Vector field F is given by F(r) = (x, y, z + 1) and surface A is the locus of points satisfying z + x² + y² + 2√√x² +y² = 8 for z > 0. (Note that the positive square root appears in this expression.) (b) Evaluate directly (using cylindrical coordinates or otherwise) the surface integral IS A I = where the flux is evaluated in the direction with positivez component. (c) Evaluate directly the volume-integral J = (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 F.dS W K = [[ B F.dS, 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
Chapter9: Systems Of Equations And Inequalities
Section: Chapter Questions
Problem 39RE
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Vector field F is given by
F(r) = (x, y, z + 1)
and surface A is the locus of points satisfying z + x² + y² + 2√x² + y² = 8 for z > 0. (Note
that the positive square root appears in this expression.)
(b) Evaluate directly (using cylindrical coordinates or otherwise) the surface integral
JS
I
J
=
where the flux is evaluated in the direction with positivez component.
(c) Evaluate directly the volume-integral
=
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
K=
=
|| F
B
F·ds,
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.
Transcribed Image Text:Vector field F is given by F(r) = (x, y, z + 1) and surface A is the locus of points satisfying z + x² + y² + 2√x² + y² = 8 for z > 0. (Note that the positive square root appears in this expression.) (b) Evaluate directly (using cylindrical coordinates or otherwise) the surface integral JS I J = where the flux is evaluated in the direction with positivez component. (c) Evaluate directly the volume-integral = 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 K= = || F B F·ds, 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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