Verify Stoke's theorem if $\mathbf{v} = z\mathbf{i} + x\mathbf{j} + y\mathbf{k}$ is taken over the hemispherical surface $x^2 + y^2 + z^2 = 1$ , $ z > 0$
Stoke's theorem states the following: $\int_C \mathbf{v} \cdot d\mathbf{r} = \int \int_S (\nabla \times \mathbf{v}) \cdot \mathbf{n} \, dS $
By conducting the line integral (Left hand side), I obtained an answer of $2\pi$ On the right hand side, I got a funky answer. First I calculated $\mathbf{n} = \nabla{\phi}/|\nabla{\phi}| $ where $\phi = z - (1 - x^2 - y^2)^{1/2}$ I then calculated $\nabla \times v$ and then calculated $(\nabla \times \mathbf{v}) \cdot \mathbf{n}$
Next I calculated $\int \int_S (\nabla \times \mathbf{v}) \cdot \mathbf{n} \, dS$ and I believe I get stuck here. Perhaps I did something wrong in an earlier step.
The value of $\phi$ should be $\phi=x^2+y^2+z^2=$constant. I think this is the mistake you made. Try solving the problem now.