# Green's theorem

• Apr 14th 2010, 07:59 AM
Tekken
Green's theorem
$\int_{C}[f(x,y)dy - g(x,y)dx] = \int_{A}\int[\frac{\partial f}{\partial x} , \frac{\partial g}{\partial {y}}] dA$

Im trying to verify this theorem for the following line integral

$\int_{C}(x + y^{2})dy + (xy^{2}-y)dx$ where $C$ is the triangle with vertices $(1,1), (3,1), (3,3)$

I had no problem calculating the right hand side of green's theorem. However im struggling on the left hand side of the theorem.

Im thinking i need to get the line integral in terms of either x or y only ...?
• Apr 14th 2010, 08:52 AM
dedust
Quote:

Originally Posted by Tekken
$\int_{C}[f(x,y)dy - g(x,y)dx] = \int_{A}\int[\frac{\partial f}{\partial x} , \frac{\partial g}{\partial {y}}] dA$

Im trying to verify this theorem for the following line integral

$\int_{C}(x + y^{2})dy + (xy^{2}-y)dx$ where $C$ is the triangle with vertices $(1,1), (3,1), (3,3)$

I had no problem calculating the right hand side of green's theorem. However im struggling on the left hand side of the theorem.

Im thinking i need to get the line integral in terms of either x or y only ...?

Let $A=(1,1), B=(3,1), D=(3,3)$
along the line segment AB, $y=1$ and $x=t$ with $1 \leq t \leq 3$, hence $dy = 0$ and $dx = dt$, calculate the integral.

do the same thing for the line segment BD and DA

regard

DD
• Apr 14th 2010, 10:28 AM
Tekken
Quote:

Originally Posted by dedust
Let $A=(1,1), B=(3,1), D=(3,3)$
along the line segment AB, $y=1$ and $x=t$ with $1 \leq t \leq 3$, hence $dy = 0$ and $dx = dt$, calculate the integral.

do the same thing for the line segment BD and DA

regard

DD

Thanks alot,

I think i may have calculated the right hand side of green's theorem incorrectly.

i let $f(x,y) = x + y^{2}$ and $g(x,y) = xy^{2} - y$

this gave me $\frac{\partial f}{\partial x} = 1$and $\frac{\partial g}{\partial y} = 2xy - 1$

so filling these into the equation i got

$\int^{3}_{1}\int^{3}_{1}[(x+y^{2})-(xy^{2}-y)]dx.dy$

However im not sure if these limits are correct, aren't the inside limits of a double integral normally $x's$ or $y's$?

If you could answer this, it would be great