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Math Help - [SOLVED] Line integral

  1. #1
    MHF Contributor arbolis's Avatar
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    [SOLVED] Line integral

    I must evaluate I=\int_C F dr where C is described by r(t)=(t^2,t^3), 0\leq t\leq 1. F(x,y)=(e^x,xy).
    My attempt : |r'(t)|=\sqrt{4t^2+9t^4}.
    I=\int_0^1 (e^{t^{2}}+t^5)\sqrt{4t^2+9t^4}dt. I don't think I've made it right... Seems hard to solve.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by arbolis View Post
    I must evaluate I=\int_C F dr where C is described by r(t)=(t^2,t^3), 0\leq t\leq 1. F(x,y)=(e^x,xy).
    My attempt : |r'(t)|=\sqrt{4t^2+9t^4}.
    I=\int_0^1 (e^{t^{2}}+t^5)\sqrt{4t^2+9t^4}dt. I don't think I've made it right... Seems hard to solve.
    i believe you want \oint_C F \cdot dr, as in the dot product of the vector field F and the vector dr
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  3. #3
    Super Member Random Variable's Avatar
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    F(x,y) is a vector field, not a function.
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    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by Random Variable View Post
    F(x,y) is a vector field, not a function.
    Ah ok, then I should read how to proceed.

    Quote Originally Posted by Jhevon View Post
    i believe you want \oint_C F \cdot dr, as in the dot product of the vector field F and the vector dr
    I think you're right although the exercise didn't precise any \oint but an \int_C.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by arbolis View Post
    Ah ok, then I should read how to proceed.


    I think you're right although the exercise didn't precise any \oint but an \int_C.
    yes, i misused the symbol here. textbooks might not put that integral sign unless they want to specify it is a closed curve/contour, but i like the symbol
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  6. #6
    Super Member Random Variable's Avatar
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     \int_{C} F \cdot dr = \int_{a}^{b} \Big(F(r(t)) \cdot r'(t)\ \Big)dt

     \int_{0}^{1} \Big((e^{t^2}, t^{5}) \cdot (2t, 3t^2)\Big)dt

     \int_{0}^{1} (2te^{t^{2}} + 3t^{7}) \ dt
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