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?

  • $\begingroup$ 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 $\endgroup$ Jul 25, 2015 at 0:29
  • $\begingroup$ then why they choose /$int xy dx + (x^2 + y^2) dy for? $\endgroup$
    – slowmonk
    Jul 25, 2015 at 0:48
  • $\begingroup$ 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 $\endgroup$ Jul 25, 2015 at 0:52
  • $\begingroup$ 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 $\endgroup$
    – slowmonk
    Jul 25, 2015 at 0:59
  • $\begingroup$ 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 $ $\endgroup$ Jul 25, 2015 at 1:06


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