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Divergence Theorem
In this section we are going to relate surface integrals to triple integrals
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Divergence Theorem
Let E be a simple solid region and S is the boundary surface of E with positive orientation
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Then,
∫∫ F idS = ∫∫∫ div F dV
S
E
Let’s see an example of how to use this theorem
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Solution
Let’s start this off with a sketch of the surface
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Note that
cylindrical coordinates would be a perfect coordinate system for this region
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0 ≤ z ≤ 4 − 3r 2
0 ≤ r ≤1
0 ≤ θ ≤ 2π
We’ll also need the divergence of the vector field so let’s get that