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
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
![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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F58070b55-d199-45de-ae72-3a0bef1f5e72%2F28f5ebc2-8767-40f4-abc9-1d496d8b2aa3%2Fckk8a6q_processed.jpeg&w=3840&q=75)
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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