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closed path integrals

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.
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;

∮C=∫C1+∫C2+∫C3\displaystyle \oint_C = \int_{C_1} + \int_{C_2} + \int_{C_3}

where CC is the closed triangular path, and;

C1C_1 is the path from (0,0) to (-1.-1)

C2C_2 is the path from (-1,-1) to (1,-1)

C3C_3 is the path from (1,-1) to (0,0)
(edited 3 years 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; ∮C=∫C1+∫C2+∫C3\displaystyle \oint_C = \int_{C_1} + \int_{C_2} + \int_{C_3}

where CC is the closed triangular path, and;

C1C_1 is the path from (0,0) to (-1.-1)

C2C_2 is the path from (-1,-1) to (1,-1)

C3C_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.
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 C1:{x=−ty=−tC_1: \begin{cases} x = -t \\ y = -t \end{cases} for 0≤t≤10 \leq t \leq 1.

Then the LHS for this side becomes

∫C1Pdx+Qdy=∫t=0t=1(Pdxdt+Qdydt)dt\displaystyle \int_{C_1} P dx + Q dy = \int_{t=0}^{t=1} \left( P \dfrac{dx}{dt} + Q \dfrac{dy}{dt} \right) dt
(edited 3 years ago)

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