Pre-Calc Homework Solutions 433

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

Ask a homework question - tutors are online