# Divergence Theorem when Surface isn't closed

So we essentially want to evaluate $$\iint_S \vec{F} \cdot d\vec{S},$$ where $\vec{F} = \langle 2x+y, x^2+y, 3z \rangle$ and $S$ is the cylinder $x^2+y^2=4$, between the surfaces $z=0$ and $z=5$.

We have that the cylinder is open at the top and the bottom. Therefore, we cannot readily apply Stoke's theorem. We need to subtract the contributions given by the flux through the top and the bottom, from the volume integral. If we let the closed surface of the cylinder be represented by $S$, the bottom surface represented by $S_1$ and the top surface by $S_2$, we have that $$\iint_S \vec{F} \cdot n dS + \iint_{S_1} \vec{F} \cdot ndS + \iint_{S_2} \vec{F} \cdot nds = \iiint_V \nabla \cdot \vec{F} dV.$$ Computing the RHS gives us that $$\iiint_V \nabla \cdot \vec{F} dV = 120 \pi.$$ How do we compute $$\iint_{S_1} \vec{F} \cdot ndS$$ and $$\iint_{S_2} \vec{F} \cdot ndS ?$$ Perhaps $$\iint_{S_1} \vec{F} \cdot ndS = \iint_{S_1} \vec{F} \cdot \langle 0, 0, -1 \rangle dS = \iint_{S_1} -3z dS?$$

At the bottom of the cylinder, $z=0$. Since $F_z(x,y,0)=0$ we find

\begin{align} \int_{S_1} \vec F \cdot \hat n\, dS&=\int_{S_1}\vec F(x,y,0)\cdot \hat z\,dS\\\\ &=0 \end{align}

and there is no flux contributed.

At the top of the cylinder, $z=5$. Since $F_z(x,y,5)=15$ we find

\begin{align} \int_{S_2} \vec F \cdot \hat n\, dS&=\int_{S_2}\vec F(x,y,5)\cdot \hat z\,dS\\\\ &=\int_0^{2\pi}\int_0^2 (15)\,\rho\,d\rho\,d\phi\\\\ &=60\pi \end{align}

Putting it all together, we find

\begin{align} \int_S \vec F\cdot \hat n \,dS&=\int_V \nabla \cdot \vec F\,dV-\int_{S_2} \vec F \cdot \hat n\, dS-\int_{S_1} \vec F \cdot \hat n\, dS\\\\ &=120\pi-60\pi-0\\\\ &=60\pi \end{align}

• What do you mean by $F_z(x,y,z)$? Why is there a subscript $z$? – user319128 Jun 4 '16 at 4:07
• Jordan, $F_z=\vec F \cdot \hat z$ is the axial (i.e., $z$) component of the vector field $\vec F$. – Mark Viola Jun 4 '16 at 14:05