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Divergence theorem
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== Example == [[File:Vector Field on a Sphere.png|thumb|The vector field corresponding to the example shown. Vectors may point into or out of the sphere.]] [[File:SurfacesWithAndWithoutBoundary.svg|right|thumb|200px|The divergence theorem can be used to calculate a flux through a [[closed surface]] that fully encloses a volume, like any of the surfaces on the left. It can ''not'' directly be used to calculate the flux through surfaces with boundaries, like those on the right. (Surfaces are blue, boundaries are red.)]] Suppose we wish to evaluate :{{oiint | intsubscpt = <math>\scriptstyle S</math> | integrand = <math>\mathbf{F}\cdot\mathbf{n} \, \mathrm{d}S,</math> }} where {{mvar|S}} is the [[unit sphere]] defined by :<math>S = \left \{ (x,y, z) \in \mathbb{R}^3 \ : \ x^2+y^2+z^2 = 1 \right \},</math> and {{math|'''F'''}} is the [[vector field]] :<math>\mathbf{F} = 2x\mathbf{i}+y^2\mathbf{j}+z^2\mathbf{k}.</math> The direct computation of this integral is quite difficult, but we can simplify the derivation of the result using the divergence theorem, because the divergence theorem says that the integral is equal to: :<math>\iiint_W (\nabla \cdot \mathbf{F})\,\mathrm{d}V = 2\iiint_W (1 + y + z)\, \mathrm{d}V = 2\iiint_W \mathrm{d}V + 2\iiint_W y\, \mathrm{d}V + 2\iiint_W z\, \mathrm{d}V,</math> where {{mvar|W}} is the [[unit ball]]: :<math>W = \left \{ (x,y, z) \in \mathbb{R}^3 \ : \ x^2+y^2+z^2\leq 1 \right \}.</math> Since the function {{mvar|y}} is positive in one hemisphere of {{mvar|W}} and negative in the other, in an equal and opposite way, its total integral over {{mvar|W}} is zero. The same is true for {{mvar|z}}: :<math>\iiint_W y\, \mathrm{d}V = \iiint_W z\, \mathrm{d}V = 0.</math> Therefore, :{{oiint | intsubscpt = <math>\scriptstyle S</math> | integrand = <math>\mathbf{F}\cdot\mathbf{n}\,\mathrm{d}S = 2\iiint_W\, dV = \frac{8\pi}{3},</math> }} because the unit ball {{mvar|W}} has [[volume]] {{math|{{sfrac|4''Ο''|3}}}}.
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