Problem with Curl of $M\hat{i} + N\hat{j}$

Question

When I take the curl of above vector I get $\partial N/\partial x$ - $\partial M/\partial y$ . But if I do take the curl in 3×3 matrix then something different comes back (As I'm in mobile I'm unable to write this in mathjax). Then what is real curl of the vector? Where is the mistake I have made.

Answer 1

Using the definition from the link that you presented in the comments, one gets $$ \text{curl}(F) = \left(\partial_x N - \partial_y M\right) \hat{k}. $$ Here it is assumed that $M=M(x,y)$ and $N=N(x,y)$ are independent of $z$.

On the other hand, I'm assuming that the $3 \times 3$ matrix (or rather, determinant) you are referring to is $$ \text{curl}(F) = \begin{vmatrix} \hat i & \hat j & \hat k \\ \partial_x & \partial_y & \partial_z \\ F_i & F_j & F_k \end{vmatrix}, $$ which, for $F = M\hat i + N \hat j + 0\hat k$, becomes \begin{align} \text{curl}(F) &= \begin{vmatrix} \hat i & \hat j & \hat k \\ \partial_x & \partial_y & \partial_z \\ M & N & 0 \end{vmatrix} = (0 - \partial_z N)\hat i - (0-\partial_z M)\hat j + (\partial_x N -\partial_y M)\hat k = (\partial_x N -\partial_y M)\hat k, \end{align} where $\partial_z M = \partial_z N = 0$ since $M$ and $N$ were assumed to be independent of $z$.

Answer 2

Remember that curl of a vector field in $\mathbb{R}^3$ is a vector in $\mathbb{R}^3$ and not a scalar. Now, if we assume both M and N are not functions of $z$ but only of $x$ and $y$ we get your solution but there is a $\hat{k}$ missing. If you evaluate the curl matrix (3x3) taking that your z-component of the vector field is zero and that both M and N are not functions of $z$ but only of $x$ and $y$ then you should get back the same, as rest of the terms go to zero!