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 Suppose we have a volume V in three dimensions that has piecewise locally planar 
boundaries. (This means that its boundary can be broken up into a finite number 
of pieces each of which looks planar at small distances. We have defined an additive 
integral of a function f over it (and can do the same for a function over "hyper"-volume 
in any number of dimensions). Can we define an integral over its boundary that is additive on volumes? The 
answer is yes. The surface of V at any point at which the boundary is locally 
planar can be characterized by an 'outward normal' direction. If we integrate 
the outward normal component of a vector v over this surface, and join together 
adjacent volumes, what is the outward normal to one will be the inward normal 
to the other where they meet, and contributions from mutual parts of the boundary 
will cancel out. This means that the integral of the outward normal of 
a continuous vector field v over the boundary,  
 and call it the outward 'flux' of v through the surface of V. We can now address the same question we examined for areas in the last section: 
if we let V be an infinitesimal axis parallel rectangular volume, with sides dx 
dy and dz, and lower corner at (x, y, z), what will this integral be?  The rectangular volume V has six outer surfaces which comprise its front and 
back in each of the three axis directions. The outward normals point positively 
on the front faces and negatively on the back ones. Exactly as in the two dimensional case, the contribution from the two faces 
normal to the x direction becomes a difference of v between its values at x + 
dx and its value at x, integrated with respect to y and z variables on these faces. 
This contribution,  We find then, that for a small volume V as indicated, we have 
 Again, the same result can be obtained for tilted parallelepiped regions or 
prisms, and again we can use additivity to deduce the same result for any volume 
with piecewise locally planar boundary and any piecewise continuous vector field 
defined on it.  This result is again extremely important. It is called the Divergence Theorem, and also is known as Gauss's Theorem. It is more or less obvious that, with suitable definitions, there is a similar theorem in any higher dimension. Again it represents a higher dimensional version of the Fundamental Theorem of Calculus: the derivative in the divergence on the right can be integrated over, yielding differences at extreme points of the integration which convert the three dimensional integral on the right into a two dimensional one over the boundary of V on the left. Exercises:22.1 Consider the vector field E defined by  22.2 What is the integral of the flux of E around a circle of radius 1 centered about the point (0, 0, 2)?  | 
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