Pre-Calc Homework Solutions 433

# Pre-Calc Homework Solutions 433 - Section 10.6 dr d 6 sin...

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Unformatted text preview: Section 10.6 dr d 6 sin , so cos )2 /2 0 433 35. 41. 62 cos )2 62 sin2 d , which using (1 cos )4 [ 1.5, 1.5] by [ 1, 1] 3 (1 Length (1 NINT evaluates to 6.887. 2 (Note: the integrand can simplify to 3 sec dr 36. d 2 sin , so (1 cos )2 /2 .) 2r dr d dr d 2 sin 2 sin 2 cos 2 . Use the curve's symmetry and note that r is defined for 22 sin2 d , which using (1 cos )4 Length (1 22 cos )2 0 /4 4 : Surface area cos 2 sin d /4 NINT evalutes to 2.296. 3 2 0 /4 2 0 cos 2 sin2 2 d cos 2 (Note: the integrand can simplify to csc dr d 2 ). 2 4 42. For a 2 sin cos 0 (4 2 2) 3.681 37. cos2 3 /4 sin cos6 cos4 3 3 3 , so cos4 d /4 0 8 3 3 1: Length 0 /4 0 sin2 3 d 1 2 dr 38. d cos 2 1 sin 2 2 3 2 sin 4 3 . dr d [ 3, 3] by [ 2, 2] 2a sin , so surface area /2 , so cos2 2 1 sin 2 cos 2 1 2 2 0 2 (2a cos ) cos /2 0 4a2 cos2 /2 4a2 sin2 d Length 0 2 0 1 sin 2 d 2 16a2 16a2 43. dx 2 d 1 2 cos2 d 4a2 2 0 2 (1 sin 2 ) d sin 2 1 sin 2 4 2d 0 2 . sin 2 c os 2 2 dy 2 d ( f ( ) cos , so f ( ) sin )2 f ( ) cos )2 ( f ( ) sin )2 ( f ( ) sin )2 39. dr d 1 2 /4 ( sin 2 )(2) ( f ( ) sin ( f ( ) cos )2 sin 2 c os 2 cos 2 Surface area 2 0 /4 cos 2 cos cos cos2 2 d 2 0 ( cos 2 )2 sin2 2 d ( f ( ) cos )2 ( f ( ))2(cos2 sin2 ) sin2 ) r2 cos ) d a 2 /2 d 2 0 /4 ( f ( ))2(cos2 ( f ( ))2 ( f ( ))2 2 2 0 cos /4 dr 2 d a 2 2 2 dr d sin 2 2 /2 4.443. 44. (a) 1 2 1 2 0 1 ( 0 a(1 0 2 sin 0 a 40. e /2 , so surface area /2 2 (b) /2 2 ad 0 a 0 2 0 /2 2e sin 5d ( 2e ) 2 2 2 (e /2 2 ) d (c) /2 2 e sin 0 /2) a cos /2 d a /2 sin /2 2a 2 5 5 (e 1 e (sin 2 /2 /2 cos ) 0 45. If g( ) 2f ( ), then (g( )2 g ( )2) 2 ( f ( )2 g is 2 times the length of f. f ( )2), so the length of 1) 40.818 ...
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