Kaia19
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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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RDKGames
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(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;

\displaystyle \oint_C = \int_{C_1} + \int_{C_2} + \int_{C_3}

where C is the closed triangular path, and;

C_1 is the path from (0,0) to (-1.-1)

C_2 is the path from (-1,-1) to (1,-1)

C_3 is the path from (1,-1) to (0,0)
Last edited by RDKGames; 4 weeks ago
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Kaia19
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(Original post by RDKGames)
You need to integrate along each of the sides of the triangle, going in a counter clockwise direction.

So we have that; \displaystyle \oint_C = \int_{C_1} + \int_{C_2} + \int_{C_3}

where C is the closed triangular path, and;

C_1 is the path from (0,0) to (-1.-1)

C_2 is the path from (-1,-1) to (1,-1)

C_3 is the path from (1,-1) to (0,0)
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.
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RDKGames
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(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 C_1: \begin{cases} x = -t \\ y = -t \end{cases} for 0 \leq t \leq 1.

Then the LHS for this side becomes

\displaystyle \int_{C_1} P dx + Q dy = \int_{t=0}^{t=1} \left( P \dfrac{dx}{dt} + Q \dfrac{dy}{dt} \right) dt
Last edited by RDKGames; 4 weeks ago
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