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Math Help - calculate line integral and draw curve C

  1. #1
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    calculate line integral and draw curve C

    I am truly astonished to the amount of help I received, I'd like to thank Scott, Mr. Fantastic, Chris, and TheEmptySet for all this help you guys have given me. I now feel like I'm no longer alone...

    This one is also a tuffy, perhaps someone may want to show me the light?


    Question:

    Let C be the closed curve C1 + C2 , where C1 is given by

    r1 (t) = t i + t^2 j, o <= t <= 1

    r2 (t) = (2 - t) i + (2 - t) j , 1 <= t <= 2



    a) Draw a sketch of the curve C, with arrows to indicate orientation (im not sure if it's possible to do this online, but a description will be very very much appreciated)

    b) Calculate the line integral \int_C <br />
\mathbf{F}\cdot\,dr , where F is the vector field: F ( x, y ) = x^2 i + x y j
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  2. #2
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    Quote Originally Posted by mohammadmurtaza View Post
    I am truly astonished to the amount of help I received, I'd like to thank Scott, Mr. Fantastic, Chris, and TheEmptySet for all this help you guys have given me. I now feel like I'm no longer alone...

    This one is also a tuffy, perhaps someone may want to show me the light?


    Question:

    Let C be the closed curve C1 + C2 , where C1 is given by

    r1 (t) = t i + t^2 j, o <= t <= 1

    r2 (t) = (2 - t) i + (2 - t) j , 1 <= t <= 2



    a) Draw a sketch of the curve C, with arrows to indicate orientation (im not sure if it's possible to do this online, but a description will be very very much appreciated)

    b) Calculate the line integral \int_C <br />
\mathbf{F}\cdot\,dr , where F is the vector field: F ( x, y ) = x^2 i + x y j
    C_1 is the parabola y = x^2 starting at (0,0) and going to (1,1).  C_2 is a straight line starting at (1,1) and going back to (0,0). I'll do the first and you the second.

    If x =t and y = t^2 then dx = dt and dy = 2t\,dt. The line integral becomes

     <br />
\int \limits_{c_1} x^2 dx + xy\,dy = \int_0^1 t^2\, dt + t \cdot t^2 \cdot2t\,dt = \int_0^1 r^2 + 2t^4 \, dt = \frac{11}{15}<br />
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  3. #3
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    Thanks Danny!!!!!

    Thanks Danny, ur the MANN!!! I really appreciate it However I am stuck on the second one. Since I don't know how to write in the correct notation, can you show me the second one as well?
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  4. #4
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    Quote Originally Posted by mohammadmurtaza View Post
    Thanks Danny, ur the MANN!!! I really appreciate it However I am stuck on the second one. Since I don't know how to write in the correct notation, can you show me the second one as well?
    Quote Originally Posted by Danny View Post
    C_1 is the parabola y = x^2 starting at (0,0) and going to (1,1).  C_2 is a straight line starting at (1,1) and going back to (0,0). I'll do the first and you the second.

    If x =t and y = t^2 then dx = dt and dy = 2t\,dt. The line integral becomes

     <br />
\int \limits_{c_1} x^2 dx + xy\,dy = \int_0^1 t^2\, dt + t \cdot t^2 \cdot2t\,dt = \int_0^1 r^2 + 2t^4 \, dt = \frac{11}{15}<br />
    Here x = 2 - t, y = 2 - t, t = 1 \to 2 so dx = dt, dy=dt so the line integral becomes

     <br />
\int_1^2 (2-t)^2 dt + (2-t)(2-t)dt<br />

    although what I might suggest is to follow the line y = x ( dy=dx) from x = 1 \to 0 so

     <br />
\int_1^0 2x^2dx = - \frac{1}{3}<br />
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  5. #5
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    I love you!!!!

    Thanks soo much danny!!!
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