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(2). Consider the double integral 1= ||. 72+ y? dA where R is the region of the XY plane, ven in the attached graph. (x+ 2)² + y² = 4 a? + y? = 4 R hen transforming the previous integral applying change of variable in polar coordinates, we otain: A) I = sin e drde + T/2 sin e drde px/2 I = Jo r2 r2m/3 -2 sin e drde + sin 0 drde /2 /2 CL 5%/6 C) I = sin 0 drde + sin 0 drde 4 cos e */2 2m/3 -2 D) I = sin 0 drde + sin 0 drde /2 4 Cos 6

(2). Consider the double integral 1= ||. 72+ y? dA where R is the region of the XY plane, ven in the attached graph. (x+ 2)² + y² = 4 a? + y? = 4 R hen transforming the previous integral applying change of variable in polar coordinates, we otain: A) I = sin e drde + T/2 sin e drde px/2 I = Jo r2 r2m/3 -2 sin e drde + sin 0 drde /2 /2 CL 5%/6 C) I = sin 0 drde + sin 0 drde 4 cos e */2 2m/3 -2 D) I = sin 0 drde + sin 0 drde /2 4 Cos 6

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter11: Topics From Analytic Geometry
Section11.5: Polar Coordinates
Problem 98E
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T/2 J4 cos 6 11 (2). Consider the double integral I = || dA where R is the region of the XY plane, 1² + y? given in the attached graph. (x + 2)² + y² = 4 a² + y? = 4 R When transforming the previous integral applying change of variable in polar coordinates, we obtain: px/2 -4 A) I = Jo sin e drde + sin e drde /2 -2m/3 B) I = sin e drde + sin 0 drde -4 cos 0 x/2 C) I = . sin 0 drde + sin e drde Jo 4 cos e x/2 2m/3 -2 D) I = sin 0 drd0+ sin 0 drde T
Transcribed Image Text:T/2 J4 cos 6 11 (2). Consider the double integral I = || dA where R is the region of the XY plane, 1² + y? given in the attached graph. (x + 2)² + y² = 4 a² + y? = 4 R When transforming the previous integral applying change of variable in polar coordinates, we obtain: px/2 -4 A) I = Jo sin e drde + sin e drde /2 -2m/3 B) I = sin e drde + sin 0 drde -4 cos 0 x/2 C) I = . sin 0 drde + sin e drde Jo 4 cos e x/2 2m/3 -2 D) I = sin 0 drd0+ sin 0 drde T
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