2 4. Surface S is the locus of points satisfying the equation (x² + y²)² + z = 1 with z ≥ 0 and vector field A is given in Cartesian coordinates by A=(y-xz, -x - yz, 1+z2). (a) Sketch surface-S then evaluate the integral Y -X2+1 d L = S A dS, S -x-Y2 -42-x -2-X using surface elements dS with positive z component. [15]
2 4. Surface S is the locus of points satisfying the equation (x² + y²)² + z = 1 with z ≥ 0 and vector field A is given in Cartesian coordinates by A=(y-xz, -x - yz, 1+z2). (a) Sketch surface-S then evaluate the integral Y -X2+1 d L = S A dS, S -x-Y2 -42-x -2-X using surface elements dS with positive z component. [15]
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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![2
4. Surface S is the locus of points satisfying the equation (x² + y²)² + z = 1 with z ≥ 0
and vector field A is given in Cartesian coordinates by A=(y-xz, -x - yz, 1+z2).
(a) Sketch surface-S then evaluate the integral
Y
-X2+1
d
L =
S
A dS,
S
-x-Y2
-42-x
-2-X
using surface elements dS with positive z component.
[15]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3011c556-643e-4a01-b0e0-55d8cf24eddf%2Fec90fa80-6b86-40ce-aa77-91a6a5dc0366%2Fd59w8eh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:2
4. Surface S is the locus of points satisfying the equation (x² + y²)² + z = 1 with z ≥ 0
and vector field A is given in Cartesian coordinates by A=(y-xz, -x - yz, 1+z2).
(a) Sketch surface-S then evaluate the integral
Y
-X2+1
d
L =
S
A dS,
S
-x-Y2
-42-x
-2-X
using surface elements dS with positive z component.
[15]
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