If $$\vec{F}=-y\vec{i}+x\vec{j}+xyz\vec{k}$$ Let $S$ be the part of sphere $x^2+y^2+z^2=25$ below the plane $z=4$ and oriented positively. Find $$\int\int_{S}curl(F).\vec{dS}$$
I have found the outward unit normal to the sphere giving: $$\vec{n}=\frac{x\vec{i}+y\vec{j}+z\vec{k}}{5}$$
$$curl(F)=xz\vec{i}-yz\vec{j}+2\vec{k}$$
Now using parametric form of the sphere we have: $x=5\cos u\cos v$, $y=5\sin u\cos v$, $z=5\sin v$
we have $|\vec{n}|=25\cos v$
so $$\int\int_{S}curl(F).\vec{dS}=\int\int_{D(u,v)}z(x^2-y^2+2)|\vec{n}|dudv$$
so $$\int\int_{S}curl(F).\vec{dS}=\int_{u=0}^{2\pi}\int_{v=0}^{\pi}125\sin v\cos v(25\cos^2 v\cos 2u+2)dudv=250\pi$$
But ans is $-18\pi$ where i went wrong
You could also use Stokes' theorem: $$\iint_{S}\operatorname{curl}{F}\cdot \vec{dS}=\oint_{C}F\cdot\vec{dr}$$ where $C$ is the boundary curve of the surface, given by the parametrization $\vec{r}(t)=(3\cos t,3 \sin t,4)$ (check that this is indeed the boundary) oriented clockwise, $t\in[0,2\pi].$
Then the integral boils down to computing $$\oint_{C}F\cdot\vec{dr}=-\int_{0}^{2\pi}F(\vec r(t))\cdot\vec{r}'(t)dt$$