# Thread: Circulation on a velocity field

1. ## Circulation on a velocity field

I have this exercise which asks me to calculate the circulation of a velocity field on V over the indicated path, this is it:

$\displaystyle \vec{V}=xy^2 \hat{i}+xe^{xy} \hat{j},y=x^2,x=0,y=1$

The thing is I don't know if I must use the path $\displaystyle y=x^2$ from (0,0) to (1,1), or if it reefers to three different paths with $\displaystyle y=x^2,x=0,y=1$ In that case I wouldn't know which extremes to use in the line integral. And the exercise that follows its similar, it gives a path, but it don't tells from where to where, but I think that in that case could be because the path is closed.

Bye and thanks!

2. Yes, "circulation" must occur on a closed path. This is actually a single path with three different parts. Draw the three graphs given and you should see which closed path this is referring to: the flow along $\displaystyle y= x^2$ from (0, 0) to (1, 1), along y= 1 from (1, 1) to (0, 1) and along x= 0 from (0, 1) to (0, 0) again.

You could also use Green's theorem to do this. In fact, it would be good idea to do it both ways as a check!

3. Originally Posted by Ulysses
$\displaystyle y=x^2,x=0,y=1$

The above expression is ill defined. Perhaps someone meant:

"The piece $\displaystyle \gamma$ of the parabola $\displaystyle y=x^2$ from $\displaystyle x=0$ to $\displaystyle x=1$".

Or equivalently:

$\displaystyle \gamma:\; \begin{Bmatrix} x=t\\y=t^2\end{matrix}\quad t\in{[0,1]}$

Edited: Sorry: I didin't read "circulation". My observation was only to the way of expressing the curve.

Regards.