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    "Evaluate the line integral

     \displaystyle\oint_C 2xy dy - x^{2} dx where the C is the perimeter of the triangle with the vertices (0,0), (1,0) and (1,1). Hence use double integration to verify Green's theorem."

    Now from the line integral I get the answer 1/3, the double integral with both limits going from 0 to 1.

    The partial differential being 2y and integrating this w.r.t y first gives y^2, hence 1.....then w.r.t x this gives x. Thus the final answer is 1.......where am I going wrong?
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    Thing about what both limits going from 0 to 1 actually represents: it represents integration over the square [0,1] x [0,1]. But you want to integrate over a triangle instead, whereby our integral limits are not so easy. Now, take our triangle, and fix x. For a fixed x, where does y range? It shouldn't be too hard to convince yourself that y ranges from 0 to x. Thus our integral becomes \displaystyle \int_0^1 \int_0^x \text{crap} \, \text{d}y  \text{d}x.
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    (Original post by Glutamic Acid)
    Thing about what both limits going from 0 to 1 actually represents: it represents integration over the square [0,1] x [0,1]. But you want to integrate over a triangle instead, whereby our integral limits are not so easy. Now, take our triangle, and fix x. For a fixed x, where does y range? It shouldn't be too hard to convince yourself that y ranges from 0 to x. Thus our integral becomes \displaystyle \int_0^1 \int_0^x \text{crap} \, \text{d}y  \text{d}x.
    That's amazing! is this always the case when it comes to a triangle?
 
 
 
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Updated: March 20, 2011
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