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Use the Divergence Theorem to evaluate $$ \iint_{S} \vec{F} \cdot \hat{n}\,dS, $$ where $ \vec{F} = \langle 4x, 2y^2, z^2 \rangle, S $ is the boundary of the region defined by $ x^2 + y^2 + z^2 \leq 4,\,\, 0 \leq z \leq 3 $ and $\hat{n} $ is the unit outward normal

Attempt: The word 'boundary' is confusing me a little. So the surface is the curved part of a cylinder from $z=0$ to $z=3$. In cylindrical coordinates, $ x = 2\cos\theta, y = 2\sin\theta, z = z$. Using the Div. Thm gives $\text{div} \vec{F} = 4 + 4y + 2z $, so I have $$ \iiint_{E} 4+4y+2z\,dV,$$ Am I correct to write this as $$ \int_{0}^{2\pi} \int_{0}^{2} \int_{0}^{3} (4 + 8\sin\theta + 2z)\,r\,dz\,dr\,d\theta? $$ I am not sure because the question wanted the surface to be just the boundary but here I am integrating over the whole surface. Then again, I think this could be a feature of the div theorem. If we are just integrating on the boundary, then what would dV be? Many thanks

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The region described is not a cylinder. It is a sphere of radius 4 cut by the planes $z=0,3$. Spherical coordinates is most appropriate for this problem.

When you apply the divergence theorem, you establish an equality between a volume integral and a surface integral with a very particular relationship in what they integrate over. Any volume must have some boundary surface(s). The volume integral is over some region, and the corresponding surface integral in the divergence theorem is only over the boundary surfaces of that region.

If the region were a cube, the volume integral would be over the whole cube. The surface integral would be over the 6 faces of the cube.

In this case, the region is (part of) a sphere. The volume integral is over that region, and the surface integral is over two planar surfaces and part of a spherical surface.

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Many thanks for your reply. I understand now. (Oh, and the addition of $ z^2 $ was a typo - it should have been $x^2 +y^2 \leq 4$) – CAF Nov 21 '12 at 16:57
One more question: In general, are surface integrals always over the boundary of some region? – CAF Nov 21 '12 at 17:00
No, surface integrals may be over "open" surfaces (think for example, a square patch of a plane). Open surfaces themselves have boundaries, however, so there is an analogue to the divergence theorem: the integral of $\nabla \times F \; dS$ on the open surface is equal to the integral of $F \cdot dr$ on the boundary curve. There is a whole broad class of these theorems--the fundamental theorem of calculus, the divergence theorem, Stokes' theorem, and so on. They all represent the same basic concept: integral of a derivative over a region equals integral of the function over the boundary. – Muphrid Nov 21 '12 at 17:05
Ok thanks. The triple integral I wrote in my first post gives me the correct answer to the Q, but to get to that integral, I had to take $y = 2\sin\theta $. I.e r takes the value 2. But then I sub that into divF and integrate over all r in the triple integral? So this is an example of what you said above right? We parametrize the surface boundary and then in the triple integral, we integrate over the whole region? – CAF Nov 21 '12 at 18:22
For the purposes of taking divergence and doing the volume integral, $y=r\sin \theta$. You can only set $r=2$ when performing the surface integration, not in the volume integration. – Muphrid Nov 21 '12 at 18:44

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