# Circulation

• May 2nd 2009, 11:43 AM
Apprentice123
Circulation
Calculate the circulation of vector field F arrond $\alpha$

1) $F(x,y) = (x-y)i + (x+y)j$ and $\alpha$ is the closed polygonal ABCDA traveled once in anticlockwise joining the points $A(1,0)$, $B(1,2)$, $C(0,2)$ and $D(0,1)$.

My solution:
Using green i have 2
The circulation is $\frac{1}{2}(A1+A2) = \frac{3}{4}$
A1 = $1*1=1$
A2 = $\frac{1*1}{2}$

Is correct ?

2) $F(x,y) = (x+2xy)i + (y+x^2)j$ and $\alpha : |x| + |y| = 1$
• May 2nd 2009, 12:05 PM
TheEmptySet
Quote:

Originally Posted by Apprentice123
Calculate the circulation of vector field F arrond $\alpha$

1) $F(x,y) = (x-y)i + (x+y)j$ and $\alpha$ is the closed polygonal ABCDA traveled once in anticlockwise joining the points $A(1,0)$, $B(1,2)$, $C(0,2)$ and $D(0,1)$.

My solution:
Using green i have 2
The circulation is $\frac{1}{2}(A1+A2) = \frac{3}{4}$
A1 = $1*1=1$
A2 = $\frac{1*1}{2}$

Is correct ?

2) $F(x,y) = (x+2xy)i + (y+x^2)j$ and $\alpha : |x| + |y| = 1$

I think that your $\frac{1}{2}$ is supposed to be a 2

for number 2

$\frac{\partial }{\partial x}(y+x^2)-\frac{\partial }{\partial y} =2x-2x=0$

Since the above is a closed curve (a Dimond) Greens theorem says it is zero.
• May 2nd 2009, 12:21 PM
Apprentice123
Quote:

Originally Posted by TheEmptySet
I think that your $\frac{1}{2}$ is supposed to be a 2

for number 2

$\frac{\partial }{\partial x}(y+x^2)-\frac{\partial }{\partial y} =2x-2x=0$

Since the above is a closed curve (a Dimond) Greens theorem says it is zero.

If green = 2 of the area nos is $\frac{1}{2}$ ?
• May 2nd 2009, 02:59 PM
TheEmptySet
Quote:

Originally Posted by Apprentice123
Calculate the circulation of vector field F arrond $\alpha$

1) $F(x,y) = (x-y)i + (x+y)j$ and $\alpha$ is the closed polygonal ABCDA traveled once in anticlockwise joining the points $A(1,0)$, $B(1,2)$, $C(0,2)$ and $D(0,1)$.

$\iint \left( \frac{\partial }{\partial x}(x+y)-\frac{\partial }{\partial y}(x-y) \right)dA=\iint 2 dA=2\iint dA$

The area of the region you calculated is correct $A=\frac{3}{2}$

$\iint 2 dA=2\iint dA=2\cdot \frac{3}{2}=3$
• May 2nd 2009, 03:03 PM
Apprentice123
Quote:

Originally Posted by TheEmptySet
$\iint \left( \frac{\partial }{\partial x}(x+y)-\frac{\partial }{\partial y}(x-y) \right)dA=\iint 2 dA=2\iint dA$

The area of the region you calculated is correct $A=\frac{3}{2}$

$\iint 2 dA=2\iint dA=2\cdot \frac{3}{2}=3$

I calculate $\frac{1}{2}(1+\frac{1}{2}) = \frac{3}{4}$
• May 2nd 2009, 03:08 PM
TheEmptySet
Quote:

Originally Posted by Apprentice123
I calculate $\frac{1}{2}(1+\frac{1}{2}) = \frac{3}{4}$

How??

Attachment 11235

Where does the 1/2 come from??
• May 2nd 2009, 03:10 PM
Apprentice123
Quote:

Originally Posted by TheEmptySet
How??

Attachment 11235

Where does the 1/2 come from??

Ohhh yes, thank you