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Math Help - Integral

  1. #1
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    Integral

    Calculate the integral independent of the path:
    \int_C tg(y) dx + ysec^2(y) dy, C -> (1,0) to (2, \frac{\pi}{4})



    Answer:
    2
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  2. #2
    Super Member Random Variable's Avatar
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    Parametrize the curve.

     r(t) = (1-t)(1,0) + t(2, \frac {\pi}{4})= (1+t, \frac {\pi t}{4}), \ 0 \le t \le 1

     \int_{C} tg(y) \, \mathrm{d}x + ysec^{2}(y) \ \mathrm{d}y = \int_{C} tg(y) \ \mathrm{d}x + \int_{C} sec^{2}(y) \ \mathrm{d}y

     \int^{1}_{0} tg( \frac {\pi t}{4})\ \mathrm{d}t + \frac {\pi}{4}\int^{1}_{0} sec^{2}(\frac { \pi t}{4}) \ \mathrm{d}t

    But what's the function g?
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  3. #3
    Moo
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    Quote Originally Posted by Random Variable View Post
    Parametrize the curve.

     r(t) = (1-t)(1,0) + t(2, \frac {\pi}{4})= (1+t, \frac {\pi t}{4}), \ 0 \le t \le 1

     \int_{C} tg(y) \, \mathrm{d}x + ysec^{2}(y) \ \mathrm{d}y = \int_{C} tg(y) \ \mathrm{d}x + \int_{C} sec^{2}(y) \ \mathrm{d}y

     \int^{1}_{0} tg( \frac {\pi t}{4})\ \mathrm{d}t + \frac {\pi}{4}\int^{1}_{0} sec^{2}(\frac { \pi t}{4}) \ \mathrm{d}t

    But what's the function g?
    It's tg, and it's the tangent function
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  4. #4
    Super Member Random Variable's Avatar
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    Oh.

    Anyways, since the line integral is independent of the path taken, you're integrating along a curve in a gradient vector field.

    And I'm going to go out on a limb and say you meant  xsec^{2}y \ \mathrm{d}y . This would mean that the vector field is the gradient of the function f(x,y) = x \ tan(y) .

    Therefore the line integral is equal to  f(2, \frac {\pi}{4}) -f(1,0) = 2*1 - 0*0 = 2
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  5. #5
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    Ok. Thank you
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