Green's Theorem

**Aryth** · July 2nd 2008, 05:43 PM

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.

**mr fantastic** · July 2nd 2008, 06:53 PM

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

**Aryth** · July 3rd 2008, 06:57 AM

Would you mind showing the iterations for the double integral?

It's just for my verification.

**Moo** · July 3rd 2008, 07:05 AM

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 :

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

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