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enter image description hereVerify Green’s Theorem for the vector field F = x i + y j and the region Ω being the part below the diagonal y = 1 − x of the unit square with the lower left corner at the origin.

i) Sketch the region. Indicate the appropriate orientation of the boundary curve.

ii) Compute the line integral and the double integral given in Green’s Theorem and compare them.

****For part (ii) I could not understand how the line integral was computed. I broke it down to the 3 lines but I do not know hoe to set up the integral for the diagonal line.Would greatly appreciate your help!****

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  • $\begingroup$ can you include your working? $\endgroup$ – danimal May 12 '15 at 10:54
  • $\begingroup$ Welcome to MSE! Please see here for how to type math formulas: meta.math.stackexchange.com/questions/5020/… $\endgroup$ – KittyL May 12 '15 at 10:59
  • $\begingroup$ I put the answer document with the question. If you could help me understand the second part of the line integral, it would be greatly appreciated. $\endgroup$ – user239904 May 13 '15 at 5:37
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You want a diagonal line that starts at the point $(x,y)=(1,0)$ and ends at the point $(x,y)=(0,1)$. In the answer given, we take the parametrisation $x(t)=t$ and $y(t)=1-t$ for which if we want to start at $(1,0)$ and end at $(0,1)$ we must run $t$ from $1$ to $0$, rather than from $0$ to $1$ (this is a slightly odd choice, it might have been clearer to take $x(t)=1-t$, $y(t)=t$, and $t$ goes from $0$ to $1$, but it's really no big deal). With this parametrisation, $dx/dt=dt/dt=1$ and $dy/dt=d(1-t)/dt=-1$ so that $dx=dt$ and $dy=-dt$. Thus, by direct substitution, if we call the diagonal line starting at $(1,0)$ and ending at $(0,1)$ $\xi$ the line integral along the diagonal line is $$\int_{\xi}\left(xdx+ydy\right)=\int_1^0(tdt+(1-t)(-dt))=\int_1^0(2t-1)dt=\int_0^1(1-2t)dt=[t-t^2]^1_0=0,$$ where in the fourth to last equality we switch the limits of integration, introducing a minus sign in the integrand.

Does that clear things up?

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  • $\begingroup$ Definitely! Thank you so much for your help. $\endgroup$ – user239904 May 13 '15 at 6:33

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