14.7 The Divergence Theorem

Transcription

1 14.7 The Divergence Theorem The divergence of a vector field is a derivative of a sort that measures the rate of flow per unit of volume at a point. A field where such flow doesn't occur is called 'divergence free', that is the flow per unit of volume across any point is zero. If a vector field is described as F( x, y, z) = M i + N j + P k, where M, N and P are differentiable functions of x, y and z, then M the divergence is given by div( F) = + x N P + y z The divergence theorem is a three dimensional version of Green's theorem from lesson 4. Here we must concern ourselves with the condition that a surface S is closed meaning that it is the complete boundary of a solid Q. The theorem gives a relationship between the integral over a solid region Q and the surface integral over the surface S of Q. Theorem The Divergence Theorem (Gauss's Theorem) Let Q be a solid region bounded by a closed surface S oriented by a unit vector pointing outward from Q. If F is a vector field whose component functions have continuous partial derivatives in Q, then Ex 1 Verify the divergence theorem by evaluating as both a surface integral and as a triple integral where S is the sphere x 2 + y 2 + z 2 = a 2 and the vector field is F( x, y, z) = z k. I'll do the triple integral first. Note div( F) = 1, so we have

2 To evaluate as a surface integral, note the sphere consists of the upper surface z = a 2 x 2 y 2 and lower surface z = a 2 x 2 y 2. Constructing G( x, y, z) = z a 2 x 2 y 2, the normal vector is orthogonal to the level surface. Thus for the upper surface we have outward normal vector n = x i + y j + k, a 2 x 2 y 2 a 2 x 2 y 2 and so the integral is where the region R is the circle of radius a centered at the origin. A switch to polar coordinates allows us to evaluate the integral as follows: giving a value of 2πa3. In a similar fashion, the lower surface yields the same result, so the 3 π 3

3 solution is 4πa3. 3 Note how much easier it was to solve the problem using the divergence theorem. Ex 2 Let S be the surface of the cylindrical solid bounded by x 2 +y 2 =9, the xy plane and the plane z=2, oriented by the outward unit normal vector N. Let a vector field be defined by F( x, y, z) = x 3 i + y 3 j + z 2 k. Use the divergence theorem to evaluate Note div( F) ( ) ( ) ( ) x x3 + y y3 = + z z2 = 3x 2 + 3y 2 + 2z so from the divergence theorem Conversion to cylindrical coordinates gives us the following: 2π r 2 2π 3 2π ( + 2z) r dz dr dθ 6r = + 4r dr dθ = dθ = 279π Ex 3 Use the Divergence Theorem to evaluate where F( x, y, z) = 2x i + 2y j + 2z k and S is the surface z = 4 x 2 y 2, z = 0 Note that the surface is a hemisphere of radius 2. We apply the Divergence Theorem and get div( F) = = 6 so our integral is Note that is just 6 times the volume of the hemisphere, so the result is π = 32π.

4 Flux and the Divergence Theorem The flux (flow) of a vector field across a surface per time unit is a measure of the fluid flow across the surface durning that time unit. It is found by summing the fluid flow across small bits of the surface, and that was how the integral came to ne used for it. (Recall the formula came from adding the column of height F. N and area S.) The triple integral measures this same amount of fluid flow, but in a different way. It calculates the amount of fluid flow (divergence) into (when negative) or out of (when positive) small cubes of volume V i. Note that the flux of the small cube will be about the same as the measure of divergence at some point within the small cube. (The flow across the small surface will be about the same as the flow out of some point in the small cube.) What happens in the solid itself is that the only fluid flow not cancelled by adjacent cubes happens on the outside edge of the surface, hence matches the fluid flow across the surface. Def source, sink, incompressible A point in a vector field (x 0 ) is classified as follows: 1. A source if div(f(x 0 )) > A sink if div(f(x 0 )) < Incompressible at (x 0 ) if div(f(x 0 )) = 0. A source is a place where fluid is entering the vector field, a sink is where fluid is exiting a vector field, and the field is incompressible at a point if fluid is neither entering nor exiting at that point. Imagine you are filling a tub. The place where the water stream from the faucet enters the water already in the tub would be a source. The place where the water runs down the drain would be a sink. Ex 4 Use the divergence theorem to show that The volume of a solid bounded by a surface S can be given by

5 We know that our divergence of the vector field must be 1, div( F) i + f j + f k y z F = x i and let x = f ( y, z). Then N = and 1 + f + f y z M N P = + + = 1 so choose x y z ds = 1 + f + f da = 1 + f + f dy dz. Now the integrand is F. NdS or y z y z i + f j + f k y z ( x i) 1 + ( f y ) 2 + ( f z ) 2 dy dz 1 + f + f y z fact that = x dy dz. The divergence theorem gives us the and the expression on the right is the volume of the solid Q, which is the region bounded by the surface S. Note that the expressions and can also be shown to give the volume of the solid Q.

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