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A) I = = B) I = C) I= D) I= Consider the double integral xy - SS₁ √²+ + 7² d √√x² Where R is the region of the Plane XY, given by the graph: When transforming the integral applying variable change to polar coordinates we get: 2/3 4 sin 8 4 sin 8 r sin cos drd0 + r sin cos drde 2+4 cos 8 4 sin 8 r² sin 0 cos 0 drd0 + r² sin cos 0 drdo √2+4 cos 0 4 sin 0 r sin cos drd0 + r sin cos drdo √2+4 cos 0 r² sin 0 cos 0 drd0 + 7² sin cos 0 drd0 Je₁ Sot JO₁ 2x/3 5/6 JO₁ 5/6 4 sin 0 Love JO₁ I= J2n/3 J0 2/3 J0 S/S 5л/6 f. 4 sin 4 sin 0 -4 sin 0 dA x² + (y-2)² = 4 R r = 2+4 cos 0

A) I = = B) I = C) I= D) I= Consider the double integral xy - SS₁ √²+ + 7² d √√x² Where R is the region of the Plane XY, given by the graph: When transforming the integral applying variable change to polar coordinates we get: 2/3 4 sin 8 4 sin 8 r sin cos drd0 + r sin cos drde 2+4 cos 8 4 sin 8 r² sin 0 cos 0 drd0 + r² sin cos 0 drdo √2+4 cos 0 4 sin 0 r sin cos drd0 + r sin cos drdo √2+4 cos 0 r² sin 0 cos 0 drd0 + 7² sin cos 0 drd0 Je₁ Sot JO₁ 2x/3 5/6 JO₁ 5/6 4 sin 0 Love JO₁ I= J2n/3 J0 2/3 J0 S/S 5л/6 f. 4 sin 4 sin 0 -4 sin 0 dA x² + (y-2)² = 4 R r = 2+4 cos 0

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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A) I = = B) I = C) I= D) I= Consider the double integral xy - SS₁ √ 2 ² + y ² d Where R is the region of the Plane XY, given by the graph: When transforming the integral applying variable change to polar coordinates we get: 2/3 4 sin 8 4 sin 8 r sin cos drd0 + r sin cos drd0 2+4 cos 8 4 sin 8 r² sin cos 0 drd0 + 7² sin cos 0 drd0 √2+4 cos 0 4 sin 0 r sin cos 0 drd0 + r sin cos drdo √2+4 cos 0 r² sin 0 cos 0 drd0 + Jo₁ So 0₁ 2/3 5/6 JO₁ South JO₁ 5/6 4 sin 0 I= J2n/3 JO 2/3. St. Jo 5л/6 f. 4 sin 4 sin -4 sin 0 dA ² sin 0 cos 0 drd0 x² + (y-2)² = 4 R r=2+4 cos 0
Transcribed Image Text:A) I = = B) I = C) I= D) I= Consider the double integral xy - SS₁ √ 2 ² + y ² d Where R is the region of the Plane XY, given by the graph: When transforming the integral applying variable change to polar coordinates we get: 2/3 4 sin 8 4 sin 8 r sin cos drd0 + r sin cos drd0 2+4 cos 8 4 sin 8 r² sin cos 0 drd0 + 7² sin cos 0 drd0 √2+4 cos 0 4 sin 0 r sin cos 0 drd0 + r sin cos drdo √2+4 cos 0 r² sin 0 cos 0 drd0 + Jo₁ So 0₁ 2/3 5/6 JO₁ South JO₁ 5/6 4 sin 0 I= J2n/3 JO 2/3. St. Jo 5л/6 f. 4 sin 4 sin -4 sin 0 dA ² sin 0 cos 0 drd0 x² + (y-2)² = 4 R r=2+4 cos 0
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