# C by adding the solutions to parts a and c together

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Unformatted text preview: on D through a small region in its boundary of area dA is −(κ∇u) · NdA, where N is the unit normal which points out of D. The total rate at which heat leaves D is given by the ﬂux integral − (κ∇u) · NdA, ∂D where ∂D is the surface bounding D. It follows from the divergence theorem that Rate at which heat leaves D = − ∇ · (κ∇u)dxdydz. (5.2) D From formulae (5.1) and (5.2), we conclude that ρσ D ∂u dxdydz = ∂t ∇ · (κ∇u)dxdydz. D This equation is true for all choices of the region D, so the integrands on the two sides must be equal: ρ(x, y, z )σ (x, y, z ) ∂u (x, y, z, t) = ∇ · (κ∇u)(x, y, z, t). ∂t Thus we ﬁnally obtain the heat equation ∂u 1 = ∇ · (κ(x, y, z )(∇u)) . ∂t ρ(x, y, z )σ (x, y, z ) In the special case where the region D is homogeneous , i.e. its properties are the same at every point, ρ(x, y, z ), σ (x, y, z ) and κ(x, y, z ) are constants, and the heat equation becomes ∂u κ ∂2u ∂2u ∂2u = + 2+ 2 . ∂t ρσ ∂x2 ∂y...
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## This document was uploaded on 01/12/2014.

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