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# img018pv0 - 2 33 V = 7T J[2"‘z]2 dx = 1.9686 0 2 35...

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Unformatted text preview: 2 33. V = 7T J. [2"‘z]2 dx = 1.9686 0 2 35. V = 17] [ex/7' + e‘X/ZP dx z 49.0218 —1 37. A == 3 Matches (a) 39. (a) R(x) = 4x — X2. '0‘) = 0 V 4 77] (4x - x2)2dx o 4 = 77] (16.152 — 8x3 +x4)dx o = ——3— — —‘—‘_ x 2x4+5 16 x5]4 51217 0 15 Section 6.2 Volume: The Disc Method 3 34. V = 77] [In x]2 dx z 3.2332 1 5 36. V = WI [2 arctan (0.2x)]2 dx -~ 15.4115 0 #Iu 38. A m Matches (1)) 19 (b) Completing the square we have 4x # x2 = 4 A (x2 — 4x + 4) = 4 — (x - 2)? Thus, y = 4 — x2 has the same volume as in part (3) since the solid has been translated only horizontally. 41. R(x) = gx, r(x) = O 61 V: ‘rrL szdx 5 = 1 a = [12x lo 18’” Note: V = %7rr2h grams = l8’lT (b) 42. R(x) = Ex, r(x) = 0 I! 2 r V=7r0 ﬁxzdx ,2” h _r2_7r 3_1 2 _3h2h ~37Trh ...
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