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Multivariable integration problem

  1. Feb 3, 2012 #1
    1. The problem statement, all variables and given/known data

    Vector calculus

    It can be shown that the area of the surface described by the vector valued function
    r(s; t) between the limits a ≤ s≤ b and c ≤ t ≤ d is given by

    A=∫(from a to b) ∫(from c to d ) ‖(∂r/∂s)×(∂r/∂t)‖ dtds


    Find the surface area of the bowl described by
    r(s; t) = s cos(t)i + s sin(t)j + s^(2)k; 0 ≤ s ≤ 1; 0 ≤ t ≤ 2π:

    2. Relevant equations



    3. The attempt at a solution

    Ok so first off I've solved this problem but am unsure if I am correct. The final answer I came to is roughly 50. However my friend thinks its roughly 5.33. I'm sure he is incorrect because in the last few steps the integral required a u substitution, where he didn't change the limits of integration in this case instead of 1 to 0 the new limits became 5 to 1.

    if someone could check this I would be greatly appreciative.

    thanks for your help.
     
    Last edited: Feb 3, 2012
  2. jcsd
  3. Feb 3, 2012 #2
    Or is a change of the limits of integration even necessary in this problem? This is my only concern otherwise I'm confident my initial answer is correct.
     
  4. Feb 3, 2012 #3
    So if I use a u-substitution to solve a definite integral do i need to change the limits?
     
  5. Feb 3, 2012 #4

    SammyS

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    If no one else checks this I'll try to get to it after I get home.

    As for the question in the quotes text, YES, if you do a u-substitution to solve a definite integral, you should also change the limits of integration to be consistent with the substitution.

    Alternatively, use u-substitution to find the anti-derivative, of course changing back to the original variable before evaluating the definite integral.
     
  6. Feb 3, 2012 #5
    Ok, thanks Sam.

    Danny
     
  7. Feb 3, 2012 #6

    Dick

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    The norm of the two derivatives are both certainly not much more than 1. You are integrating over an area of 2*pi. 50 is certainly wrong. It's pretty easy to ballpark it that much.
     
  8. Feb 4, 2012 #7
    mmm I must have made a careless error.

    Danny
     
  9. Feb 4, 2012 #8

    Dick

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    Doing it in detail, I agree with your friend's 5.33.
     
  10. Feb 4, 2012 #9
    Hey Dick, did you have to use a u-substitution for the last step to this problem?
     
  11. Feb 4, 2012 #10

    Dick

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    Sure I did. Used a trig identity too. What did you do? I think it's about time you told us.
     
  12. Feb 5, 2012 #11
    well the last step is where i seem to be going wrong i think. The second equation I get when i integrate is 2pi S(4S^(2)+1)^(1/2)

    Which i then integrate using a u-substitution. still getting wrong answer
     
  13. Feb 6, 2012 #12

    Dick

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    That is the correct integral. But how are we supposed to know what is going wrong if you don't tell us what you got and how you got it?? I get around 5.33, your friend got around 5.33. Why don't you?
     
    Last edited: Feb 6, 2012
  14. Feb 6, 2012 #13
    Ok so integrating that integral i set

    u =4s^(2)+1
    du=8s ds
    (1/8)du = sds

    subbing that into the original integral and integrating i get pi/6 (4s^(2)+1)^(3/2
    and now i have to change my limits? From 1 to 0.
     
  15. Feb 6, 2012 #14
    so new limits will be

    u=4(1)^(2)+1 = 5
    and
    u=4(0)^(2)+1 = 1

    and using these new limits i get the wrong answer.
     
  16. Feb 6, 2012 #15
    however if i don't change the limits and keep them the same at 1 to 0 i get 5.33 :S, what am i missing?
     
  17. Feb 6, 2012 #16

    Dick

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    If you had written the answer in terms of u as pi/6 u^(3/2) then you would need to use the new u-limits. Since you changed the integral back to s, pi/6 (4s^2+1)^(3/2) you should use the original s limits, 0 to 1.
     
  18. Feb 6, 2012 #17
    Ah!! thanks for the help :)
     
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