1. Vector field F is given by F(r) = (x, y, and surface A is the locus of points satisfying ≈ + x² + y² + 2√x² + y² = 8 for z > 0. (Note that the positive square root appears in this expression.) (a) Sketch by hand (not by computer) surface A. (b) Evaluate directly (using cylindrical coordinates or otherwise) the surface integral I J = (x, y, z + 1) = - where the flux is evaluated in the direction with positivez component. (c) Evaluate directly the volume-integral JS K= F.ds JIJ (V.F) dV where W is the solid region lying between surface A and the (x, y)-plane. (d) Hence, without further integration, determine the value of = JS F·ds, B where B is the disc of radius 2 in the (x, y)-plane, centred on the origin, with normal dS in the positive z direction and justify your answer.
1. Vector field F is given by F(r) = (x, y, and surface A is the locus of points satisfying ≈ + x² + y² + 2√x² + y² = 8 for z > 0. (Note that the positive square root appears in this expression.) (a) Sketch by hand (not by computer) surface A. (b) Evaluate directly (using cylindrical coordinates or otherwise) the surface integral I J = (x, y, z + 1) = - where the flux is evaluated in the direction with positivez component. (c) Evaluate directly the volume-integral JS K= F.ds JIJ (V.F) dV where W is the solid region lying between surface A and the (x, y)-plane. (d) Hence, without further integration, determine the value of = JS F·ds, B where B is the disc of radius 2 in the (x, y)-plane, centred on the origin, with normal dS in the positive z direction and justify your answer.
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter11: Topics From Analytic Geometry
Section: Chapter Questions
Problem 18T
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Transcribed Image Text:1. Vector field F is given by
F(r) = (x, y,
and surface A is the locus of points satisfying ≈ + x² + y² + 2√x² + y² = 8 for z > 0. (Note
that the positive square root appears in this expression.)
(a) Sketch by hand (not by computer) surface A.
(b) Evaluate directly (using cylindrical coordinates or otherwise) the surface integral
I
J
= (x, y, z + 1)
=
-
where the flux is evaluated in the direction with positivez component.
(c) Evaluate directly the volume-integral
JS
K=
F.ds
JIJ (V.F) dV
where W is the solid region lying between surface A and the (x, y)-plane.
(d) Hence, without further integration, determine the value of
=
JS F·ds,
B
where B is the disc of radius 2 in the (x, y)-plane, centred on the origin, with normal dS in
the positive z direction and justify your answer.
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