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Unformatted text preview: 1176 CHAPTER 14 . . Vector Calculus 1462 65. Show that if G = ∇ × H , for some vector field H with contin uous partial derivatives, then ∇ · G = 0. 66. Show the converse of exercise 65; that is, if ∇ · G = 0, then G = ∇ × H for some vector field H . Hint: Let H ( x , y , z ) = , x G 3 ( u , y , z ) du , − x G 2 ( u , y , z ) du . EXPLORATORY EXERCISES 1. In some calculus and engineering books, you will find the vec tor identity ∇ × ( F × G ) = ( G · ∇ ) F − G ( ∇ · F ) − ( F · ∇ ) G + F ( ∇ · G ) . Which two of the four terms on the righthand side look like they should be undefined? Write out the lefthand side as completely as possible, group it into four terms, identify the two familiar terms on the righthand side and then de fine the unusual terms on the righthand side. (Hint: The nota tion makes sense as a generalization of the definitions in this section.) 2. Prove the vector formula ∇ × ( ∇ × F ) = ∇ ( ∇ · F ) − ∇ 2 F . As in exercise 1, a major part of the problem is to decipher an unfamiliar notation. 3. Maxwell’s laws relate an electric field E ( t ) to a magnetic field H ( t ). In a region with no charges and no current, the laws state that ∇ · E = , ∇ · H = , ∇ × E = − µ H t and ∇ × H = µ E t . From these laws, prove that ∇ × ( ∇ × E ) = − µ 2 E tt and ∇ × ( ∇ × H ) = − µ 2 H tt . 14.6 SURFACE INTEGRALS Whether it is the ceiling of the Sistine Chapel, the dome of a college library or the massive roof of the Toronto SkyDome, domes are impressive architectural structures, in part because of their lack of visible support. This feature of domes worries architects, who must be certain that the weight is properly supported. A critical part of an architect’s calculation is the mass of the dome. How would you compute the mass of a dome? You have already seen how to use double integrals to compute the mass of a twodimensional lamina and triple integrals to find the mass of a threedimensional solid. However, a dome is a threedimensional structure more like a thin shell (a surface) than a solid. We hope you’re one step ahead of us on this one: if you don’t know how to find the mass of a dome exactly, you can try to approximate its mass by slicing it into a number of small sections and estimating the mass of each section. In Figure 14.40, we show a curved surface that has been divided into a number of pieces. If the pieces are small enough, notice that the density of each piece will be approximately constant. S i ( x i , y i , z i ) FIGURE 14.40 Partition of a surface So, first subdivide (partition) the surface into n smaller pieces, S 1 , S 2 ,..., S n . Next, let ρ ( x , y , z ) be the density function (measured in units of mass per unit area). Further, for each i = 1 , 2 ,..., n , let ( x i , y i , z i ) be a point on the section S i and let S i be the surface area of S i . The mass of the section S i is then given approximately by ρ ( x i , y...
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 Spring '10
 CARTER
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