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14) where ^r, '^ and ^z are unit vectors in the direction of increasing values of r, ' and z respectively. 2) the divergence is the ux per unit volume. Let CHAPTER 4. THE DIVERGENCE OF A VECTOR FIELD 36 r } . 2: De nition of the geometric variables for the computation of the divergence in cylinder coordinates. 2). Let us rst consider the ux of v through the surface elements perpendicular to ^r. The size of this surface is rd'dz and (r + dr)d'dz respectively at r and r + dr. The normal components of v through these surfaces are vr (r ' z ) and vr (r + dr ' z ) respectively.

17) for free space (where = 0). However, if we want to nd the relation between the mass and the resulting gravitational eld we must also use the eld equation ( g) = 4 G at places where the mass is present. More speci cally, we have to integrate the eld equation in order to nd the total e ect of the mass. The theorem of Gauss gives us an expression for the volume integral of the divergence of a vector eld. 2) it was shown that the divergence is the ux per unit volume. 4) gives us the outward ux d through an in nitesimal volume dV d = ( v)dV .

In one dimension the vector v has only one component vx , hence ( v) = @x vx . A \volume" in one dimension is simply a line, let this line run from x = a to x = b. 1) is the di erence of the function vx at its endpoints. 2) CHAPTER 6. THE THEOREM OF GAUSS 52 This expression will be familiar to you. 2) to derive the theorem of Stokes. Problem a: Compute the ux of the vector eld v(x y z) = (x + y + z)^z through a sphere with radius R centered on the origin by explicitly computing the integral that de nes the ux.

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