**Calc2 Help Needed** Find the area of the surface generated by revolving around the curve y=x^3 / 2 from x= 0 to 2 about the x-axis.

- anonymous

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- anonymous

\[2\pi \int\limits x^3/2 \sqrt ( (4+9x^4)/4\]

- dumbcow

|dw:1328069467445:dw|
to find surface area, we add up all the circumferences times dx
\[SA = 2 \pi \int\limits_{a}^{b}R dx\]

- anonymous

Gotcha.

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## More answers

- dumbcow

oops i forgot something...where did the sqrt come from?

- anonymous

From the formula

- dumbcow

haha i have forgotten the SA formula, im just doing this on the fly
is it that sqrt(1+f'(x)^2)

- anonymous

yep

- anonymous

well

- anonymous

I'm trying to type it out.

- anonymous

integral of 2pi * original line * sqrt( 1 + (derivative of original line)^2

- dumbcow

you can simplify the integral to
\[\pi/2\int\limits_{0}^{2}x^{3}\sqrt{4+9x^{4}} dx\]
set u = 4+9x^4
du = 36x^3 dx
\[\rightarrow \pi/72 \int\limits_{0}^{2}\sqrt{u} du\]

- anonymous

what happened to the 2 under x^3 and the 4 under the sqrt?

- dumbcow

i combined them
sqrt(4) = 2
1/2 *1/2 = 1/4
multiplied by 2pi
2pi*1/4 = pi/2

- anonymous

Oh sweet.. Gotcha!

- dumbcow

you can take it from here, integrate sqrt(u) then plug in 4+9x^4 back in for u

- anonymous

ok so what do I bring back front from the u? 1/36?

- dumbcow

not sure what you mean
i multiplied the 1/36 by the pi/2

- anonymous

On the u-substituion.. Shouldn't something be brought out front like 1/36 + 4? From the du = 36x^3 + 4

- dumbcow

oh no the derivative of a constant is 0
du = 36x^3 + 0

- dumbcow

du = 36x^3 dx
dx = du/36x^3
--> x^3 sqrt(u) du / 36x^3 = sqrt(u) du/36

- anonymous

So would the final answer be.. 15.41599pi.. My hw says incorrect :(

- dumbcow

i get 52.14

- anonymous

didn't work :/

- anonymous

You changed the limits right? I had 0 to 152

- dumbcow

oh no i kept the limits as 0 to 2, just plugged the x back in after integrating

- anonymous

Wouldn't they need to be changed since we used u-substitution? I meant 0 to 148 btw

- dumbcow

it depends, you can definitely do it that way but i have found it it more confusing to change the limits rather than just substituting back the original variable
the new limits really should be 4 to 148
if x=0, u = 4

- dumbcow

http://www.wolframalpha.com/input/?i=integrate+pi%2F2*x%5E3*sqrt%284%2B9x%5E4%29+dx+from+0+to+2

- anonymous

Yeah not working.. wtf

- dumbcow

must have set it up wrong i guess ...maybe i missed something
hold on i'll check out some resources on surface area integrals

- anonymous

Ok thank you.

- anonymous

Hopefully you have more luck than me..

- dumbcow

is it 40.8285 ??

- anonymous

Nope :(

- anonymous

Sorry, I know it's irritating. So thanks so much for the help.

- dumbcow

ok, well i don't see anything wrong with 1st integral
your welcome

- anonymous

It might be an error on the teacher's part..

- anonymous

dumbcow please answer my question

- dumbcow

haha sure...can you figure out what we did wrong on this question

- anonymous

no im not in calc yet... still working on inishing precal

- anonymous

i looked at it but its a foreign language to me

- anonymous

I wish I was in pre-cal so bad..

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