# Green's theorem: what does closed path integral mean?

I'm studying green's theorem and having brief idea about this theorem. but little bit confused with first example

http://tutorial.math.lamar.edu/Classes/CalcIII/GreensTheorem.aspx (goes to example 1)

if you look at the example it said evaluate $\int xy dx + x^2+y^2 dy$ where this equation came? and how this related with bottom picture?

• Exactly what is your question? The example simply uses green theorem to compute a line integral over a triangular path as a double integral over the region bounded by the path. The equation $(xy)dx + (x^2+y^2) dy$ has nothing to do with the picture, the picture only shows the path or region of integration Jul 25, 2015 at 0:29
• then why they choose /$int xy dx + (x^2 + y^2) dy for? Jul 25, 2015 at 0:48 • Well it is an example, they could have choosen anything of the form$\int P(x,y)dx+Q(x,y)dy$, it happens that with that particular example, by applying Greens theorem, the double integral that results is very easy to compute Jul 25, 2015 at 0:52 • are they choose$\int xy dx + x^2+y^2 dy$from the picture? I'm very confuse how they extract this equation and why they use it. i do not understand they can choose anything. probably i'm missing something Jul 25, 2015 at 0:59 • No, the picture is only to display the contour of integration, it has nothing to do with the integrand. Let$C$be the path define by the triangle with vertices$(0,0); (1,0)$and$ (1,2)$, if$C^o$denotes the region bounded by$C$, by Green's theorem $$\int_C P(x,y) dx + Q(x,y) dy = \iint_{C^o} \left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} \right)dA$$ Since$C^o$is the interior of the triangle$C$, in this case$\iint_{C^o} dA = \int_0^1\int_0^{2x} dy dx$. In your example you have that$P(x,y) = xy$and$Q(x,y) =x^2+y^2 \$ Jul 25, 2015 at 1:06