How to find this double integral? Given $\vec F=y\hat i+(x-2xz)\hat j-xy\hat k$ evaluate
$$\iint_R(\nabla \times\vec F)\cdot \vec n dS $$
Where $S$ is surface represented by $x^2+y^2+z^2=a^2$ for $z\ge 0$
My attempt:
i found curl
$$\nabla\times \vec F=\left|\begin{matrix} \hat i & \hat j & \hat k\\ \frac{\partial }{\partial x}&\frac{\partial }{\partial y}&\frac{\partial }{\partial z} \\ y& x-2xz&-xy\end{matrix}\right|=(x,y,-2z)$$
normal vector
$$\vec n=\frac{\nabla g}{|\nabla g|}=\frac{(2x, 2y, 2z)}{2a}=(x/a, y/a, z/a)$$
used Stokes theorem andtaking projection on the XY plane: $z=0$
$$\iint_R(\nabla \times\vec F)\cdot \vec n dS=\iint (x,y,-2z)\cdot (x/a, y/a, z/a)dxdy$$
$$=\frac1a\int \int \left(x^2+y^2-2z^2\right)dxdy$$
putting limits of $x=-a$ to $x=a$ and $y=-\sqrt{a^2-x^2}$ to $y=\sqrt{a^2-x^2}$ and $z=0$
$$=\frac1a\int_{-a}^{a} \int_{\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}} \left(a^2-0\right)dxdy$$
$$=\pi a^3$$
I am not sure where I made mistake. please correct me and help solve this integration. Thanks
 A: As a direct computation your flux should be
$$\iint_S(\nabla \times\vec F)\cdot \vec n dS=\iint_{x^2+y^2\leq a^2} (x,y,-2f)\cdot (x/a, y/a, f/a)\sqrt{1+f_x^2+f_y^2}dxdy$$
where $z=f(x,y)=\sqrt{a^2-x^2-y^2}$.
Another way (easier): by applying Stokes' Theorem, we have
$$\iint_S(\nabla \times\vec F)\cdot \vec n dS=\int_{\gamma}\vec F\cdot d\vec s=\int_0^{2\pi}(y(t)x'(t)+x(t)y'(t)+0)dt=[x(t)y(t)]_0^{2\pi}=0$$
where $\gamma$ is the circle $x^2+y^2=a^2$ in the plane $z=0$ counterclockwise oriented, i.e. $\gamma(t)=(a\cos(t),a\sin(t),0)$ with $t\in [0,2\pi]$.
A: You can use Stokes' theorem twice:
$$I=\iint_S(\nabla\times \vec F)\cdot \vec n dS=\oint_B\vec F\cdot d\vec l=\iint_{S_1}(\nabla\times \vec F)\cdot \vec n_1 dS_1$$
Here $S$ and $S_1$ share the same boundary $B$. In your problem $B$ is the circle of radius $a$, centered in the origin, in the $z=0$ plane. If you want $S_1$ to be the disk of radius $a$, centered in the origin, in the $z=0$ plane, then $\vec n_1=\hat k$.
Then $$I=\iint_{S_1}(x,y,0)\cdot(0,0,1)dS_1=0$$
