Two small Divergence Theorem questions

Question

$1)$Use the divergence theorem to evaluate the flux $\int_S \underline{F} \cdot \underline{dS} $where the vector field $\underline{F} = xz\, e_1 + 3x\, e_2 - 2z \,e_3$

i)$S$ is the closed cylinder bounded by the surface $x^2 + y^2 = 1$ and the planes $z=0$ and $z=3$.

ii)$S$ is the open curved cylindrical surface $x^2 + y^2 = 1,\,\, 0 \leq z \leq 3$

i) is fine, but I am unsure about how top use the div theorem for ii). What I did was compute $\int_V \nabla \cdot F\,dV$ for both the top and bottom circles and subtract their contributions from i). This gives an answer of $3 \pi/2$. My worry is that when I do the actual surface integral I get something completely different.

My working for the surface integral: On the curved surface; $\underline{dS} = cos \phi\, e_1 + sin \phi\,e_2\,d\phi dz $ So, $F \cdot dS = (cos^2 \phi z + 3 cos\phi sin \phi)d \phi dz. $ Integrating with $z \in [0,3]$ and $\phi \in [0,2\pi]$ gives $9\pi/2.$

Similarly, on top surface I get $F \cdot dS = -6 d\phi d\rho$ which gives $-12 \pi$ and $0$ contribution from the bottom surface. Adding these do not give the $-3\pi/2$ attained in i)

Many thanks.

Christian Blatter · Accepted Answer · 2013-03-19T21:12:37.630

2

The divergence theorem applies to a body $B\subset{\mathbb R}^3$ and its outward oriented boundary $\partial B=:S$. It says that for any field ${\bf F}$ defined in a neighborhood of $B$ one has $$\int_S {\bf F}\cdot d\vec \omega=\int_B{\rm div}({\bf F})\ dV\ .$$ (i) Computing the divergence we obtain $${\rm div}({\bf F}(x,y,z)=z-2\ .$$ Here $S=\partial B$ and therefore $$\int_S {\bf F}\cdot d\vec \omega=\int_B{\rm div}({\bf F})\ dV=\left({3\over2}-2\right){\rm vol}(B)=-{3\pi\over2}\ .$$ (ii) The boundary $\partial B$ of $B$ consists of three pieces: The bottom surface $S_1$, the top surface $S_2$, and the cylindrical surface $S_3$. We are told to compute the flow integral $\int_S {\bf F}\cdot d\vec \omega$ for $S:=S_3$.

According to the divergence theorem $$\int_{S_3} {\bf F}\cdot d\vec \omega= \int_B{\rm div}({\bf F})\ dV-\int_{S_1} {\bf F}\cdot d\vec \omega-\int_{S_2} {\bf F}\cdot d\vec \omega\ .$$

On the bottom surface $S_1$ the outward normal is $(0,0,-1)$, and the scalar surface element is ${\rm d}\omega={\rm d}(x,y)$, i.e., the area element on the unit disk $D$. Therefore $$\int_{S_1}{\bf F}\cdot d\vec \omega=\int_D(0,3x,0)\cdot(0,0,-1)\ {\rm d}(x,y)=0\ .$$ On the top surface $S_2$ the outward normal is $(0,0,1)$, Therefore $$\int_{S_2}{\bf F}\cdot d\vec\omega=\int_D(3x,3x,-6)\cdot(0,0,1)\ {\rm d}(x,y)=-6 \int_D {\rm d}(x,y)=-6\pi\ .$$ It follows that $$\int_{S_3} {\bf F}\cdot d\vec \omega=-{3\pi\over2}-0-(-6\pi)={9\pi\over2}\ .$$

edited Mar 19 '13 at 21:12

answered Mar 17 '13 at 14:50

Christian Blatter

226,825

Surely for the top surface the angle goes from $0$ to $2\pi$, and so you get $-12\pi$? – CAF Mar 17 '13 at 18:42
Or to be more precise, on top surface $\underline{dS} = e_3 d\rho d \phi $ and so $F \cdot dS = -6 d \rho d \phi$. Hence $-6 \int_0^{2\pi} \int_0^1 d\rho d \phi = -12\pi$ – CAF Mar 17 '13 at 18:48
@CAF: See my edit. Hope, it's clearer now. – Christian Blatter Mar 17 '13 at 19:07
Are you taking the angle as just $\pi$? Do we not integrate over the whole of the top surface so r is from 0 to 1, while the angle is from 0 to $2\pi$? – CAF Mar 17 '13 at 19:18
@CAF: The area of the top surface is equal to the area of a unit circle, which is $\pi$. – Christian Blatter Mar 17 '13 at 19:21
I see. So what is wrong with the integral I did? – CAF Mar 17 '13 at 20:07
@CAF: Your ${\rm d}\omega$ on the top surface $S_2$ is wrong. When you use polar coordinates then ${\rm d}\omega=r\ {\rm d}(r,\phi)$. – Christian Blatter Mar 17 '13 at 20:32
How would I do part ii) solely using the divergence theorem? – CAF Mar 19 '13 at 20:33
@CAF: See my edit. That's my last word on this matter. – Christian Blatter Mar 19 '13 at 21:13

Two small Divergence Theorem questions

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