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The general proof of the divergence theorem nvdA = f V vdV
Chapter 13, Problem 13.37(choose chapter or problem)
The general proof of the divergence theorem nvdA = f V vdV (13.63) is fairly complicated and not especially illuminating. However, there are a few special cases where it is reasonably simple and quite instructive. Here is one: Consider a rectangular region bounded by the six planes x = X and X + A, y = Y and Y B, and z = Z and Z C, with total volume V = ABC. The surface S of this region is made up of six rectangles that we can call S1 (in the plane x = X), S2 (in the plane x = X + A), and so on. The surface integral on the left of (13.63) is then the sum of six integrals, one over each of the rectangles S1, S2, and so forth. (a) Consider the first two of these integrals and show that li+B n vdA + f nv dA = f dy rc dz [vx(X A, y, z) vx(X, y, z)]. fsi s2 (b) Show that the integrand on the right can be rewritten as an integral of avxlax over x running from x = X to x = X + A. (c) Substitute the result of part (b) into part (a), and write down the corresponding results for the two remaining pairs of faces. Add these results to prove the divergence theorem (13.63).
Questions & Answers
QUESTION:
The general proof of the divergence theorem nvdA = f V vdV (13.63) is fairly complicated and not especially illuminating. However, there are a few special cases where it is reasonably simple and quite instructive. Here is one: Consider a rectangular region bounded by the six planes x = X and X + A, y = Y and Y B, and z = Z and Z C, with total volume V = ABC. The surface S of this region is made up of six rectangles that we can call S1 (in the plane x = X), S2 (in the plane x = X + A), and so on. The surface integral on the left of (13.63) is then the sum of six integrals, one over each of the rectangles S1, S2, and so forth. (a) Consider the first two of these integrals and show that li+B n vdA + f nv dA = f dy rc dz [vx(X A, y, z) vx(X, y, z)]. fsi s2 (b) Show that the integrand on the right can be rewritten as an integral of avxlax over x running from x = X to x = X + A. (c) Substitute the result of part (b) into part (a), and write down the corresponding results for the two remaining pairs of faces. Add these results to prove the divergence theorem (13.63).
ANSWER:Step 1 of 5
(a) As discussed, the surface integral on the left side of Eq. (13.63) has the form \(I_{1}+\cdots+I_{6}\), where \(I_{1}=\int_{S_{1}} \mathbf{n} \cdot \mathbf{v} d A\) and so on.
Now, on the surface \(S_{1}\) the outward normal is \(\mathbf{n}=-\hat{\mathbf{x}}\), while on \(S_{1}\), \(\mathbf{n}=\hat{\mathbf{x}}\).