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Introduction to Electrodynamics
Found in: Page 55
Introduction to Electrodynamics

Introduction to Electrodynamics

Book edition 4th edition
Author(s) David J. Griffiths
Pages 613 pages
ISBN 9780321856562

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Short Answer

Compute the line integral of

v=(r cos2θ)r^-(r cosθsinθ)θ^+3rϕ^

around the path shown in Fig. 1.50 (the points are labeled by their Cartesian coordinates).Do it either in cylindrical or in spherical coordinates. Check your answer, using Stokes' theorem. [Answer: 3rr /2]

The line integral is evaluated to be 3π2 . The left and right side of the Stokes theorem gives same result. Hence strokes theorem is verified.

See the step by step solution

Step by Step Solution

Step 1: Describe the given information

The given vector is, v=(r cos2θ)r^-(r cosθsinθ)θ^+3rϕ^ . The line integral of the given vector has to be evaluated over the path drawn as follows:

Step 2: Define the Stokes theorem

The integral of curl of a function f (x,y,z) over an open surface area is equal to the line integral of the function (×v).ds= b v.dl.

Step 3: Compute the left side of strokes theorem

The formula of curl of a vector in spherical coordinates is

The curl of the vector v=(r cos2θ)r^-(r cosθsinθ)θ^+3rϕ^ is obtained as

×v=1r sinθ(sinθ(3r))θ-rcosθsinθϕr^+1r1sinθr cos2θϕ-(r(3r))rθ^+1r(r(r cosθsinθr-(r cos2θ)θϕ^

=1rsinθ(3r)cosθ r^+1r(-6r)θ^+0=3 cotθ r^-6^θ

The differential elemental area is da=rdrdθ ϕ^ . Substitute role="math" localid="1654664744003" 3cot r^- 6^θ for ×v , into the stokes theorem (×v).dτ=-v.da

role="math" localid="1654665787372" (×v).dτ=0+01 6rdr 0π2 dϕ =6r2210ϕ 0π2 =(3)π2 =3π2...................(1)

Step 4: Compute the right side of strokes theorem 

The differential length vector is given bydl=dr r^+rdθ θ^+r sinθ ϕ^ . Along the path, localid="1654672580270" θ=π2,θ=0 0 and r varies from 0 to 1.Hence the line integral becomes,

localid="1654672141711" v.dl=(r cos2 θr^-(r cosθ sinθ)θ^+3r ϕ^)(dr r^+ θ^+r sinθdϕϕ^) =(r cos2θ)dr-(r2 cosθ sinθ)+3r2sinθdϕ =r cos2π2dr-r2 cosπ2sin 3r2 sinπ23r2sinπ2 =0 dr-0+3r2

Simplify further as

localid="1654672150719" v.dl=3r2 =3r2ϕ =3r20 =0

Along the path (ii), θ=π2 ,r=1, and ϕ varies from 0 tolocalid="1654667564073" π2 .Hence the line integral becomes,

localid="1654672163671" v.dl=((r cos2 θ)r^-(r cosθ sinθ)θ^+3r ϕ^)(dr r^+ θ^+r sinθdϕϕ^) =(r cos2θ)dr-(r2 cosθ sinθ)+3r2sinθdϕ =(1cos2π2)dr-((12) cosπ2sin π23r2 sinπ2)+3r2sin(π2) =3

Simplify further as,

v.dl=0π23 =3(ϕ)0π2 =3π2

Along the path (iii), θvaries from π2 to localid="1654668630697" tan112, localid="1654668661756" ϕ=π2 and localid="1654668647962" r=1sinθ, such that localid="1654668615795" dr=-11sin2θcosθdθ Hence the line integral becomes,

localid="1654672691611" v.dl=((r cos2 θ)r^-(r cosθ sinθ)θ^+3r ϕ^)(dr r^+dr r^+rdθ θ^ r sinθdϕϕ^) =(r cos2θ)dr-(r2 cosθ sinθ)+3r2sinθdϕ =(1sinθcos2θ)(-1sinθcosθdθ)-(1sinθ2cosθsinθ)dθ+0

Simplify further as,

localid="1654670347504" v.dl=-cos3θsin3θ+cosθsinθ =cosθsinθcos2θ+sin2θsin2θ =π2tan-1122cosθsin3θ ...........(2)

Let localid="1654669903166" x=sin θ, then dx=cos2θdθ.Substitute x for localid="1654672592461" sinθ and dx for cos θdθ into equation (2)

v.dl=-v.dl=12sin2θ =12x2

Substitute back for into above result as,

localid="1654672710295" v.dl=12sin2 θ

Evaluate the limit as,

localid="1654672130422" v.dl=12sin2θx2tan-112 =12sin2tan-112-12sin2π2 =12(0.2)-12 =2

Along the path (iv), θ=tan-112,ϕ=π2 and r varies from localid="1654671277385" 5 to 0, Hence the line integral becomes,

localid="1654672112340" v.dl=rcos2θr^-(r cosθsinθ)θ^+3rϕ^)(dr r^+rdθ θ^ +rsinθdϕϕ^ =r cos2tan-112dr =500.8rdr =0.8r2205

Simplify further as,

v.d=0.80-522 =-2

The integral of all the four parts are added to give:

v.dl=0+3π2+2+-2 =3π2 .........(3)

From equation (1) and (3), the left and right side gives same result. Hence strokes theorem is verified.

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