5. Consider the double integral rydA r=4+4 cos(0) = 25 Where R is the region of the XY plane, i1 -io given in the attached graph: By transforming the above integral by applying variable change in polar coordinates, one obtains: -4+4 cos(@) A) (n* cos(0) sin(0))drd0 + / 1. I ( cos(0) sin(0))drd0 , con 0, = arc cos () 4+4 cos(8) rdrde + rdrd0, con 01 = arc cos () ra+4 cos(0) C) I (* cos(0) sin(0))drd0 + TT(* cos(0) sin(0))drd0, con 0, = arc cos () r4+4 cos(8) D) I( cos(0) sin(@))drd0 + (12 cos(0) sin(0))drd®, con 61 (): = are coS
5. Consider the double integral rydA r=4+4 cos(0) = 25 Where R is the region of the XY plane, i1 -io given in the attached graph: By transforming the above integral by applying variable change in polar coordinates, one obtains: -4+4 cos(@) A) (n* cos(0) sin(0))drd0 + / 1. I ( cos(0) sin(0))drd0 , con 0, = arc cos () 4+4 cos(8) rdrde + rdrd0, con 01 = arc cos () ra+4 cos(0) C) I (* cos(0) sin(0))drd0 + TT(* cos(0) sin(0))drd0, con 0, = arc cos () r4+4 cos(8) D) I( cos(0) sin(@))drd0 + (12 cos(0) sin(0))drd®, con 61 (): = are coS
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:5. Consider the double integral
rydA
r=4+4 cos(0)
= 25
Where R is the region of the XY plane,
i1 -o
given in the attached graph:
By transforming the above integral by applying variable change in polar coordinates, one obtains:
-4+4 cos(@)
A)
(p cos(0) sin(0))drd0 + | (cos(0) sin(0))drd0, con 0, = arc cos ()
4+4 cos(8)
rdrde +
rdrdo, con 01 = arc cos ()
ra+4 cos(0)
C)
IP cos(0) sin (0)drdo +
TI(* cos(0) sin(0))drd0, con 0, = arc cos ()
r4+4 cos(8)
D)
T [* cos(0) sin(0))drd0 +
(r2 cos(0) sin(0))drd®, con 61
():
= are coS
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