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

• In the end you need to take $z=\sqrt{a^2-x^2-y^2}$ not zero. Feb 21, 2022 at 16:11
• You have two errors. One is that $ds\ne dx dy$. The surface element is on the sphere, not in the $z=0$ plane. Then you mention Stokes, but I don't see where you are using it. It would transform your surface integral into a line integral. Feb 21, 2022 at 16:12

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 $$I=\iint_{S_1}(x,y,0)\cdot(0,0,1)dS_1=0$$
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 $$t\in [0,2\pi]$$.