Solution to HWP05.01

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Transcript of Solution to HWP05.01

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Hence we get the x- and y-components of the surface paramtzn fctn as:

05.01-1HWP 05.01Introduction: Constructing the Surface Parameterization Function

05.01-2

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2

05.01-3

2

(a) Infinitesimal increment vectors

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:

05.01-5+x2 ]

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(c) 05.01-5

05.01-6

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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.

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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

05.02-2

05.02-3

05.02-2

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05.02-4

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Substitute: u := s/R, -> s = Ru, ds = Rdu

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see HWP5.01 (e)

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