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Unformatted text preview: Section 10.6
22. 25. 431 [ 4.7, 4.7] by [ 3.1, 3.1] [ 6, 6] by [ 4, 4] Use the symmetries of the graphs: the shared area is
/2 The area in question is half the circle minus two lobelike regions:
1 (2)2 2
/2 4
0 1 [2(1 2/2 cos )]2 d 2 cos
1 2 2
0 /2 8
0 (1 cos 2 )d
/2 1 [2(1 2 sin )]2 d 4 sin2 ) d
1 2 1 sin 2 4
/2 8 23. For a 2 sin 1: 1 sin 2 4 2 6 16 2 26.
0 (4 4 8 sin 2 cos 0 8
0 [ 3, 3] by [ 2, 2] [ 3, 3] by [ 2, 2] (a) To find the integration limits, solve The curves intersect at the origin and when 3a cos 2 cos
3 2 cos
2 . 3 1 0 a(1 1 . cos ) Because of the curve's symmetry, the area inside the outer loop is
2 /3 Use the symmetries of the curves: the area in question is
/3 2
0 1 (2 cos 2 1)2 d 4 cos 4 sin
0 2 /3 2
0 1 [(3a cos )2 2
/3 0 a (1 1 2 cos ) ] d cos2 ) d 1) d
/3 2 (4 cos2 sin 2 1) d
2 /3 0 a2 a
2
0 (9 cos2 (8 cos
2 2 cos 2
3 2 3 /3 2 cos 2 sin 2 a2 4 24. 2 sin 2 a2 0 (b) Again, use the curve's symmetry. The inner loop's area is 2
1
2 /3 2 (2 cos sin 2 1)2 d 4 sin
2 /3 2
[ 3, 3] by [ 2, 2] 3 3 . 2 Subtract this from the answer in (a) to get 3 3 . The curves intersect when 6 cos 2
6
/6 3 or
5 . 6
/6 Use the symmetries of the curves. The area in question is 4
0 1 (6 cos 2 2 3) d 6 sin 2
0 3 3 . ...
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This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.
 Spring '08
 GERMAN

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