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Math Help - Using symmetry for curve integrals...

  1. #1
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    Using symmetry for curve integrals...

    Problem: Use symmetry considerations to find the following. (Picture of curves attached.)

    (a) Let C be the polygonal curve shown in Diagram (a).

    Compute \oint (e^{x^2} - 2xy)dx + (2xy - x^2)dy.

    (b) Let C be the curve shown in Diagram (b); you might visualize as a racetrack with two semicircular ends.

    Compute \oint (4x^3y - 3y^2)dx + (x^4 + e^{\sin(y)})dy.

    For (a), my method was to split the curve in five curves, C_1 \to C_5, where
    C_1 goes from (-1,1) to (3,1),
    C_2 goes from (3,1) to (4,2),
    C_3 goes from (4,2) to (3,3),
    C_4 goes from (3,3) to (-1,3), and
    C_5 goes from (-1,3) to (-1,1).

    Then I parametrized each of the curves as follows (where 0 \leq t \leq 1 for each curve):
    C_1: x = 4t - 1, y = 1, dx = 4dt, dy = 0,
    C_2: x = t + 3, y = t + 1, dx = dt, dy = dt,
    C_3: x = 4 - t, y = t + 2, dx = -dt, dy = dt,
    C_4: x = 3 - 4t, y = 3, dx = -4dt, dy = 0, and
    C_5: x = -1, y = 3 - 2t, dx = 0, dy = -2dt.

    Then the next step would be to split the integral into five integrals, one for each curve, and add them together. But the problem is the e^{x^2} term that hinders my integration attempts after the parametrization, and I can't evaluate the integral.

    I assume that there must be a better approach to doing this problem since I haven't actually used symmetry as it asks. But I'm not exactly sure how to apply symmetry to curve integrals.
    Attached Thumbnails Attached Thumbnails Using symmetry for curve integrals...-problem.jpg  
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  2. #2
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    I'm pretty sure you just want to simplify the region by cutting in in half over its axis of symmetry and then double the result.


    It's been awhile since I took Calculus III though so I could be wrong.
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  3. #3
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    \oint{e^{x^2}}dx\:=\:0 - due to symmetry
    \int_{-1}^{4}{e^{x^2}}dx\:+\int_{4}^{-1}{e^{x^2}}dx\:=\:0
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  4. #4
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    So if I use that concept in looking at my first four curves (since dx = 0 in the fifth curve), I can apply it as follows:

    \int_{-1}^3 {(e^{x^2})dx} + \int_{3}^4 {(e^{x^2})dx} + \int_{4}^3 {(e^{x^2})dx} + \int_{3}^{-1} {(e^{x^2})dx}

    = \left( \int_{-1}^3 {(e^{x^2})dx} + \int_{3}^{-1} {(e^{x^2})dx} \right)  + \left(\int_{3}^4 {(e^{x^2})dx} + \int_{4}^3  {(e^{x^2})dx} \right)

    = \left(\int_{-1}^3 {(e^{x^2})dx} - \int_{-1}^{3} {(e^{x^2})dx} \right)  + \left(\int_{3}^4  {(e^{x^2})dx} - \int_{3}^4  {(e^{x^2})dx} \right) = 0

    So the problem then can be solved as I had originally set it up, with the only difference being that I need to now solve the closed integral:

    \oint{(-2xy)dx + (2xy - x^2)dy}

    instead of the original, correct?
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  5. #5
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    Could I use the exact same logic to also show that \oint{(-2xy)dx} = 0 by symmetry to simplify the integral further?
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  6. #6
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    Upon further reflection, I figured that it can't work with respect to -2xy since the integral has a y in it and I am only integrating with respect to x for symmetrical purposes. So then y would be in the final answer, which isn't good.

    However, after even more thinking, the y in the final answer is actually going to end up being irrelevant because of symmetry, making the sum total zero for that term.

    (Sorry for the three posts in a row. I'm not trying to artificially bump the thread. I just wanted to share my thoughts on the rest of the problem.)
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