# Thread: Green's Theorem

1. ## Green's Theorem

Use Green's Theorem to evaluate the line integral along the given positively oriented curve.

$\int_C e^y ~dx + 2xe^y ~dy$

$C$ is the square with sides x=0, x=1, y=0, and y=1.

2. Originally Posted by Aryth
Use Green's Theorem to evaluate the line integral along the given positively oriented curve.

$\int_C e^y ~dx + 2xe^y ~dy$

$C$ is the square with sides x=0, x=1, y=0, and y=1.
Green's Theorem: $\oint_C f(x, y) \, dx + g(x, y) \, dy = \int \int_{R} \left( \frac{\partial g}{\partial x} - \frac{\partial f}{\partial y} \right) \, dx \, dy$ where R is the region bounded by C.

In your case, $f(x, y) = e^y \Rightarrow \frac{\partial f}{\partial y} = e^y$ and $g(x, y) = 2x e^y \Rightarrow \frac{\partial g}{\partial x} = 2 e^y$.

Therefore $\oint_C e^y ~dx + 2xe^y ~dy = \int_{y=0}^{1} \int_{x = 0}^{1} e^y \, dx \, dy = \int_{y=0}^{1} e^y \, dy = e - 1$.

3. Would you mind showing the iterations for the double integral?

It's just for my verification.

4. Originally Posted by Aryth
Would you mind showing the iterations for the double integral?

It's just for my verification.
You mean this one :

$\int_{y=0}^{1} \int_{x = 0}^{1} e^y \, dx \, dy$

?

Because $e^y$ doesn't contain any x, we can write it this way :

$=\underbrace{\left(\int_{y=0}^1 e^y \, dy\right)}_{=e-1} \cdot \underbrace{\left(\int_{x=0}^1 \, dx\right)}_{=1}$