# closed path integrals

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#1
I have this question to do but we never covered how to go about closed path integrals over a triangular region, absolutely stuck please help.

Closed path integral of P(x, y) dx + Q(x, y) dy = Greens theorem integral(∂Q/∂x − ∂P/∂y )dx dy.
By evaluating the integrals on both sides verify that the theorem holds in the case of the triangular region R with vertices at (0, 0), (−1, −1) and (1, −1) and the functions
P and Q defined by P(x, y) = xy and Q(x, y) = xy^2.
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4 weeks ago
#2
(Original post by Kaia19)
I have this question to do but we never covered how to go about closed path integrals over a triangular region, absolutely stuck please help.

Closed path integral of P(x, y) dx + Q(x, y) dy = Greens theorem integral(∂Q/∂x − ∂P/∂y )dx dy.
By evaluating the integrals on both sides verify that the theorem holds in the case of the triangular region R with vertices at (0, 0), (−1, −1) and (1, −1) and the functions
P and Q defined by P(x, y) = xy and Q(x, y) = xy^2.
You need to integrate along each of the sides of the triangle, going in a counter clockwise direction.

So we have that; where is the closed triangular path, and; is the path from (0,0) to (-1.-1) is the path from (-1,-1) to (1,-1) is the path from (1,-1) to (0,0)
Last edited by RDKGames; 4 weeks ago
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#3
thats what I did the first time but was unsure if it was right because there are 2 sloped sides so do they need to be handled differently than a strictly vertical or horizontal side? thanks.
0
4 weeks ago
#4
(Original post by Kaia19)
thats what I did the first time but was unsure if it was right because there are 2 sloped sides so do they need to be handled differently than a strictly vertical or horizontal side? thanks.
For strictly horizontal/vertical sides you have that either dx or dy is zero so your LHS simplifies nicely.

For the sloped side, e.g. let us take the path for .

Then the LHS for this side becomes Last edited by RDKGames; 4 weeks ago
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