anonymous
  • anonymous
**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.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[2\pi \int\limits x^3/2 \sqrt ( (4+9x^4)/4\]
dumbcow
  • 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
  • anonymous
Gotcha.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

dumbcow
  • dumbcow
oops i forgot something...where did the sqrt come from?
anonymous
  • anonymous
From the formula
dumbcow
  • dumbcow
haha i have forgotten the SA formula, im just doing this on the fly is it that sqrt(1+f'(x)^2)
anonymous
  • anonymous
yep
anonymous
  • anonymous
well
anonymous
  • anonymous
I'm trying to type it out.
anonymous
  • anonymous
integral of 2pi * original line * sqrt( 1 + (derivative of original line)^2
dumbcow
  • 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
  • anonymous
what happened to the 2 under x^3 and the 4 under the sqrt?
dumbcow
  • dumbcow
i combined them sqrt(4) = 2 1/2 *1/2 = 1/4 multiplied by 2pi 2pi*1/4 = pi/2
anonymous
  • anonymous
Oh sweet.. Gotcha!
dumbcow
  • dumbcow
you can take it from here, integrate sqrt(u) then plug in 4+9x^4 back in for u
anonymous
  • anonymous
ok so what do I bring back front from the u? 1/36?
dumbcow
  • dumbcow
not sure what you mean i multiplied the 1/36 by the pi/2
anonymous
  • anonymous
On the u-substituion.. Shouldn't something be brought out front like 1/36 + 4? From the du = 36x^3 + 4
dumbcow
  • dumbcow
oh no the derivative of a constant is 0 du = 36x^3 + 0
dumbcow
  • dumbcow
du = 36x^3 dx dx = du/36x^3 --> x^3 sqrt(u) du / 36x^3 = sqrt(u) du/36
anonymous
  • anonymous
So would the final answer be.. 15.41599pi.. My hw says incorrect :(
dumbcow
  • dumbcow
i get 52.14
anonymous
  • anonymous
didn't work :/
anonymous
  • anonymous
You changed the limits right? I had 0 to 152
dumbcow
  • dumbcow
oh no i kept the limits as 0 to 2, just plugged the x back in after integrating
anonymous
  • anonymous
Wouldn't they need to be changed since we used u-substitution? I meant 0 to 148 btw
dumbcow
  • 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
  • dumbcow
http://www.wolframalpha.com/input/?i=integrate+pi%2F2*x%5E3*sqrt%284%2B9x%5E4%29+dx+from+0+to+2
anonymous
  • anonymous
Yeah not working.. wtf
dumbcow
  • 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
  • anonymous
Ok thank you.
anonymous
  • anonymous
Hopefully you have more luck than me..
dumbcow
  • dumbcow
is it 40.8285 ??
anonymous
  • anonymous
Nope :(
anonymous
  • anonymous
Sorry, I know it's irritating. So thanks so much for the help.
dumbcow
  • dumbcow
ok, well i don't see anything wrong with 1st integral your welcome
anonymous
  • anonymous
It might be an error on the teacher's part..
anonymous
  • anonymous
dumbcow please answer my question
dumbcow
  • dumbcow
haha sure...can you figure out what we did wrong on this question
anonymous
  • anonymous
no im not in calc yet... still working on inishing precal
anonymous
  • anonymous
i looked at it but its a foreign language to me
anonymous
  • anonymous
I wish I was in pre-cal so bad..

Looking for something else?

Not the answer you are looking for? Search for more explanations.