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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 x-axis.

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  1. anonymous
    • 4 years ago
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    \[2\pi \int\limits x^3/2 \sqrt ( (4+9x^4)/4\]

  2. dumbcow
    • 4 years ago
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    |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
    • 4 years ago
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    Gotcha.

  4. dumbcow
    • 4 years ago
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    oops i forgot something...where did the sqrt come from?

  5. anonymous
    • 4 years ago
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    From the formula

  6. dumbcow
    • 4 years ago
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    haha i have forgotten the SA formula, im just doing this on the fly is it that sqrt(1+f'(x)^2)

  7. anonymous
    • 4 years ago
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    yep

  8. anonymous
    • 4 years ago
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    well

  9. anonymous
    • 4 years ago
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    I'm trying to type it out.

  10. anonymous
    • 4 years ago
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    integral of 2pi * original line * sqrt( 1 + (derivative of original line)^2

  11. dumbcow
    • 4 years ago
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    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
    • 4 years ago
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    what happened to the 2 under x^3 and the 4 under the sqrt?

  13. dumbcow
    • 4 years ago
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    i combined them sqrt(4) = 2 1/2 *1/2 = 1/4 multiplied by 2pi 2pi*1/4 = pi/2

  14. anonymous
    • 4 years ago
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    Oh sweet.. Gotcha!

  15. dumbcow
    • 4 years ago
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    you can take it from here, integrate sqrt(u) then plug in 4+9x^4 back in for u

  16. anonymous
    • 4 years ago
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    ok so what do I bring back front from the u? 1/36?

  17. dumbcow
    • 4 years ago
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    not sure what you mean i multiplied the 1/36 by the pi/2

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

  19. dumbcow
    • 4 years ago
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    oh no the derivative of a constant is 0 du = 36x^3 + 0

  20. dumbcow
    • 4 years ago
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    du = 36x^3 dx dx = du/36x^3 --> x^3 sqrt(u) du / 36x^3 = sqrt(u) du/36

  21. anonymous
    • 4 years ago
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    So would the final answer be.. 15.41599pi.. My hw says incorrect :(

  22. dumbcow
    • 4 years ago
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    i get 52.14

  23. anonymous
    • 4 years ago
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    didn't work :/

  24. anonymous
    • 4 years ago
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    You changed the limits right? I had 0 to 152

  25. dumbcow
    • 4 years ago
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    oh no i kept the limits as 0 to 2, just plugged the x back in after integrating

  26. anonymous
    • 4 years ago
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    Wouldn't they need to be changed since we used u-substitution? I meant 0 to 148 btw

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

  29. anonymous
    • 4 years ago
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    Yeah not working.. wtf

  30. dumbcow
    • 4 years ago
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    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
    • 4 years ago
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    Ok thank you.

  32. anonymous
    • 4 years ago
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    Hopefully you have more luck than me..

  33. dumbcow
    • 4 years ago
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    is it 40.8285 ??

  34. anonymous
    • 4 years ago
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    Nope :(

  35. anonymous
    • 4 years ago
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    Sorry, I know it's irritating. So thanks so much for the help.

  36. dumbcow
    • 4 years ago
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    ok, well i don't see anything wrong with 1st integral your welcome

  37. anonymous
    • 4 years ago
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    It might be an error on the teacher's part..

  38. anonymous
    • 4 years ago
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    dumbcow please answer my question

  39. dumbcow
    • 4 years ago
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    haha sure...can you figure out what we did wrong on this question

  40. anonymous
    • 4 years ago
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    no im not in calc yet... still working on inishing precal

  41. anonymous
    • 4 years ago
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    i looked at it but its a foreign language to me

  42. anonymous
    • 4 years ago
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    I wish I was in pre-cal so bad..

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