Multivariate Calculus - Bruce E. Shapiro
Multivariate Calculus - Bruce E. Shapiro Multivariate Calculus - Bruce E. Shapiro
196 LECTURE 24. FLUX INTEGRALS & GAUSS’ DIVERGENCE THEOREM Definition 24.2 The flow through an oriented surface A is the volume of water that passes through that surface per unit time, flow = A‖v‖ cos θdt = v · Adt Definition 24.3 The flux through an oriented surface A is the rate of flow through A, namely flux = v · A Definition 24.4 The flux of a vector field F through an oriented surface element dA is flux = F · dA Definition 24.5 The flux of a vector field F through an oriented surface A or flux integral is n∑ flux = lim F i · dA i = F · dA n→∞ i=1 A Figure 24.2: The geometry for example 24.1. Example 24.1 Find the flux integral of the vector field F = (2, 3, 5) through a rectangle parallel to the yz-plane (i.e., normal to the x-axis) with corners as illustrated in figure 24.2 (1, 0, 0), (1, 1, 0), (1, 1, 2), (1, 0, 2) Solution. The dimensions of the rectangle are 1×2 so the area is 2. A normal vector is A = (2, 0, 0) hence the flux is flux = A · F = (2, 0, 0) · (2, 3, 5) = 4 Revised December 6, 2006. Math 250, Fall 2006
LECTURE 24. FLUX INTEGRALS & GAUSS’ DIVERGENCE THEOREM 197 Example 24.2 Find the flux integral of through a disk of radius 2 on the plane in the upward direction. F = (2, 0, 0) x + y + z = 2 Solution. A normal vector to the plane that is oriented upward is N = (1, 1, 1) which has magnitude √ 3 and hence a unit normal vector is n = 1 √ 3 (1, 1, 1) The area of the disk is A = π × 2 2 = 4π. Hence an area vector is Therefore the flux is A = nA = 4π √ 3 (1, 1, 1) flux = F · A = 4π √ 3 (2, 0, 0) · (1, 1, 1) = 8π √ 3 Example 24.3 Find the outward flux integral of the vector field F = (2 − x, 0, 0) over the face of the cube with one corner at the origin, entirely in the first octant, with each side of length 3. Solution. Since the vector field is entirely parallel to the x-axis, of the six surfaces of the cube, four of them have normals that are perpendicular to the vector field. Since the dot product of these vectors with the vector field is zero, they produce no contribution to the flux. The remaining two surfaces are those perpendicular to the x axis, one at x = 3 and the other at x = 0. Each of these two surfaces has an area of A = 9. At the x = 3 surface, n = (1, 0, 0) and F = (−1, 0, 0), hence flux = F · nA = (1, 0, 0) · (−1, 0, 0) × 9 = −9 At the x = 0 surface, n = (−1, 0, 0) and F = (2, 0, 0), hence flux = F · nA = (−1, 0, 0) · (2, 0, 0) × 9 = −18 So the total flux is −9 − 18 = −27. We have previously defined the divergence as ∇F = M x + N y + P z when F = (M, N, P ). We can use flux integrals to give an alternate definition of the divergences, which gives a more physical description of why it is called “divergence.” In fact, the limit in the following theorem was the original definition of the divergence and it was not until sometime later that the derivative formula was derived. Math 250, Fall 2006 Revised December 6, 2006.
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196 LECTURE 24. FLUX INTEGRALS & GAUSS’ DIVERGENCE THEOREM<br />
Definition 24.2 The flow through an oriented surface A is the volume of water<br />
that passes through that surface per unit time,<br />
flow = A‖v‖ cos θdt = v · Adt<br />
Definition 24.3 The flux through an oriented surface A is the rate of flow through<br />
A, namely<br />
flux = v · A<br />
Definition 24.4 The flux of a vector field F through an oriented surface element<br />
dA is<br />
flux = F · dA<br />
Definition 24.5 The flux of a vector field F through an oriented surface A or<br />
flux integral is<br />
n∑ <br />
flux = lim F i · dA i = F · dA<br />
n→∞<br />
i=1<br />
A<br />
Figure 24.2: The geometry for example 24.1.<br />
Example 24.1 Find the flux integral of the vector field F = (2, 3, 5) through a<br />
rectangle parallel to the yz-plane (i.e., normal to the x-axis) with corners<br />
as illustrated in figure 24.2<br />
(1, 0, 0), (1, 1, 0), (1, 1, 2), (1, 0, 2)<br />
Solution. The dimensions of the rectangle are 1×2 so the area is 2. A normal vector<br />
is<br />
A = (2, 0, 0)<br />
hence the flux is<br />
flux = A · F = (2, 0, 0) · (2, 3, 5) = 4 <br />
Revised December 6, 2006. Math 250, Fall 2006