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Math Help - [SOLVED] Calc III Line Integrals

  1. #1
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    [SOLVED] Calc III Line Integrals

    Hello, I can't figure this one out. Can someone please help me with this problem?


    Evaluate the integral

    <br />
\int \ {(3x + 2y)} dx + {(2x - y)} dy<br />

    along the stated curve

    a.) y = sin((pi*x)/2) from (0,0) to (1,1)

    b.) x = y^3 from (0,0) to (1,1)
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  2. #2
    MHF Contributor Calculus26's Avatar
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    See attachment
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    Thanks for the quick response.

    I did that for a. but the answer is supposed to be 3.
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  4. #4
    MHF Contributor Calculus26's Avatar
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    Note the fied is conservative with

    phi = 3/2*x^2 +2xy -y^2/2

    so we have path independence

    The answer to both a and b are 3
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  5. #5
    MHF Contributor Calculus26's Avatar
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    There must be a mistake in your integration as even working out the integral in my first post yields 3 as well.

    Oh man... I don't get how you got 3.

    Quote Originally Posted by Calculus26 View Post
    Note the fied is conservative with

    phi = 3/2*x^2 +2xy -y^2/2

    so we have path independence

    The answer to both a and b are 3

    How did you get phi? and that we have path independence?
    Last edited by mr fantastic; July 20th 2009 at 07:18 PM. Reason: Merged posts
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  6. #6
    MHF Contributor Calculus26's Avatar
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    See attachment for details of integration.

    To answer your second question have you studied Path Independence Yet?

    Let the field be F = f i + g j

    if df/dy = dg/dx F is the gradient of a scalar function phi.

    The line integral is then simply phi(endpoint) - phi(initialpoint)

    If you haven't studied this you will including how to find phi -- note for the phi I gave grad(phi) = F

    and phi(1,1)- phi(0,0) = 3

    For a complete discussion of path independence and conservative fields

    see my web site Line Integrals
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    Last edited by mr fantastic; July 20th 2009 at 07:19 PM. Reason: Merged posts
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  7. #7
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    No, we haven't yet. But I'm taking Calc III in summer school and everything moves so fast. I'm sure I'll be learning it tomorrow. Thank you sooo much. You were very helpful!
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