I think I've made a huge mistake...
So from my understanding I can evaluate the area to the left of a curve using Green's theorem which states taht the line integral is = to this special double integral right?
Evaluate $\int_c F(r) \cdot dr$ counterclockwise around the boundary.
- $F = [x^{2}e^{y}, y^{2}e^{x}$ and R the rectangle with vertices: (0,0), (2,0), (2,3), (0,3)
So clearly the rectangle has an area of 6, unsure how this will play later.
so $$F1 = x^{2}e^{y} $$
$$F2 = y^{2}e^{x} $$
because $\frac{d}{dx} e^{y} = e^{y}$ and so treating $x^{2}$ as constant: $x^{2} \frac{d}{dy}e^{y} = x^{2}e^{y}$
$$\frac{d}{dy}F1 = x^{2}e^{y}$$ $$\frac{d}{dx}F2 = y^{2}e^{x}$$
$\therefore$
$$\int \int_R (x^{2}e^{y} - y^{2}e^{x} dx dy$$
but this doesn't look elegant or nice....