## 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 x-axis.

1. anonymous

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

2. dumbcow

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

3. anonymous

Gotcha.

4. dumbcow

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

5. anonymous

From the formula

6. dumbcow

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

7. anonymous

yep

8. anonymous

well

9. anonymous

I'm trying to type it out.

10. anonymous

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

11. 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 $\rightarrow \pi/72 \int\limits_{0}^{2}\sqrt{u} du$

12. anonymous

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

13. dumbcow

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

14. anonymous

Oh sweet.. Gotcha!

15. dumbcow

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

16. anonymous

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

17. dumbcow

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

18. anonymous

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

19. dumbcow

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

20. dumbcow

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

21. anonymous

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

22. dumbcow

i get 52.14

23. anonymous

didn't work :/

24. anonymous

You changed the limits right? I had 0 to 152

25. dumbcow

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

26. anonymous

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

27. 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

28. dumbcow
29. anonymous

Yeah not working.. wtf

30. 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

31. anonymous

Ok thank you.

32. anonymous

Hopefully you have more luck than me..

33. dumbcow

is it 40.8285 ??

34. anonymous

Nope :(

35. anonymous

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

36. dumbcow

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

37. anonymous

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

38. anonymous

39. dumbcow

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

40. anonymous

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

41. anonymous

i looked at it but its a foreign language to me

42. anonymous

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