1} dx\]
what about now?

Answer 19

i plugged it into wolfram and got this
constant+pi sqrt(2 x^2+1) x+(pi sinh^(-1)(sqrt(2) x))/sqrt(2)

Answer 20

but it doesnt look right to me

Answer 21

try a trig substiution
recall
\[\tan^2(\theta)+1=\sec^2(\theta)\]

Answer 22

ok, srry but i have to go, i have class

Answer 23

\[2x^2+1 =(\sqrt{2}x)^2+1 \\ \text{ comparing } \\ (\sqrt{2} x)^2+1 \text{ with } (\tan(\theta))^2+1 \\ \text{ you should want to make the substitution } \sqrt{2} x=\tan(\theta)\]