# Surface integral via divergence theorem and Stokes' theorem

I want to compute $\iint_{\Sigma}(\nabla\times F)\cdot \omega dS$ where $\Sigma:=\{(x,y,z)\in\mathbb{R}^3:\sqrt{x^2+y^2}=z, 2+\frac{x}{4}\leq z\leq 4\}$, $F=(y,z,x)$ and $\omega (-3,0,3)=\frac{\sqrt{2}}{2}(1,0,1)$.

1) Applying the divergence theorem I got $\iint_{\Sigma}(\nabla\times F)\cdot \omega dS=\iiint_{V(\Sigma)}\nabla\cdot (\nabla\times F)dV-\iint_{z=4\cap cone}(\nabla\times F)\cdot \omega dS-\iint_{z=2+x/2\cap cone}(\nabla\times F)\cdot \omega dS=0-0-(-16\pi )=16\pi$

2) To check the result I've also applied Stokes's theorem taking as C the circumference $\sqrt{x^2+y^2}=4$ and I get $\oint_C F\cdot dr=\int_{0}^{2\pi}(4\sin (t),4,4\cos (t))\cdot (-4\sin (t), 4\cos (t),0)dt- \iint_{z=2+x/2\cap cone} (\nabla\times F)\cdot \omega dS =16\pi -0=16\pi$

So, is what I did correct?

• $\Sigma$ is not a closed surface. When applying the divergence theorem, you need to subtract the flux through the planes $z=4$ and $z=2+x/4$. – Kuifje Jan 21 '17 at 18:50
• @Kuifje You are right, I applied the divergence theorem again taking into account what you said and I got $-16\pi$ as a result so I guess there's something wrong with the line integral. – lorenzo Jan 21 '17 at 19:52
• Your line integral is not necessarily wrong. But you have to compute a second one, the one at at the intersection between the plane $z=2+x/4$ and the cone. – Kuifje Jan 21 '17 at 19:59
• @Kuifje I subtracted $\iint_{S} (\nabla\times F) \cdot \omega dS=0$ (where $S$ is the intesection between the cone and the plane $z=2+x/4$) from $\oint_{C} F\cdot dr$ (since I understand that was equal to $\iint$ on the region of interest $+ \iint$ on the previously mentioned intersection of cone and plane) and I got $16\pi$, which is equal from what I got using the divergence theorem, so I think what I did is correct. – lorenzo Jan 22 '17 at 10:45

The contour $x^2+y^2=4$ does bound the open surface given by $z=\sqrt{x^2+y^2}$ with $2+x/4\le z\le 4$ plus the open (ellipse) surface $\sqrt{x^2+y^2}\le 2+x/4=z$. It also bounds the disk $x^2+y^2\le 4$, $z=4$.
But in applying Stokes's Theorem, the line integral either (i) does not include the part of the closed surface $\sqrt{x^2+y^2}\le 4$, $z=4$ or (ii) only accounts for the surface $\sqrt{x^2+y^2} \le 4$, $z=4$.