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Thread: Line integral

  1. #1
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    Line integral

    Please check is correct

    Calculate the line integral along the curve C

    1) \int_C (x+2y)dx + (x-y)dy
    C: x = 2cost; y=4sent; (0 \leq t \leq \frac{ \pi}{4})

    My solution

    \int_0^{\frac{ \pi}{4}} (2cost,8sent).(2cost,-4sent)dt
    \int_0^{\frac{ \pi}{4}} (4cos^2t-32sin^2t)dt
    .
    .
    .
    = \frac{-7 \pi}{4} + \frac{9 \sqrt{2}}{8}

    2) \int_C -ydx + xdy
    C: y^2 = 3x
    (3,3) to (0,0)

    My solution

    \int_0^3 (- \sqrt{3t},t).(1,\frac{3}{2 \sqrt{3t}})dt
    .
    .
    .
    = -3
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  2. #2
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    Quote Originally Posted by Apprentice123 View Post
    Please check is correct

    Calculate the line integral along the curve C

    1) \int_C (x+2y)dx + (x-y)dy
    C: x = 2cost; y=4sent; (0 \leq t \leq \frac{ \pi}{4})

    My solution

    \int_0^{\frac{ \pi}{4}} (2cost,8sent).(2cost,-4sent)dt
    \int_0^{\frac{ \pi}{4}} (4cos^2t-32sin^2t)dt
    .
    .
    .
    = \frac{-7 \pi}{4} + \frac{9 \sqrt{2}}{8}
    This doesn't look right at all to me. If x = 2\cos t and y = 4\sin t then dx = -2\sin t\,dt and dy = 4\cos t\,dt. The integral is
    . . . . . . \begin{aligned}\int_0^{\pi/4}\bigl((2\cos t&{} + 8\sin t)(-2\sin t) + (2\cos t - 4\sin t)(4\cos t)\bigr)\,dt \\ &= \int_0^{\pi/4}(8\cos^2 t - 20\cos t\sin t - 16 \sin^2 t)\,dt \\ &= \ldots = 1-\pi.\end{aligned}

    Quote Originally Posted by Apprentice123 View Post
    2) \int_C -ydx + xdy
    C: y^2 = 3x
    (3,3) to (0,0)

    My solution

    \int_0^3 (- \sqrt{3t},t).(1,\frac{3}{2 \sqrt{3t}})dt
    .
    .
    .
    = -3
    That looks correct, except that if the path goes from (3,3) to (0,0) then the integral should be \int_3^0\!\!\!\ldots rather than \int_0^3\!\!\!\ldots , giving the answer +3 instead of 3.
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  3. #3
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    Thank you.
    1) I forgot to make the derivative.

    2) How do I know the limits of integral having two points?
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