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By Harley Flanders

Moment direction in Calculus

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If xyz 5* 0, then by Ex. 7, cube c * ) ( > ) ( * ) < i.. \x + y + z) \x + y + z) \x + y + z) 10. q If x > 0, y > 0, or z > 0, apply Ex. 9 to x/(x 27 9 + y + zq)1,q, etc. 12. f, at (£, -£, — -£). 14. 272. [Suppose x = y. Then z2z = 1 and 2x2 + z2 = \7-, hence 423 — 17z + 8 = 0, that is, tz — 17t + 16 = 0, where Z = 2z. By inspection, 2 = 1 is a root, so t2 - 17* + 16 = (t - 1) (t2 + t - 16). ] 16. / ( * , 2/, 0) = a? + y2 > 0 = /(0, 0, 0). SAMPLE TEST 1. Find the second degree Taylor polynomial of x 1— x+ # at (3, 1).

12. 6, hence rh Izz = f rh Kfcrr2 cfe) (r2) = 7r5a4 / ( ^ + l Y dz Ma2 (3h4 + 10h2c2 + 15c4) 10c2 (h2 + 3c2) 16. 18. 20. 22. /** = Iyy = JAfa2 gm-cm2, 7ZZ = \Ma2 gm-cm2 \M(Za2 + SaL + L2) Ixx = Iyy = |ik/a 2 , Izz = Ma2 722 = Ma2, Ixx = / w = |M(3a 2 + 2/*2) SAMPLE TEST 1. Evaluate / / / xy dxdy dz over the prism bounded above by the plane 2. 3. x + y + z = 5 and below by the triangle with vertices (0, 0, 0), (1, 0, 0), and ( 0 , 2 , 0 ) . Compute the volume of the region bounded by the sphere x2 + y2 + z2 = 2 and the paraboloid z = x2 + 2/2.

Y By a little algebra, r(a:) = (x-1)2 s(y) = (y - 2) 2 42/ Multiply: — xy =2J - 2o (* ~ X) " 47 (2/ "" 2) + e(*' y)> where each term in the numerator of e(x, y) involves (x — l ) 2 or (y — 2) 2 , or (a: — 1) (y — 2), and the denominator is xy. Since (x — l ) 2 / l x ~" (1? 2)| >0, (2/ — 2) 2 /|x — (1,2)| >0, and (x - 1) (2/ - 2)/|x - (1, 2)| ► 0, we have |e(z, 2/) |/|x — (1, 2) | > 0. 30 2. 3. 4. Chapter 8 (a) (b) (c) Yes 2j - 3k F(o\ + b\ + ck) = (a + 6)J + (6 + 2c) k ^ i + j for any choice of a, b, c.

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