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Math Help - Question about Green's Theorem

  1. #1
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    Question about Green's Theorem

    I am working on a take home quiz for my Calculus III class, and one of the problems says "Verify Green's Theorem for the line integral ∫xy dx + x dy, where C is the triangle with vertices (0,0) (1,1) (2,0)."

    Ok so this might be kind of a dumb question, but since it says "Verify" instead of "evaulate" am I supposed to do something different rather than find what the integral is equal to? I know how to evaluate this but I just am not sure if that means the same thing as verify. I don't want to misinterpret the problem and end getting a completely wrong answer.
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  2. #2
    Junior Member bondesan's Avatar
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    The Green's Theorem says that:

    \int_{\partial R} P dx + Q dy = \int\int_{R}\left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)dxdy

    In other words, it tells you that you can either calculate a double integral over the region R, or you can calculate a line integral over the contour wich encloses R. So to verify this theorem you must evaluate both and show that they're equal.

    I hope it helps.
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  3. #3
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    Thank you bondesan, I solved the double integral portion and got an answer of 1. However, I have no idea how to solve the ∫P dx + Q dy part. Could someone help out with this please?
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  4. #4
    Junior Member bondesan's Avatar
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    You can use the definition

    \displaystyle{\int_C F\cdot dr}

    Where C is a picewise smooth curve described by the parametric equation r(t) = x(t) \overset{\rightharpoonup }{i} + y(t) \overset{\rightharpoonup }{j} and F is a bidimensional vector field F(x,y)=P(x,y)\overset{\rightharpoonup }{i} +Q(x,y)\overset{\rightharpoonup }{j}

    By doing \dfrac{dr}{dt} = r'(t) we can rewrite the line integral as:

    \displaystyle{\int_C F(r(t))\cdot r'(t)~dt}

    Where we have a dot product. So for the problem you have F(x,y) = xy \overset{\rightharpoonup }{i}  + x^2 \overset{\rightharpoonup }{j} and r(t) will depend, because you should take three line integrals over the three lines that closes the triangle, and each one will have a different parametrization.

    So, let P1=(0,0),~P2=(2,0),~P3=(1,1).

    We can set the first line L1 as the one who links P1 to P2, L2 links P2 to P3, and L3 links P3 to P0. We followed a counter-clockwise direction.

    Now we should parametrize L1, L2, L3.

    For L1:

    You can see that this line has y=0 and 0 \leq x \leq 2, so maybe the parametrization r_1(t)=(2t,0) with t\in[0,1] should be nice for us.

    For L2:

    This line begins at the P2 then go to P3, so we want that for some r(t), when t=a we will be at P2and when t=b we will be at P3. You can use the formula r(t)=(1-t)P2 + tP3. It follows that r_2(t)=(2-t,t) with t\in[0,1].

    For L3:

    We can use the same formula again, so r(t)=(1-t)P3 + tP0 give us r_3(t)=(1-t,1-t) with t\in[0,1]



    Now that we have the three parametric equations for each line we can evaluate:

    \displaystyle{\int_{Li} F(r_i(t))\cdot r_i'(t)~dt} for i = 1,2,3

    Finally, we sum each value to get the final answer.

    I apologize if I did any mistake, I don't have any book around here by now and I had to remember the whole process - that's why it lacks some deep explanations. I did my best and I hope it helps. Maybe the other guys around here can add some thoughts about it.
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  5. #5
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    Your integral should be  - \int \int_D ( Q_x - P_y )~dydx because the direction of the path is clockwise , in this case ...

    Then it should be  - \int \int_D x dydx

    Note that the x-coordinates of the centroid of a region is defined as  \frac{  \int \int_D x dydx }{  \int \int_D  dydx }

    Therefore ,   - \int \int_D x dydx  = -  ( \text{ x-centroid }  ) ~ \cdot ( \text{ area })

     = - ( \frac{ 0 +  1 + 2 }{3} ) ( 1 ) = -1
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