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Verify the divergenece theorem to $$\mathbf{F }=4xi-2y^2j+z^2k$$ for the region bounded by $x^2+y^2=4$ , $z=0$, $z=3$

I've already done the triple integral for the divergence $\iiint_R \operatorname{div}\bar F\;dV$ and the result I got is $84\pi$, but I'm having trouble solving it by surface integrals. I've defined $S_1$ as $x^2+y^2-4$ and the normal vector I got was $(2,2,0)$ but as I continue, the answer I keep getting is $4\pi$

Can you guys help me out with this?

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  • $\begingroup$ You might want to put your whole calculation in. Otherwise it's very hard to tell where you went wrong. Did you remember to include the endcaps of the cylinder? Did you remember to normalize the normal vector? $\endgroup$ – spaceisdarkgreen May 22 '17 at 2:57
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This region is a cylinder with radius $2$ around the $z$ axis, and so it can be parameterized as $r(\theta, z) = (2\cos \theta, 2 \sin \theta, z)$, where $\theta \in [0, 2\pi]$ and $z\in [0,3]$. Then $$r_\theta = (-2\sin \theta, 2\cos \theta, 0)$$ $$r_z = (0,0,1)$$ $$\vec{N} = r_\theta \times r_z = (2\cos \theta, 2\sin \theta, 0)$$ (note that $\vec N$ points outwards). Now you need to evaluate $$I_1 = \int_{0}^{ 2\pi}\int_{0}^3 \vec{F}(r(\theta, z))\cdot \vec{N}\, dz\, d\theta$$ Now you need to integrate over the top and bottom on the integral. Since $z$ is constant, these integrals are equivalent to $$I_2=-\iint_S F(x,y,0)\, dA$$ $$I_3=\iint_S F(x,y,3)\, dA$$ where $S\subseteq \mathbb R^2$ is the disc centered at the origin with radius $2$. The negative sign in $I_2$ is due to the fact that the normal vector points outwards, which is downward on that part of the surface. Then your final answer is $I_1+I_2+I_3$, which should be equal to what you got using the divergence theorem.

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Parametrize the side of the cylinder of interest by

$$r(t,z)=\langle 2\cos (t), 2 \sin (t),z \rangle$$

With $t \in [0,2\pi]$ and $z \in [0,3]$. An outward pointing unit normal vector for the side is $\langle \cos(t), \sin(t),0 \rangle$ and also $dS=2dzdt$ hence we get that,

$$\iint_{S_1} \vec F \cdot \vec n dS$$

$$=2 \int_{0}^{2\pi} \int_{0}^{3} \langle 8\cos (t), -8\sin^2 (t), z^2 \rangle \cdot \langle \cos t, \sin t,0 \rangle dz dt$$

$$=16 \int_{0}^{2\pi} \int_{0}^{3} (\cos^2(t)-\sin^3(t)) dz dt$$

$$=48\pi$$


Next parametrize the top of the cylinder by $r(x,y)=\langle x,y,3 \rangle$ with $x^2+y^2 \leq 4$. An obvious outward pointing unit normal is $\langle 0,0,1 \rangle$. Furthermore it is easy to see that $dS=dA$ at the top of the cylinder, because the surface is flat parallel to the $xy$ plane.

From here,

$$\iint_{S_2} \vec F \cdot \vec n dS=\iint_{D} \langle ..,...,9 \rangle \cdot \langle 0,0,1 \rangle dA$$

$$=9\text{Area}({D})=9(\pi (2^2))=36 \pi$$


The bottom piece is much similar. It evaluates to zero. Really because $\vec F=\langle..,...,0 \rangle$ there and $\vec n=\langle 0,0,-1 \rangle$ there. Therefore $\vec F \cdot \vec n=0$ on the bottom surface.


Hence indeed we get,

$$\iint_{S} \vec F \cdot d\vec S=48\pi+36\pi=84\pi$$

As the divergence theorem claims.

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