✓ )² + y² = 4 In polar coordinates we have x² + y² = 2² and x = rcos(8), so the boundary circle becomes r² = 4r cos (0) (x - 2 D = {(r, 0) | -π/2 ≤ 0 π/2, 0≤ r ≤ 4cos(0)} 4 cos (0) V= // L = = 48 -T/2 = 48 /2 = 48 = 96 , or r = 4cos(0). Thus the disk D is given by (3x² + 3y²)dA = 3,4 4 X X X T/2 =126² 5²¹² (cos(0))+de 1 cos(20) 14cos (0) Jo 2 1/2 Jo = 37² de = 192 -)²de /2 -π/2 /2 [1 + 2cos(20) + (1 + cos(40))]de r dr de (cos(0))+de
✓ )² + y² = 4 In polar coordinates we have x² + y² = 2² and x = rcos(8), so the boundary circle becomes r² = 4r cos (0) (x - 2 D = {(r, 0) | -π/2 ≤ 0 π/2, 0≤ r ≤ 4cos(0)} 4 cos (0) V= // L = = 48 -T/2 = 48 /2 = 48 = 96 , or r = 4cos(0). Thus the disk D is given by (3x² + 3y²)dA = 3,4 4 X X X T/2 =126² 5²¹² (cos(0))+de 1 cos(20) 14cos (0) Jo 2 1/2 Jo = 37² de = 192 -)²de /2 -π/2 /2 [1 + 2cos(20) + (1 + cos(40))]de r dr de (cos(0))+de
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter11: Topics From Analytic Geometry
Section: Chapter Questions
Problem 52RE
Related questions
Question
![)² + y² = 4
In polar coordinates we have x² + y² = r² and x = rcos(8), so the boundary circle becomes ² =
4r cos (0)
(x - 2
D = {(r, 0)| -π/2 ≤ 0 π/2,0 ≤ r ≤ 4cos(0)}
π/2
4 cos(0)
V
1 = √ √ (²x
[]
=
/2
[TR]
= 48
= 48
= 48
or r = 4cos(0). Thus the disk D is given by
= 96
(3x² + 3y²)dA
3µ4
4
X
X
X
[5/²
5
0
=
(cos(0))+de
T/2 1 + cos(20)
0
2
17/²
14 cos (6) de = 192
372
-)²de
π/2
-π/2
π/2
[1 + 2cos(20) + ½(1 + cos(40))]de
r dr de
(cos(0))+de](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe489eec6-06b4-4f41-8536-03174f230880%2F974d240d-61ef-4ea2-a92a-2fe3f388706b%2F38mnmydd_processed.png&w=3840&q=75)
Transcribed Image Text:)² + y² = 4
In polar coordinates we have x² + y² = r² and x = rcos(8), so the boundary circle becomes ² =
4r cos (0)
(x - 2
D = {(r, 0)| -π/2 ≤ 0 π/2,0 ≤ r ≤ 4cos(0)}
π/2
4 cos(0)
V
1 = √ √ (²x
[]
=
/2
[TR]
= 48
= 48
= 48
or r = 4cos(0). Thus the disk D is given by
= 96
(3x² + 3y²)dA
3µ4
4
X
X
X
[5/²
5
0
=
(cos(0))+de
T/2 1 + cos(20)
0
2
17/²
14 cos (6) de = 192
372
-)²de
π/2
-π/2
π/2
[1 + 2cos(20) + ½(1 + cos(40))]de
r dr de
(cos(0))+de
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