hw910ans - Math 20C Prof. Fitzsimmons Answers to...

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Unformatted text preview: Math 20C Prof. Fitzsimmons Answers to even-numbered exercises from sections 15.2-15.4 §15.2 (4) (a) D = {(x, y ) : 0 ≤ y ≤ 1 − x2 , 0 ≤ x ≤ 1}; the corresponding iterated integral is 1 0 0 1−x2 xy dy dx whose value is 1 . 12 √ (b) D = {(x, y ) : 0 ≤ x ≤ 1 − y, 0 ≤ y ≤ 1}; the corresponding iterated integral is 1 0 0 √ 1−y xy dx dy whose value is also (30) (36) [ln(2)]2 42 2x 00 1 . 12 √ x2 + y dy dx = 16 2/3 (58) (e − 1)2 §15.4 (2) (a) 0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π , −1 ≤ z ≤ 2. (b) r ≥ 0, 0 ≤ θ ≤ π/2, −3 ≤ z ≤ 1 (8) (a) On the intersection of the two boundary surfaces, we have 4 − z + z 2 = 6, so z = −1 or z = 2. The value z = 2 is the appropriate one for the problem (see Fig. 19, p. 919), resulting in x2 + y 2 = 6 − 22 = 2. √ √ 2 2π (b) Volume is 0 0 (4 − r 2 ) − 6 − r 2 r dr dθ = 6π − (14) (18) π /2 3 2 r dr dθ = 9π/2 0√ 0 arctan 3 2/ cos θ 2 r cos θ dr dθ π/4 0 2 π 3 /2 (6 3 − 8) ≈ 4.82 = 8 3 arctan π/4 √ 3 √ 8 sec2 θ dθ = 3 ( 3 − 1) ...
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