## The Area of a Triangle Using Greens Theorem

Green's Theorem says
$\oint_C P \: dx + Q \: dy = \int_R \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} dx \: dy$
The contour below
$C$
is made up of the three paths
$C_1 , \: C_2 , \: C_3$
which form the sides of the triangle.

The equation of
$C_1$
is
$(x,t)=(at, 0), \: (dx,dy) = (a dt, 0) , \: 0 \leq t \leq 1$

The eqquation of
$C_2$
is
$(x,t) = (b+(a-b)t, ht), \: (dx, dy) =( (a-b) dt, h dt ), \: 0 \leq t \leq 1$

The eqquation of
$C_3$
is
$(x,t) = (a-at,h ht), \: (dx, dy) = (-a dt, -bdt), \: 0 \leq t \leq 1$

Because
$y=dy=0$
on
$C_1$
the integral along
$C_1$
does not contribution to the contouur integral.
Take
$P=0, \: Q=x$
. Along
$C_2$

$\oint x \:dy =\int^1_0 (b+(a-b)t) h dt =[bht+(a-b)h \frac{t^2}{2} ]^1_0 =hb+(a-b) \frac{h}{2} =(a+b) \frac{h}{2}$

Along
$C_3$
,
$\oint x \: dy =\int^1_0 (a-at) (-h) dt = [-aht + \frac{aht^2}{2}]^1_0 =-ah+ \frac{ah}{2} = - \frac{ah}{2}$

The integral, and the area are the sum of thesed
$\frac{bh}{2}$