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Solution to HWP05.01
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Transcript of Solution to HWP05.01
Hence we get the x- and y-components of the surface paramtzn fctn as:
05.01-1HWP 05.01Introduction: Constructing the Surface Parameterization Function
2
2
05.01-4(b) Infinitesimal surface area vector
See drawing above, showing -vector in the F -plane, attached to a point on the paraboloid surface. The infinitesimal increment vectors are also shown at the same surface point.
Corr190401:
Corr190401:
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f ,- vec*ors
,ayq'sirr fi-dir".tinl l-rygyses fkc siJ t r i t4|.e J . S O
05.01-7
Route I:
(d)
(*)
(*) Note: After shrinking HQ to zero, the -vectors on the"flattened" paraboloid, P(HQ), are pointing in the (+z)-direction, i.e., into the volumeV, as shown in the drawing above. But the surface Q is required to have its -vectors pointing out of the volumeV. That's why the -vectors on surface Q must be reversed relative to the -vectorson the flattened P(HQ)-surface.
V
V
RouteII:Wewillusetheresultsforparameterizationofacirculardiskshowninthethein-classexamplefile,postedontheHWProblemspageofthecoursewebsite:
hw05R_P3900_SurfaceIntgr+Stokes_Pt1_CircDisk_V01.pdf.Pleasereviewthisexampleand,inparticular,payattentiontothefiguresshowninthisexample.Inthefollowing,wewillrefertothisfile,andtheexampleshowntherein,as“CD”,forshort.
ThesurfaceQisacirculardiskofradiusRinthex-y-planecenteredatthecoordinateorigin,i.e.,withacenter𝑐 = 𝑐#𝑐%𝑐& = [000].TheparameterizationfunctionforthisdisksurfacegiveninCDis:
𝑟 𝑠, 𝜙 = 𝑥 𝑠, 𝜙 , 𝑦 𝑠, 𝜙 ,𝑧 𝑠, 𝜙 = 𝑠 cos 𝜙 ,𝑠 sin 𝜙 ,0 ,withintegrationintervals
0 ≤ 𝑠 ≤ 𝑅and0 ≤ 𝜙 < 2𝜋.
Theinfinitesimalincrementvectorsof𝑟 𝑠, 𝜙 aregiveninCDas𝑑<𝑟 = 𝜕<𝑟 𝑠, 𝜙 𝑑𝑠 = cos 𝜙 ,sin 𝜙 ,0 𝑑𝑠
and𝑑>𝑟 = 𝜕>𝑟 𝑠, 𝜙 𝑑𝜙 = [−𝑠 sin 𝜙 ,𝑠 cos 𝜙 ,0]𝑑𝜙.
Wewanttheareaelementvectors𝑑𝑎topointinthe(−𝑧)-direction:thisisthedirectionpointingoutwardfromthevolumeV,enclosedbetweentheparabolicdomesurfacePandthecirculardisksurfaceQ.Therefore,asshownindetailinCD,wemustchoosethecrossproductoftheinfinitesimalincrementvectors(with𝑑𝑠 > 0and𝑑𝜙 > 0)as
𝑑𝑎 = 𝑑>𝑟 × 𝑑<𝑟 AsshowninCD,thiswillgiveus
𝑑𝑎 = 0, 0, −𝑠 𝑑𝑠𝑑𝜙whichdoesindeedpointinthe(−𝑧)-direction,asrequired.This𝑑𝑎correspondstoacrossproductofthepartialderivatives
𝜕>𝑟 𝑠, 𝜙 × 𝜕<𝑟 𝑠, 𝜙 = 0, 0, −𝑠whichwewillusebelowtocarryoutthesurfaceintegrationviatheparametrizationequation.
Compoundingthegivenintegrandvectorfield,𝐹(𝑟),withthetheparameterizationfunction𝑟 𝑠, 𝜙 ,weget
05.01-8
𝐹 𝑟 𝑠, 𝜙 = 𝑥 𝑠, 𝜙 𝑦 𝑠, 𝜙 ,𝑦 𝑠, 𝜙 𝑧 𝑠, 𝜙 ,𝑧 𝑠, 𝜙 𝑥 𝑠, 𝜙 + 𝑥 𝑠, 𝜙 G
= 𝑠Gsin 𝜙 cos 𝜙 ,0,𝑠GcosG 𝜙 ,makinguseofthefactthat𝑧 𝑠, 𝜙 = 0everywhere.
The 𝑠, 𝜙 -integrandintheparameterizationequationforthesurfaceintegral(seebelow)isthengivenby
𝑓 𝑠, 𝜙 ≔𝐹 𝑟 𝑠, 𝜙 ∙ 𝜕>𝑟 𝑠, 𝜙 × 𝜕<𝑟 𝑠, 𝜙
= 𝐹& 𝑟 𝑠, 𝜙 𝜕>𝑟 𝑠, 𝜙 × 𝜕<𝑟 𝑠, 𝜙&
= 𝑠GcosG 𝜙 −𝑠= −𝑠KcosG 𝜙 .
Notethatthevectorfieldcomponents𝐹# 𝑟 𝑠, 𝜙 and𝐹% 𝑟 𝑠, 𝜙 donotcontributetotheintegrand𝑓 𝑠, 𝜙 ,nortoitsintegral,sincethe𝑥-and𝑦-componentsofthepartialderivativecrossproductarezero.
Bytheparameterizationequationforthesurfaceintegral,wehave𝑑𝑎 ∙ 𝐹(𝑟)
L = 𝑑𝑠MN 𝑑𝜙GO
N 𝐹 𝑟 𝑠, 𝜙 ∙ 𝜕>𝑟 𝑠, 𝜙 × 𝜕<𝑟 𝑠, 𝜙
= 𝑑𝑠MN 𝑑𝜙GO
N 𝑓 𝑠, 𝜙 =− 𝑑𝑠M
N 𝑠K 𝑑𝜙GON cosG 𝜙
Usinganintegraltable,wefind𝑑𝜙GO
N cosG 𝜙 = 𝜋.Hence,
𝑑𝑎 ∙ 𝐹(𝑟)L = −𝜋 𝑑𝑠M
N 𝑠K
=−𝜋 PQ𝑠Q
N
M
= −OQ𝑅Q
05.01-9
(e) Integral over the entire closed surface S 05.01-10
By given Concatenation Theorem for surface integrals:
Insert results for P- and Q-integral from parts (c) and (d):
q.e.d.
05.0205.02-1
the
This is illustrated in the two figures on the next page: Please review and make sure you understand them!
is again perpendicular to the F -plane, just as for the
are just concentric circles around the z-axis; and so are lines of constant s and constant z in the volume V.
. This is due to the fact that lines of constant s on the P-surface
05.02-4
, ,
Substitute: u := s/R, -> s = Ru, ds = Rdu
-> zb(s)=H(1-u2),
->(zb(s))2 = H2 (1-u2)2 = H2 (1-2u2+u4)
see HWP5.01 (e)
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