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Setup an integral for the curve

  1. Feb 2, 2013 #1
    1. The problem statement, all variables and given/known data
    for the curve x= sqrt(64-y^2), -4<=y<=4

    (identify from multiple choice
    1. setup an integral for the curve.
    2. identify the graph
    3. find the length of the curve.


    2. Relevant equations
    x = sqrt(64-y^2), -4<=y<=4


    3. The attempt at a solution
    1. dx/dy = y*sqrt(64-y^2)
    2. (dx/dy)^2 = y^2 * (64-y^2)
    3. integral (-4 to 4) sqrt(1 + 64y^2 - y^4)

    1. however, the answer for integral for the length is
    L=Integral(-4,4) 8(64-y^2)^(-1/2) dy. i'm not sure how they got that?

    2. i'm not sure how to graph this function.
     
  2. jcsd
  3. Feb 2, 2013 #2

    tiny-tim

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    hi whatlifeform! :smile:
    no, you've correctly got the 2y and the 1/2,

    but you haven't differentiated the function: the √ should become 1/2 1/√ :wink:

    (generally, ()n in the chain rule becomes n()n-1)
     
  4. Feb 2, 2013 #3
    what do you mean?
     
  5. Feb 2, 2013 #4
    i've got L = integral (-4,4) sqrt (1 + (y^2)/(64-y^2)) dy

    however, i'm not sure how to get to: L=Integral(-4,4) 8(64-y^2)^(-1/2) dy
     
  6. Feb 2, 2013 #5
    update: got it solved. but i'm still not sure how to graph this function.
     
  7. Feb 2, 2013 #6

    haruspex

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    What do you get if you square both sides to get rid of the square root? Do you recognise the form that results?
     
  8. Feb 3, 2013 #7

    tiny-tim

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    ie, what is x2 = 64 - y2 ? :smile:
     
  9. Feb 3, 2013 #8
    the graph of?
     
  10. Feb 3, 2013 #9

    dx

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    x2 + y2 = a2

    is an origin-centered circle with radius a
     
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