Verify Stokes' Theorem I am having trouble verifying stokes theorem for the following surface and vector field. Let $F$ be $z\vec{i}+xy^2\vec{j}+x^2z\vec{k}$ and let $z=x^2+y^2$ be the surface, where $0 \leq z \leq 6$. So I took the top of this bowl (in the plane $z=6$)to be a disc of radius equal to $\sqrt{6}$. Stokes theorem says that the surface integral  of $curl(F)\cdot n dS$ where n is a normal vector to the surface, is equal to the closed line integral of $F\cdot dr$, i.e. the work done around the closed loop. When I evaluate the line integral (using polar coordinates integrating w.r.t. theta) I get $9\pi$. When I evaluate the surface integral, (again using polar coordinates using $dA=dr dr\theta$) I get $18\pi$.
Can anybody out there actually verify Stokes' theorem for this surface and field, and tell me where I am going wrong!
 A: The result of the line integral is correct, so the mistake must be with your calculation of the surface integral. You have $\nabla \times \vec F = \left( 0,1-2xz,y^2 \right)$ and with a surface of the form $z=g(x,y)$, the normal $\vec n$ is given by $\left(-g_x,-g_y,1\right)$. The projection of the surface onto the $xy$-plane is a disc centered in the origin and with radius $\sqrt{6}$, so you have to integrate over this disc $D$:
$$\iint_D \left( 0,1-2xz,y^2 \right) \cdot \left(-g_x,-g_y,1\right) \,\mbox{d}A$$
where $z=g(x,y)$ with $g(x,y)=x^2+y^2$, so:
$$\iint_D \left( 0,1-2x\left(x^2+y^2\right),y^2 \right) \cdot \left(-2x,-2y,1\right) \,\mbox{d}A = 
\iint_D \left( 4 x^3 y + 4 x y^3 + y^2 - 2 y \right)
 \,\mbox{d}x\,\mbox{d}y$$
You can choose to switch to polar coordinates or not, but you should find $9\pi$ either way.
Maybe this is sufficient to find your mistake? If not, perhaps you can show us your calculations.
A: Note that we have
$$\nabla \times \vec F=\hat y(1-2xz)+\hat zy^2$$
The outer surface element $\hat n\,dx\,dy$ is given by
$$\hat n\,dx\,dy=\left( -2\hat xx-2\hat yy+\hat z\right)\,dx\,dy$$
Therefore, we find that
$$\begin{align}
\int_S \nabla \times \vec F\cdot \hat n\,dS&=\int_{-\sqrt{6}}^{\sqrt{6}}\int_{-\sqrt{6-y^2}}^{\sqrt{6-y^2}} \,\,\,(4xyz-2y+y^2)\,dx\,dy\\\\
&=\int_0^{\sqrt 6}\int_0^{2\pi}(4r^4\cos(\phi)\sin(\phi)-2r\sin(\phi)+r^2\sin^2(\phi))\,d\phi\,r\,dr\\\\
&=\int_0^{\sqrt 6}\int_0^{2\pi}(r^2\sin^2(\phi))\,d\phi\,r\,dr\\\\
&=\int_0^{2\pi}\sin^2(\phi)\,d\phi\,\int_0^{\sqrt 6}r^3\,dr\\\\
&=\pi \frac{36}{4}\\\\
&=9\pi
\end{align}$$ 
which agrees with the line integral result reported in the OP.
