Find $\int_{C}Fdr$ for the given $F$ and $C$

Question

I need to find $\int_{C}Fdr$.

$$F = (2x-y+4)i + (5y+3x-6)j$$

$C$ is the triangle with vertices $(0,0)$, $(3,0)$, $(3,2)$ traversed counterclockwise.

My first approach to doing this is splitting $C$ into $C1$, $C2$, $C3$

First is $C1$ from $(3,0)$ to $(0,0)$ so I plugged in the parametrization into the formula.

So my $F = (2(3-3t) -y +4)i + (5y + 3(3-3t) -6)j$

and $dr = (4-y)i + (5y - 6)j$

so then I have the integral...

$$\int_{0}^{1}((10 - 6t -y)i + (5y+3-9t)j) * ((4-y)i + (5y - 6)j)$$

(Note $*$ means dot product)

Solving this gives me just $C1$, I need to do this two more times to get $C$. This is seems way too long of a solution and there must be a much faster way of doing this using perhaps differential notation or something, I'm not sure.

Can anyone show me the better solution to this, I really don't think my answer is the way to go.

Answer 1

Use Green's theorem: $$\int_C Fdr = \iint_T\Big(\dfrac{\partial Q}{\partial x} - \dfrac{\partial P}{\partial y}\Big)dA,$$ where $F(x,y) = (P(x,y), Q(x,y)).$ Here, $C$ is the boundary of the triangle and $T$ is the region enclosed by the triangle.