Question

Find the circulation done by the velocity field V(x,y)= -y/[(x+1)^2 +4y^2]i + (x+1)/[(x+1)^2 +4y^2]j around the positively oriented square with vertices (2, 0), (0, 2), (−2, 0), and (0, −2).

Answer 1

The circulation of V around the given path (call it C) is simply the line integral of V along C. Since C is a closed curve, we can use Green's theorem.

`\displaystyle \int_C \vec V(x,y) \cdot d\vec r = \iint_{\mathrm{int}(C)} (\partial)/(\partial x)\left[(x+1)/((x+1)^2+4y^2)\right] - (\partial)/(\partial y)\left[-(y)/((x+1)^2+4y^2)\right] \, dx\,dy`

where int(C) is the interior or region bounded by C.

We have

`(\partial)/(\partial x)\left[(x+1)/((x+1)^2+4y^2)\right] = (4y^2 - (x+1)^2)/(((x+1)^2+4y^2)^2)`

`(\partial)/(\partial y)\left[-(y)/((x+1)^2+4y^2)\right] = (4y^2-(x+1)^2)/(((x+1)^2+4y^2)^2)`

so the double integral and hence the circulation is zero.