If curl is $xy \hat{i} + xy \hat{j} + -(x+y)z \hat{k}$ what is the vector function?

Question

If curl of a vector is $xy \hat{i} + xy \hat{j} + -(x+y)z \hat{k}$ what is the vector ? How to approach the problem. Is there specific method to get the vector function ? Only thing I get is the usual curl matrix or curl formula. How would I get the the vector by comparing $(\partial A_z/\partial y - \partial A_x/\partial z) $ = xy $(\partial A_z/\partial x - \partial A_x /\partial z)$ = xy and $(\partial A_y/\partial x - \partial A_x/\partial y )$ = -(x+y)z .

Answer 1

One way to approach this is to convert to an equivalent differential form and then apply the algorithm described in this answer to find an antiderivative of it. In detail, apply the Hodge star operator to the 1-form $A\,dx+B\,dy+C\,dz$ to get $\eta = A\,dy\wedge dz + B\,dz\wedge dx + C\,dx\wedge dy$. Steps 1 and 2 of the algorithm produce the form $$\left[tzB(tx,ty,tz)-tyC(tx,ty,tz)\right]\,dt\wedge dx + \left[txC(tx,ty,tz)-tzA(tx,ty,tz)\right]\,dt\wedge dy + \left[tyA(tx,ty,tz)-txB(tx,ty,tz)\right]\,dt\wedge dz$$ and so a vector field with curl $(A,B,C)$ is $$\int_0^1 \left[zB(tx,ty,tz)-yC(tx,ty,tz),xC(tx,ty,tz)-zA(tx,ty,tz),yA(tx,ty,tz)-xB(tx,ty,tz)\right]\,t\,dt.$$

In this problem, we have $A:(x,y,z)\mapsto xy$, $B:(x,y,z)\mapsto xy$ and $C:(x,y,z)\mapsto-(x+y)z$, so, after some simplification, $$\mathbf F(x,y,z) = \left[(2x+y)yz\,\mathbf i - (x+2y)xz\,\mathbf j - (x-y)xy\,\mathbf k\right]\int_0^1 t^3\,dt \\ = \frac14\left[(2x+y)yz\,\mathbf i - (x+2y)xz\,\mathbf j - (x-y)xy\,\mathbf k\right].$$

There are, of course, other antiderivatives. Just as you can add an arbitrary constant to the antiderivative of a single-variable scalar function to get another antiderivative, you can add any irrotational vector field to $\mathbf F$ to get another vector field with the same curl.