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anonymous
 4 years ago
**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 xaxis.
anonymous
 4 years ago
**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 xaxis.

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[2\pi \int\limits x^3/2 \sqrt ( (4+9x^4)/4\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1328069467445:dw to find surface area, we add up all the circumferences times dx \[SA = 2 \pi \int\limits_{a}^{b}R dx\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oops i forgot something...where did the sqrt come from?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0haha i have forgotten the SA formula, im just doing this on the fly is it that sqrt(1+f'(x)^2)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'm trying to type it out.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0integral of 2pi * original line * sqrt( 1 + (derivative of original line)^2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you 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
 4 years ago
Best ResponseYou've already chosen the best response.0what happened to the 2 under x^3 and the 4 under the sqrt?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i combined them sqrt(4) = 2 1/2 *1/2 = 1/4 multiplied by 2pi 2pi*1/4 = pi/2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you can take it from here, integrate sqrt(u) then plug in 4+9x^4 back in for u

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok so what do I bring back front from the u? 1/36?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0not sure what you mean i multiplied the 1/36 by the pi/2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0On the usubstituion.. Shouldn't something be brought out front like 1/36 + 4? From the du = 36x^3 + 4

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh no the derivative of a constant is 0 du = 36x^3 + 0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0du = 36x^3 dx dx = du/36x^3 > x^3 sqrt(u) du / 36x^3 = sqrt(u) du/36

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So would the final answer be.. 15.41599pi.. My hw says incorrect :(

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You changed the limits right? I had 0 to 152

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh no i kept the limits as 0 to 2, just plugged the x back in after integrating

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Wouldn't they need to be changed since we used usubstitution? I meant 0 to 148 btw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=integrate+pi%2F2*x%5E3*sqrt%284%2B9x%5E4%29+dx+from+0+to+2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah not working.. wtf

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0must have set it up wrong i guess ...maybe i missed something hold on i'll check out some resources on surface area integrals

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Hopefully you have more luck than me..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry, I know it's irritating. So thanks so much for the help.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok, well i don't see anything wrong with 1st integral your welcome

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It might be an error on the teacher's part..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dumbcow please answer my question

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0haha sure...can you figure out what we did wrong on this question

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no im not in calc yet... still working on inishing precal

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i looked at it but its a foreign language to me

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I wish I was in precal so bad..
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