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Consider the double integral: I = √(1-x² - y²) dA where R is the region of the XY plane represented in the attached graph. Y R r = 2.sen (0) y = 1 π/4 B) I = √*/²4 [2²sen(0) (1 − r²)drdo + [7/1² When transforming the previous integral applying the change of variable to polar coordinates, we obtain: X π/4 2 sen(0) π/2 csc (0) A) I = ["/ª [²(º) (r – r³)drd0 + [7/72 **) (r = 1³) drdo - π/4 rcsc(0) (°) (1 – r²³) drdo π/4 2 C) I = √7/¹4 [²sen(0) (r – r³) drd0 + 74² ² (r = r²³) drdo - - π/4

Consider the double integral: I = √(1-x² - y²) dA where R is the region of the XY plane represented in the attached graph. Y R r = 2.sen (0) y = 1 π/4 B) I = √*/²4 [2²sen(0) (1 − r²)drdo + [7/1² When transforming the previous integral applying the change of variable to polar coordinates, we obtain: X π/4 2 sen(0) π/2 csc (0) A) I = ["/ª [²(º) (r – r³)drd0 + [7/72 **) (r = 1³) drdo - π/4 rcsc(0) (°) (1 – r²³) drdo π/4 2 C) I = √7/¹4 [²sen(0) (r – r³) drd0 + 74² ² (r = r²³) drdo - - π/4

Advanced Engineering Mathematics
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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Consider the double integral: 1₂₁₁ R where R is the region of the XY plane represented in the attached graph. T/4 2 sen(0) B) I= So C) I = π/4 2 sen(0) I = When transforming the previous integral applying the change of variable to polar coordinates, we obtain: π/4 D) I = ¹² * Y π/4 2 sen(0) π/2 rcsc(0) A) I = √™¹/4 √²s(®) (r – rª)drd0 + 7/² *(*) (r = r²³) drdo /4 Jo 2 sen(0) R (1 — x² — y²) dA - r = 2.sen(0) y = 1 (1 − r²)drdo + [71² π/4 rcsc(0) *(*) (1 − r²)drdo (r = r²³) drdo + [7/1² √² (r = 1 π/4 -r³) drde π/2 sen(0) (r = r²³) drdo + √3/2² (250) Jπ/4 Jcsc(0) (r-r³) drde
Transcribed Image Text:Consider the double integral: 1₂₁₁ R where R is the region of the XY plane represented in the attached graph. T/4 2 sen(0) B) I= So C) I = π/4 2 sen(0) I = When transforming the previous integral applying the change of variable to polar coordinates, we obtain: π/4 D) I = ¹² * Y π/4 2 sen(0) π/2 rcsc(0) A) I = √™¹/4 √²s(®) (r – rª)drd0 + 7/² *(*) (r = r²³) drdo /4 Jo 2 sen(0) R (1 — x² — y²) dA - r = 2.sen(0) y = 1 (1 − r²)drdo + [71² π/4 rcsc(0) *(*) (1 − r²)drdo (r = r²³) drdo + [7/1² √² (r = 1 π/4 -r³) drde π/2 sen(0) (r = r²³) drdo + √3/2² (250) Jπ/4 Jcsc(0) (r-r³) drde
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