# Theorem of Green

• Apr 26th 2009, 02:12 PM
Apprentice123
Theorem of Green
If C is the triangle of vertices (0,0), (1,0) and (1,1) oriented in direction anti-clockwise

$\displaystyle \int_C 2xydx + (x^2+x)dy = ?$

Solution:

$\displaystyle \frac{ \partial (x^2+x)}{ \partial x} - \frac{ \partial (2xy)}{ \partial y} = 1$

$\displaystyle 1 \int \int_D dxdy = \int_0^1 \int_0^1 dxdy = 1$

Correct ????
• Apr 26th 2009, 02:24 PM
Calculus26
Not quite

Green's Thm --- Line Integral = integral(dg/dx-df/dy) over the region enclosed by the C

you computed dg/dx-df/dy correctly = 1

so the line integral is the area of the triangle then which is 1/2
• Apr 26th 2009, 02:28 PM
Apprentice123
Quote:

Originally Posted by Calculus26
Not quite

Green's Thm --- Line Integral = integral(dg/dx-df/dy) over the region enclosed by the C

you computed dg/dx-df/dy correctly = 1

so the line integral is the area of the triangle then which is 1/2

$\displaystyle \int_a^b$
How do I determine a and b ?
• Apr 26th 2009, 02:30 PM
Jester
$\displaystyle \int_0^1\int_0^x dy\,dx$
• Apr 26th 2009, 02:33 PM
Apprentice123
Quote:

Originally Posted by danny arrigo
$\displaystyle \int_0^1\int_0^x dy\,dx$

How did you know $\displaystyle \int_0^1\int_0^x dy\,dx$ ?
• Apr 26th 2009, 02:58 PM
Calculus26
I know you know how to set up a double integral otherwise you wouldn't be doing line integrals and green's theorem.

the equation of the line from (0,0) to (1,1)

is y = x

so y varies from 0 to x as x varies from 0 to 1
• Apr 26th 2009, 03:03 PM
Apprentice123
Of course. thanks
• Apr 26th 2009, 03:09 PM
Jester
Quote:

Originally Posted by Apprentice123

How did you know $\displaystyle \int_0^1\int_0^x dy\,dx$ ?

Here is the region of interest. In order to set up

$\displaystyle \iint_R f(x,y)\,dA$ then we would need the following

$\displaystyle \int_{\text{point}}^{\text{point}} \int_{\text{curve}}^{\text{curve}} f(x,y)\,dy\,dx$.

Since the inner integral involves $\displaystyle dy$ then the curve to cuve in in the $\displaystyle y$ direction and in your case $\displaystyle y=0 \to x$ and hence the inside integral $\displaystyle \int_0^x$. Then outer integral is point to point and since the outer involves $\displaystyle dx$ then we are moving in the $\displaystyle x$ direction and from the picture of the region it $\displaystyle x = 0 \to 1$ and hence the outer integral of $\displaystyle \int_0^1.$

Hope that helps.
• Apr 26th 2009, 03:18 PM
Apprentice123
Quote:

Originally Posted by danny arrigo
Here is the region of interest. In order to set up

$\displaystyle \iint_R f(x,y)\,dA$ then we would need the following

$\displaystyle \int_{\text{point}}^{\text{point}} \int_{\text{curve}}^{\text{curve}} f(x,y)\,dy\,dx$.

Since the inner integral involves $\displaystyle dy$ then the curve to cuve in in the $\displaystyle y$ direction and in your case $\displaystyle y=0 \to x$ and hence the inside integral $\displaystyle \int_0^x$. Then outer integral is point to point and since the outer involves $\displaystyle dx$ then we are moving in the $\displaystyle x$ direction and from the picture of the region it $\displaystyle x = 0 \to 1$ and hence the outer integral of $\displaystyle \int_0^1.$

Hope that helps.

Thank you