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Surface Integral confirmation

  1. Dec 4, 2013 #1
    Hello, everyone. I've used this forum several times in the past for help with problems or just to see other vantage points on a subject. This is my first post, so please bare with me in trying to format my post correctly. I'm studying for my final exam in Calculus III next week and am working some problem types I'm sure will be on there. The first one I had a bit of difficulty working through is this surface integral.

    1. The problem statement, all variables and given/known data
    Find the area of the surface defined by the part of the cone z=√{x2+y2} that lies between the plane y=x and y=x2.



    2. Relevant equations
    ∫|rxχry|dA = ∫√{x2/(x2+y2) + y2/(x2+y2) +1}dydx


    3. The attempt at a solution

    So I converted to polar coordinates:
    x=rcosθ y=rsinθ

    0 ≤ r ≤ tanθ/cosθ
    0≤θ≤ π/4


    And actually as I was typing this out..I realized I forgot to leave the sq.rt in my integrand...*sigh*
    My answer came out to 2...but I know that's wrong!

    Have I set it up correctly so far?

    ∫√{cos2θ + sin2θ +1}r drdθ


    Actually...looking at it now, I believe the only change to my answer if I did include the sq.rt, would be that my answer is √2?

    Thank you guy for any help; I'll double check my work with the sq.rt symbol in my calculations.
    I'm actually not worried about taking the integral, just the set up.
    My professor has the formula as ∫f(x,y,z(x,y))*√{(dz/dx)2+(dz/dy)2+1}dA but working through the concepts myself, I believe I set it up correctly?

    Thanks again!



    (I tried using the "complex" symbols and obviously didn't know how to make them look good...so spent 15 minutes rewriting most of this, so let me know if something doesn't look quite right!)
     
  2. jcsd
  3. Dec 4, 2013 #2
    Oh, my new answer is √{2}/2.
     
  4. Dec 4, 2013 #3

    LCKurtz

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    Polar coordinates certainly isn't the obvious choice given that the region in the xy plane has nothing to do with circles. That being said, you have the limits right for the first octant portion. Unless you only want the first octant, there is more in the second octant.

    I worked it in rectangular coordinates and got ##\frac {\sqrt 2} 6##, so one of us is wrong. Try it in rectangular and see what you get. And if you want the surface including the second octant, double the answer.
     
  5. Dec 4, 2013 #4
    I'll rework it in rectangular coordinates, but the problem doesn't state to only use the first octant, but I don't see where the other octant comes into the picture. If the Domain in x-y is only in the region between y=x and y=x^2, then it would only be in quadrant 1..unless we use the negative z values for the cone, huh? It doesn't state it in the problem, so I'll just make a note of it.

    I'll let you know what I get in rectangular coordinates.
     
  6. Dec 4, 2013 #5
    Oh, and I did it in polar coordinates to eliminate the denominators in the integrand; looking at this in rectangular coordinates is much more work..
     
  7. Dec 4, 2013 #6

    LCKurtz

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    Yes, ignore what I said about the second quadrant. Momentary brain malfunction. But you will find rectangular coordinates is very easy.
     
  8. Dec 4, 2013 #7
    I don't see where it's easier with the square root symbol grouping every into one term..
    Could you show me what you did?
     
  9. Dec 5, 2013 #8

    LCKurtz

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    Here's the xy integral you stated:

    What happens if you add those two fractions together under that square root?

    [Edit] Also, the integral you set up in polar coordinates is set up correctly and gives the same answer I got of ##\frac{\sqrt 2} 6## if it is correcty worked.
     
    Last edited: Dec 5, 2013
  10. Dec 5, 2013 #9
    Yeah, I meant to post last night when I saw that. It was sq.rt(2)/6! Thanks for the help.
     
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