Find the circulation using equation of motion

[MaxManus]

Homework Statement

v = ui + v j
u = cos(x)sin(y)
v = -sin(x)cos(y)
Find the circulation around randa to the square defined by: x = y = [-0.5*pi,0.5*pi]

Homework Equations

Is there a rule that says which sides on the square that get i and whics gets -i when you draw the square?

 

[HallsofIvy]

The direction of flow is given by the equation of motion itself, not by any "rule".

Assuming that v is the velocity vector of a flow, taking x=0.5pi gives v = cos(0.5pi)cos(y)i- sin(0.5pi)sin(y)j= -i. Taking x= -05pi gives v= i.

 

[MaxManus]

Thanks, so it is like this?

right vertical line -j, x = 0.5*pi
left vertival line j, x = -0.5*pi
top horizontal line i, y = 0.5*pi
bottom horizontal line -i , y = -0.5*pi

 

[HallsofIvy]

Thanks, so it is like this?

right vertical line -j, x = 0.5*pi
left vertival line j, x = -0.5*pi
top horizontal line i, y = 0.5*pi
bottom horizontal line -i , y = -0.5*pi

I realized I had miscopied your formula and rewrote what I had.
At x= 0.5pi, cos(x)= 0 and sin(x)= 1 so v= -cos(y)j.
At x= -04pi, cos(x)= 0 and sin(x)= -1 so v= cos(y)j.
At y= 0.5pi, cos(y)= 0 and sin(y)= 1 so v= -cos(x)i.
At y= -0.5pi, cos(y)= 0 and sin(y)= -1 so v= cos(x)i.

In particular, notice that the flow is across the boundary of the square- at each point on the boundary, the flow is at right angles to the boundary so the total flow around the square is 0.

 

[MaxManus]

Thank you.