COSMOS Chapter 9 Solution 27 | Answer Key - Edubirdie

University:
Michigan State University
Course:
ERG 160 | Success in Science, Technology, Engineering, and Mathematics
Academic year:

2021
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494

Pages:

2
Author:

diamundgalx45p

COSMOS: Complete Online Solutions Manual Organization System Chapter 9, Solution 27. π Have: a cos 2θ A = 2∫ 04 ∫ 0 rdrdθ π = ∫ 04 a 2 cos 2 2θ dθ π 2 θ sin 4θ  4 =a  +  8 0 2 = Have: π 8 a2 J O = ∫ r 2dA π a cos 2θ = 2∫ 04 ∫ 0 r 2 ( rdrdθ ) π 1 = ∫ 04 a 4 cos 4 2θ dθ 2 Now: ( cos 4 2θ = cos 2 2θ 1 − sin 2 2θ = cos 2 2θ − ) 1 2 sin 4θ 4 π Then: 1 1   J O = a 4 ∫ 04  cos 2 2θ − sin 2 4θ  dθ 2 4   1  θ sin 4θ  1  θ sin 8θ = a 4  + −  − 2  2 8  4 2 16 = π  4   0 1 4π 1 π  a  − ×  4 4 4 4 JO = 3π 4 a 64 continued COSMOS: Complete Online Solutions Manual Organization System And: kO = = JO A 3π 4 a 64 π 8 a2 kO = a 6 4

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