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By Harley Flanders

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A ) . = (0,0,0). )*) = Wyx — Vzx + UZy — WXy>+ VXZ — UyZ = 0. 26. If / = / ( p ) , then g r a d / = /'(p) (p*, py, pz) = p" 1 /'(p)x. By the same formula, applied to p - 1 /'(p), and by Ex. 19, d i v [ g r a d / ( P ) ] = divCp-Tx] = grad(p" 1 /0 x + p" 1 /' divx = [p- 1 (p- 1 /0 , x]-x + 3p-T = p(p" 1 / , ) / + 3 p - r = p-Kpf" - / ' ) + 3P"1/' = / " + 2p~y = p-i(pf)" (or p - W ) ' ) . Section 5, page 306 2. 6, 2, - 6 4. -12, - 4 , -12 6. Vs 8. 6V2 Section 6, page 309 2. F = grad(#32/22), 1 4.

Note that 2/i + 2/o = 2A. 20. Use cylindrical coordinates. Take the cone along the z-axis, apex at 0. Take the axis of the cylinder at x = a csc a, y = 0. The solid cone is de­ scribed by — rVE

2 20. 2(e - e) 22. 6. 14. ^ (fr, | ) 16. 0, by symmetry (0, 39/76) 397 A 16. Section 6, page ^. 4375 gm 2 14. 39 & ^ A 18. 24. i f 26. iVE ff 407 3 (a2 - r2y2r 4. jj 6. T2* = ff 8. (4a/37r, 4a/37r) dr d$, 0 < r < a, [2(1 - r 2 ) 1 ' 2 - l ] r dr d0, 0 < <9 < 2TT 0 < r < V3/2, 0 < 6 < 2TT 9. Both left and right halves of the semicircular disk S have the same y, hence y for S is the same. 10. f a(sin a, 1 - cos a)/a 14. TT/10 12. (16a/157r, 16a/157r) 3 16. 4TT /3 18. Indirect solutions by symmetry: j = jj x6 = jj y\ K = jj x*y2 = jj x2y\ Now fr- = ff r7 dr dd = jj r6 cto dy = / 7 (a:2 + y2)z dx dy = J + 3X + 3K + J, J + SK = iw.

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