Solution Manual For Introduction To Electrodynamics, 4th Edition

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Solution ManualIntroduction to ElectrodynamicsFourth EditionDavid J. Griffiths2012

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2Contents1Vector Analysis42Electrostatics263Potential534Electric Fields in Matter925Magnetostatics1106Magnetic Fields in Matter1337Electrodynamics1458Conservation Laws1689Electromagnetic Waves18510 Potentials and Fields21011 Radiation23112 Electrodynamics and Relativity261c

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4CHAPTER 1. VECTOR ANALYSISChapter 1Vector AnalysisProblem 1.1CHAPTER 1. VECTOR ANALYSIS3Chapter 1Vector AnalysisProblem 1.1!"#$ABCB+C︸︷︷︸|B|cosθ1︸︷︷︸|C|cosθ2}|B|sinθ1}|C|sinθ2θ1θ2θ3(a) From the diagram,|B+C|cosθ3=|B|cosθ1+|C|cosθ2.|A||B+C|cosθ3=|A||B|cosθ1+|A||C|cosθ2.So:A·(B+C) =A·B+A·C. (Dot product is distributive)Similarly:|B+C|sinθ3=|B|sinθ1+|C|sinθ2. Mulitply by|A|ˆn.|A||B+C|sinθ3ˆn=|A||B|sinθ1ˆn+|A||C|sinθ2ˆn.Ifˆnis the unit vector pointing out of the page, it follows thatA×(B+C) = (A×B) + (A×C). (Cross product is distributive)(b) For the general case, see G. E. Hay’sVector and Tensor Analysis, Chapter 1, Section 7 (dot product) andSection 8 (cross product)Problem 1.2!A=B%C&B×C'A×(B×C)The triple cross-product isnotin general associative.For example,supposeA=BandCis perpendicular toA, as in the diagram.Then (B×C) points out-of-the-page, andA×(B×C) pointsdown,and has magnitudeABC.But (A×B) = 0, so (A×B)×C= 0!=A×(B×C).Problem 1.3!y%z(x!B"AθA= +1ˆx+ 1ˆy−1ˆz;A=√3;B= 1ˆx+ 1ˆy+ 1ˆz;A·B= +1 + 1−1 = 1 =ABcosθ=√3√3 cosθ⇒cosθ.θ= cos−1(13)≈70.5288◦Problem 1.4The cross-product of any two vectors in the plane will give a vector perpendicular to the plane. For example,we might pick the base (A) and the left side (B):c©2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist.(a) From the diagram,|B+C|cosθ3=|B|cosθ1+|C|cosθ2. Multiply by|A|.|A||B+C|cosθ3=|A||B|cosθ1+|A||C|cosθ2.So:A·(B+C) =A·B+A·C. (Dot product is distributive)Similarly:|B+C|sinθ3=|B|sinθ1+|C|sinθ2. Mulitply by|A|ˆn.|A||B+C|sinθ3ˆn=|A||B|sinθ1ˆn+|A||C|sinθ2ˆn.Ifˆnis the unit vector pointing out of the page, it follows thatA×(B+C) = (A×B) + (A×C). (Cross product is distributive)(b) For the general case, see G. E. Hay’sVector and Tensor Analysis, Chapter 1, Section 7 (dot product) andSection 8 (cross product)Problem 1.2CHAPTER 1. VECTOR ANALYSIS3Chapter 1Vector AnalysisProblem 1.1!"#$ABCB+C︸︷︷︸|B|cosθ1︸︷︷︸|C|cosθ2}|B|sinθ1}|C|sinθ2θ1θ2θ3(a) From the diagram,|B+C|cosθ3=|B|cosθ1+|C|cosθ2. Multiply by|A|.|A||B+C|cosθ3=|A||B|cosθ1+|A||C|cosθ2.So:A·(B+C) =A·B+A·C. (Dot product is distributive)Similarly:|B+C|sinθ3=|B|sinθ1+|C|sinθ2. Mulitply by|A|ˆn.|A||B+C|sinθ3ˆn=|A||B|sinθ1ˆn+|A||C|sinθ2ˆn.Ifˆnis the unit vector pointing out of the page, it follows thatA×(B+C) = (A×B) + (A×C). (Cross product is distributive)(b) For the general case, see G. E. Hay’sVector and Tensor Analysis, Chapter 1, Section 7 (dot product) andSection 8 (cross product)Problem 1.2!A=B%C&B×C'A×(B×C)The triple cross-product isnotin general associative.For example,supposeA=BandCis perpendicular toA, as in the diagram.Then (B×C) points out-of-the-page, andA×(B×C) pointsdown,and has magnitudeABC.But (A×B) = 0, so (A×B)×C= 0!=A×(B×C).Problem 1.3!y%z(x!B"AθA= +1ˆx+ 1ˆy−1ˆz;A=√3;B= 1ˆx+ 1ˆy+ 1ˆz;B=√3.A·B= +1 + 1−1 = 1 =ABcosθ=√3√3 cosθ⇒cosθ=13.θ= cos−1(13)≈70.5288◦Problem 1.4The cross-product of any two vectors in the plane will give a vector perpendicular to the plane. For example,we might pick the base (A) and the left side (B):A=−1ˆx+ 2y+0z;=−1x+0y+3z.c©2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist.The triple cross-product isnotin general associative.For example,supposeA=BandCis perpendicular toA, as in the diagram.Then (B×C) points out-of-the-page, andA×(B×C) pointsdown,and has magnitudeABC.But (A×B) =0, so (A×B)×C=06=A×(B×C).Problem 1.3CHAPTER 1. VECTOR ANALYSIS3Chapter 1Vector AnalysisProblem 1.1!"#$ABCB+C︸︷︷︸|B|cosθ1︸︷︷︸|C|cosθ2}|B|sin}|C|sinθ1θ2θ3(a) From the diagram,|B+C|cosθ3=|B|cosθ1+|C|cosθ2.|A||B+C|cosθ3=|A||B|cosθ1+|A||C|cosθ2.So:A·(B+C) =A·B+A·C. (Dot product is distributive)Similarly:|B+C|sinθ3=|B|sinθ1+|C|sinθ2. Mulitply by|A|ˆn.|A||B+C|sinθ3ˆn=|A||B|sinθ1ˆn+|A||C|sinθ2ˆn.Ifˆnis the unit vector pointing out of the page, it follows thatA×(B+C) = (A×B) + (A×C). (Cross product is distributive)(b) For the general case, see G. E. Hay’sVector and Tensor Analysis, Chapter 1, Section 7 (dot product)Section 8 (cross product)Problem 1.2!A=B%C&B×C'A×(B×C)The triple cross-product isnotin general associative.For example,supposeA=BandCis perpendicular toA, as in the diagram.Then (B×C) points out-of-the-page, andA×(B×C) pointsdown,and has magnitudeABC.But (A×B) = 0, so (A×B)×C= 0!=A×(B×C).Problem 1.3!y%z(x!B"AθA= +1ˆx+ 1ˆy−1ˆz;A=√3;B= 1ˆx+ 1ˆy+ 1ˆz;A·B= +1 + 1−1 = 1 =ABcosθ=√3√3 cosθ⇒cosθ.θ= cos−1(13)≈70.5288◦Problem 1.4The cross-product of any two vectors in the plane will give a vector perpendicular to the plane. For examwe might pick the base (A) and the left side (B):c©2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist.A= +1ˆx+ 1ˆy−1ˆz;A=√3;B= 1ˆx+ 1ˆy+ 1ˆz;B=√3.A·B= +1 + 1−1 = 1 =ABcosθ=√3√3 cosθ⇒cosθ=13.θ= cos−1(13)≈70.5288◦Problem 1.4The cross-product of any two vectors in the plane will give a vector perpendicular to the plane. For example,we might pick the base (A) and the left side (B):A=−1ˆˆˆˆˆˆ

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CHAPTER 1. VECTOR ANALYSIS5A×B=∣∣∣∣∣∣∣∣∣ˆxˆy ˆz−1 2 0−1 0 3∣∣∣∣∣∣∣∣∣= 6ˆx+ 3ˆy+ 2ˆz.This has the rightdirection, but the wrongmagnitude. To make aunitvector out of it, simply divide by itslength:|A×B|=√36 + 9 + 4 = 7.ˆn=A×B|A×B|=67ˆx+37ˆy+27ˆz.Problem 1.5A×(B×C) =∣∣∣∣∣∣∣∣∣ˆxˆyˆzAxAyAz(ByCz−BzCy) (BzCx−BxCz) (BxCy−ByCx)∣∣∣∣∣∣∣∣∣=ˆx[Ay(BxCy−ByCx)−Az(BzCx−BxCz)] +ˆy() +ˆz()(I’ll just check the x-component; the others go the same way)=ˆx(AyBxCy−AyByCx−AzBzCx+AzBxCz) +ˆy() +ˆz().B(A·C)−C(A·B) = [Bx(AxCx+AyCy+AzCz)−Cx(AxBx+AyBy+AzBz)]ˆx+ ()ˆy+ ()ˆz=ˆx(AyBxCy+AzBxCz−AyByCx−AzBzCx) +ˆy() +ˆz(). They agree.Problem 1.6A×(B×C)+B×(C×A)+C×(A×B) =B(A·C)−C(A·B)+C(A·B)−A(C·B)+A(B·C)−B(C·A) =0.So:A×(B×C)−(A×B)×C=−B×(C×A) =A(B·C)−C(A·B).If this is zero, then eitherAis parallel toC(including the case in which they point inoppositedirections, orone is zero), or elseB·C=B·A= 0, in which caseBis perpendicular toAandC(including the caseB=0.)Conclusion:A×(B×C) = (A×B)×C⇐⇒eitherAis parallel toC, orBis perpendicular toAandC.Problem 1.7r= (4ˆx+ 6ˆy+ 8ˆz)−(2ˆx+ 8ˆy+ 7ˆz) =2ˆx−2ˆy+ˆzr=√4 + 4 + 1 =3ˆr=rr=23ˆx−23ˆy+13ˆzProblem 1.8(a) ¯Ay¯By+ ¯Az¯Bz= (cosφAy+ sinφAz)(cosφBy+ sinφBz) + (−sinφAy+ cosφAz)(−sinφBy+ cosφBz)= cos2φAyBy+ sinφcosφ(AyBz+AzBy) + sin2φAzBz+ sin2φAyBy−sinφcosφ(AyBz+AzBy) +cos2φAzBz= (cos2φ+ sin2φ)AyBy+ (sin2φ+ cos2φ)AzBz=AyBy+AzBz.X(b) (Ax)2+ (Ay)2+ (Az)2= Σ3i=1AiAi= Σ3i=1(Σ3j=1RijAj) (Σ3k=1RikAk)= Σj,k(ΣiRijRik)AjAk.This equalsA2x+A2y+A2zprovidedΣ3i=1RijRik={1if j=k0if j6=k}Moreover, ifRis to preserve lengths forallvectorsA, then this condition is not onlysufficientbut alsonecessary. For supposeA= (1,0,0). Then Σj,k(ΣiRijRik)AjAk= ΣiRi1Ri1, and this must equal 1 (since wewantA2x+A2y+A2z= 1). Likewise, Σ3i=1Ri2Ri2= Σ3i=1Ri3Ri3= 1. To check the casej6=k, chooseA= (1,1,0).Then we want 2 = Σj,k(ΣiRijRik)AjAk= ΣiRi1Ri1+ ΣiRi2Ri2+ ΣiRi1Ri2+ ΣiRi2Ri1. But we alreadyknow that the first two sums are both 1; the third and fourth areequal, so ΣiRi1Ri2= ΣiRi2Ri1= 0, and soon for other unequal combinations ofj,k.XIn matrix notation:˜RR= 1, where ˜Ris thetransposeofR.c

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6CHAPTER 1. VECTOR ANALYSISProblem 1.9CHAPTER 1. VECTOR ANALYSIS5!x"y#z$%Looking down the axis:"y&x'z"z′&y′'x′()*c©2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights resprotected unLooking down the axis:CHAPTER 1. VECTOR ANALYSIS5!x"y#z$%Looking down the axis:"y&x'z"z′&y′'x′()*c©2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist.A 120◦rotation carries thezaxis into they(=z) axis,yintox(=y), andxintoz(=x). SoAx=Az,Ay=Ax,Az=Ay.R=0 0 11 0 00 1 0Problem 1.10(a)No change.(Ax=Ax,Ay=Ay,Az=Az)(b)A−→ −A,in the sense (Ax=−Ax,Ay=−Ay,Az=−Az)(c) (A×B)−→(−A)×(−B) = (A×B).That is, ifC=A×B,C−→C.Nominus sign, in contrast tobehavior of an “ordinary” vector, as given by (b). IfAandBarepseudovectors, then (A×B)−→(A)×(B) =(A×B). So the cross-product of two pseudovectors is again apseudovector. In the cross-product of a vectorand a pseudovector, one changes sign, the other doesn’t, and therefore the cross-product is itself avector.Angular momentum(L=r×p) andtorque(N=r×F) are pseudovectors.(d)A·(B×C)−→(−A)·((−B)×(−C)) =−A·(B×C). So, ifa=A·(B×C), thena−→ −a;a pseudoscalarchanges signunder inversion of coordinates.Problem 1.11(a)∇f= 2xˆx+ 3y2ˆy+ 4z3ˆz(b)∇f= 2xy3z4ˆx+ 3x2y2z4ˆy+ 4x2y3z3ˆz(c)∇f=exsinylnzˆx+excosylnzˆy+exsiny(1/z)ˆzProblem 1.12(a)∇h= 10[(2y−6x−18)ˆx+ (2x−8y+ 28)ˆy].∇h= 0 at summit, so2y−6x−18 = 02x−8y+ 28 = 0 =⇒6x−24y+ 84 = 0}2y−18−24y+ 84 = 0.22y= 66 =⇒y= 3 =⇒2x−24 + 28 = 0 =⇒x=−2.Top is3 miles north, 2 miles west, of South Hadley.(b) Putting inx=−2,y= 3:h= 10(−12−12−36 + 36 + 84 + 12) =720 ft.(c) Putting inx= 1,y= 1:∇h= 10[(2−6−18)ˆx+ (2−8 + 28)ˆy] = 10(−22ˆx+ 22ˆy) = 220(−ˆx+ˆy).|∇h|= 220√2≈311 ft/mile;direction:northwest.c

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CHAPTER 1. VECTOR ANALYSIS7Problem 1.13r= (x−x′)ˆx+ (y−y′)ˆy+ (z−z′)ˆz;r=√(x−x′)2+ (y−y′)2+ (z−z′)2.(a)∇(r2) =∂∂x[(x−x′)2+(y−y′)2+(z−z′)2]ˆx+∂∂y()ˆy+∂∂z()ˆz= 2(x−x′)ˆx+2(y−y′)ˆy+2(z−z′)ˆz= 2r.(b)∇(1r) =∂∂x[(x−x′)2+ (y−y′)2+ (z−z′)2]−12ˆx+∂∂y()−12ˆy+∂∂z()−12ˆz=−12()−322(x−x′)ˆx−12()−322(y−y′)ˆy−12()−322(z−z′)ˆz=−()−32[(x−x′)ˆx+ (y−y′)ˆy+ (z−z′)ˆz] =−(1/r3)r=−(1/r2) ˆr.(c)∂∂x(rn) =nrn−1∂r∂x=nrn−1(121r2rx) =nrn−1ˆrx, so∇(rn) =nrn−1ˆrProblem 1.14y= +ycosφ+zsinφ; multiply by sinφ:ysinφ= +ysinφcosφ+zsin2φ.z=−ysinφ+zcosφ; multiply by cosφ:zcosφ=−ysinφcosφ+zcos2φ.Add:ysinφ+zcosφ=z(sin2φ+ cos2φ) =z. Likewise,ycosφ−zsinφ=y.So∂y∂y= cosφ;∂y∂z=−sinφ;∂z∂y= sinφ;∂z∂z= cosφ. Therefore(∇f)y=∂f∂y=∂f∂y∂y∂y+∂f∂z∂z∂y= + cosφ(∇f)y+ sinφ(∇f)z(∇f)z=∂f∂z=∂f∂y∂y∂z+∂f∂z∂z∂z=−sinφ(∇f)y+ cosφ(∇f)z}So∇ftransforms as a vector.qedProblem 1.15(a)∇·va=∂∂x(x2) +∂∂y(3xz2) +∂∂z(−2xz) = 2x+ 0−2x= 0.(b)∇·vb=∂∂x(xy) +∂∂y(2yz) +∂∂z(3xz) =y+ 2z+ 3x.(c)∇·vc=∂∂x(y2) +∂∂y(2xy+z2) +∂∂z(2yz) = 0 + (2x) + (2y) = 2(x+y)Problem 1.16∇·v=∂∂x(xr3) +∂∂y(yr3) +∂∂z(zr3) =∂∂x[x(x2+y2+z2)−32]+∂∂y[y(x2+y2+z2)−32]+∂∂z[z(x2+y2+z2)−32]= ()−32+x(−3/2)()−522x+ ()−32+y(−3/2)()−522y+ ()−32+z(−3/2)()−522z= 3r−3−3r−5(x2+y2+z2) = 3r−3−3r−3= 0.This conclusion is surprising, because, from the diagram, this vector field is obviously diverging away from theorigin.How, then, can∇·v= 0?The answer is that∇·v= 0 everywhereexceptat the origin, but at theorigin our calculation is no good, sincer= 0, and the expression forvblows up. In fact,∇·visinfiniteatthat one point, and zero elsewhere, as we shall see in Sect. 1.5.Problem 1.17vy= cosφ vy+ sinφ vz;vz=−sinφ vy+ cosφ vz.∂vy∂y=∂vy∂ycosφ+∂vz∂ysinφ=(∂vy∂y∂y∂y+∂vy∂z∂z∂y)cosφ+(∂vz∂y∂y∂y+∂vz∂z∂z∂y)sinφ.Use result in Prob. 1.14:=(∂vy∂ycosφ+∂vy∂zsinφ)cosφ+(∂vz∂ycosφ+∂vz∂zsinφ)sinφ.∂vz∂z=−∂vy∂zsinφ+∂vz∂zcosφ=−(∂vy∂y∂y∂z+∂vy∂z∂z∂z)sinφ+(∂vz∂y∂y∂z+∂vz∂z∂z∂z)cosφ=−(−∂vy∂ysinφ+∂vy∂zcosφ)sinφ+(−∂vz∂ysinφ+∂vz∂zcosφ)cosφ. Soc

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8CHAPTER 1. VECTOR ANALYSIS∂vy∂y+∂vz∂z=∂vy∂ycos2φ+∂vy∂zsinφcosφ+∂vz∂ysinφcosφ+∂vz∂zsin2φ+∂vy∂ysin2φ−∂vy∂zsinφcosφ−∂vz∂ysinφcosφ+∂vz∂zcos2φ=∂vy∂y(cos2φ+ sin2φ)+∂vz∂z(sin2φ+ cos2φ)=∂vy∂y+∂vz∂z.XProblem 1.18(a)∇×va=∣∣∣∣∣∣∣∣∣ˆxˆyˆz∂∂x∂∂y∂∂zx23xz2−2xz∣∣∣∣∣∣∣∣∣=ˆx(0−6xz) +ˆy(0 + 2z) +ˆz(3z2−0) =−6xzˆx+ 2zˆy+ 3z2ˆz.(b)∇×vb=∣∣∣∣∣∣∣∣∣ˆxˆyˆz∂∂x∂∂y∂∂zxy2yz3xz∣∣∣∣∣∣∣∣∣=ˆx(0−2y) +ˆy(0−3z) +ˆz(0−x) =−2yˆx−3zˆy−xˆz.(c)∇×vc=∣∣∣∣∣∣∣∣∣ˆxˆyˆz∂∂x∂∂y∂∂zy2(2xy+z2) 2yz∣∣∣∣∣∣∣∣∣=ˆx(2z−2z) +ˆy(0−0) +ˆz(2y−2y) =0.Problem 1.19AxyzvvvvBAs we go from pointAto pointB(9 o’clock to 10 o’clock),xincreases,yincreases,vxincreases, andvydecreases, so∂vx/∂y >0, while∂vy/∂y <0.On the circle,vz= 0, and there is nodependence onz, so Eq. 1.41 says∇×v=ˆz(∂vy∂x−∂vx∂y)points in thenegativezdirection(into the page), as the righthand rule would suggest.(Pick any other nearby points on thecircle and you will come to the same conclusion.) [I’m sorry, but Icannot remember who suggested this cute illustration.]Problem 1.20v=yˆx+xˆy; orv=yzˆx+xzˆy+xyˆz; orv= (3x2z−z3)ˆx+ 3ˆy+ (x3−3xz2)ˆz;orv= (sinx)(coshy)ˆx−(cosx)(sinhy)ˆy; etc.Problem 1.21(i)∇(f g) =∂(f g)∂xˆx+∂(f g)∂yˆy+∂(f g)∂zˆz=(f∂g∂x+g∂f∂x)ˆx+(f∂g∂y+g∂f∂y)ˆy+(f∂g∂z+g∂f∂z)ˆz=f(∂g∂xˆx+∂g∂yˆy+∂g∂zˆz)+g(∂f∂xˆx+∂f∂yˆy+∂f∂zˆz)=f(∇g) +g(∇f).qed(iv)∇·(A×B) =∂∂x(AyBz−AzBy) +∂∂y(AzBx−AxBz) +∂∂z(AxBy−AyBx)=Ay ∂Bz∂x+Bz ∂Ay∂x−Az ∂By∂x−By ∂Az∂x+Az ∂Bx∂y+Bx ∂Az∂y−Ax ∂Bz∂y−Bz ∂Ax∂y+Ax ∂By∂z+By ∂Ax∂z−Ay ∂Bx∂z−Bx ∂Ay∂z=Bx(∂Az∂y−∂Ay∂z)+By(∂Ax∂z−∂Az∂x)+Bz(∂Ay∂x−∂Ax∂y)−Ax(∂Bz∂y−∂By∂z)−Ay(∂Bx∂z−∂Bz∂x)−Az(∂By∂x−∂Bx∂y)=B·(∇×A)−A·(∇×B).qed(v)∇×(fA) =(∂(f Az)∂y−∂(f Ay)∂z)ˆx+(∂(f Ax)∂z−∂(f Az)∂x)ˆy+(∂(f Ay)∂x−∂(f Ax)∂y)ˆzc

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CHAPTER 1. VECTOR ANALYSIS9=(f∂Az∂y+Az ∂f∂y−f∂Ay∂z−Ay ∂f∂z)ˆx+(f∂Ax∂z+Ax ∂f∂z−f∂Az∂x−Az ∂f∂x)ˆy+(f∂Ay∂x+Ay ∂f∂x−f∂Ax∂y−Ax ∂f∂y)ˆz=f[(∂Az∂y−∂Ay∂z)ˆx+(∂Ax∂z−∂Az∂x)ˆy+(∂Ay∂x−∂Ax∂y)ˆz]−[(Ay ∂f∂z−Az ∂f∂y)ˆx+(Az ∂f∂x−Ax ∂f∂z)ˆy+(Ax ∂f∂y−Ay ∂f∂x)ˆz]=f(∇×A)−A×(∇f).qedProblem 1.22(a) (A·∇)B=(Ax ∂Bx∂x+Ay ∂Bx∂y+Az ∂Bx∂z)ˆx+(Ax ∂By∂x+Ay ∂By∂y+Az ∂By∂z)ˆy+(Ax ∂Bz∂x+Ay ∂Bz∂y+Az ∂Bz∂z)ˆz.(b)ˆr=rr=xˆx+yˆy+zˆz√x2+y2+z2. Let’s just do thexcomponent.[(ˆr·∇)ˆr]x=1√(x∂∂x+y∂∂y+z∂∂z)x√x2+y2+z2=1r{x[1√+x(−12)1(√)32x]+yx[−121(√)32y]+zx[−121(√)32z]}=1r{xr−1r3(x3+xy2+xz2)}=1r{xr−xr3(x2+y2+z2)}=1r(xr−xr)= 0.Same goes for the other components. Hence:(ˆr·∇)ˆr=0.(c) (va·∇)vb=(x2∂∂x+ 3xz2∂∂y−2xz∂∂z)(xyˆx+ 2yzˆy+ 3xzˆz)=x2(yˆx+ 0ˆy+ 3zˆz) + 3xz2(xˆx+ 2zˆy+ 0ˆz)−2xz(0ˆx+ 2yˆy+ 3xˆz)=(x2y+ 3x2z2)ˆx+(6xz3−4xyz)ˆy+(3x2z−6x2z)ˆz=x2(y+ 3z2)ˆx+ 2xz(3z2−2y)ˆy−3x2zˆzProblem 1.23(ii) [∇(A·B)]x=∂∂x(AxBx+AyBy+AzBz) =∂Ax∂xBx+Ax ∂Bx∂x+∂Ay∂xBy+Ay ∂By∂x+∂Az∂xBz+Az ∂Bz∂x[A×(∇×B)]x=Ay(∇×B)z−Az(∇×B)y=Ay(∂By∂x−∂Bx∂y)−Az(∂Bx∂z−∂Bz∂x)[B×(∇×A)]x=By(∂Ay∂x−∂Ax∂y)−Bz(∂Ax∂z−∂Az∂x)[(A·∇)B]x=(Ax ∂∂x+Ay ∂∂y+Az ∂∂z)Bx=Ax ∂Bx∂x+Ay ∂Bx∂y+Az ∂Bx∂z[(B·∇)A]x=Bx ∂Ax∂x+By ∂Ax∂y+Bz ∂Ax∂zSo [A×(∇×B) +B×(∇×A) + (A·∇)B+ (B·∇)A]x=Ay ∂By∂x−Ay ∂Bx∂y−Az ∂Bx∂z+Az ∂Bz∂x+By ∂Ay∂x−By ∂Ax∂y−Bz ∂Ax∂z+Bz ∂Az∂x+Ax ∂Bx∂x+Ay ∂Bx∂y+Az ∂Bx∂z+Bx ∂Ax∂x+By ∂Ax∂y+Bz ∂Ax∂z=Bx ∂Ax∂x+Ax ∂Bx∂x+By(∂Ay∂x−∂Ax∂y/+∂Ax∂y/)+Ay(∂By∂x−∂Bx∂y/+∂Bx∂y/)+Bz(−∂Ax∂z/+∂Az∂x+∂Ax∂z/)+Az(−∂Bx∂z/+∂Bz∂x+∂Bx∂z/)= [∇(A·B)]x(same foryandz)(vi) [∇×(A×B)]x=∂∂y(A×B)z−∂∂z(A×B)y=∂∂y(AxBy−AyBx)−∂∂z(AzBx−AxBz)=∂Ax∂yBy+Ax ∂By∂y−∂Ay∂yBx−Ay ∂Bx∂y−∂Az∂zBx−Az ∂Bx∂z+∂Ax∂zBz+Ax ∂Bz∂z[(B·∇)A−(A·∇)B+A(∇·B)−B(∇·A)]x=Bx ∂Ax∂x+By ∂Ax∂y+Bz ∂Ax∂z−Ax ∂Bx∂x−Ay ∂Bx∂y−Az ∂Bx∂z+Ax(∂Bx∂x+∂By∂y+∂Bz∂z)−Bx(∂Ax∂x+∂Ay∂y+∂Az∂z)c

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10CHAPTER 1. VECTOR ANALYSIS=By ∂Ax∂y+Ax(−∂Bx∂x/+∂Bx∂x/+∂By∂y+∂Bz∂z)+Bx(∂Ax∂x/−∂Ax∂x/−∂Ay∂y−∂Az∂z)+Ay(−∂Bx∂y)+Az(−∂Bx∂z)+Bz(∂Ax∂z)= [∇×(A×B)]x(same foryandz)Problem 1.24∇(f /g) =∂∂x(f /g)ˆx+∂∂y(f /g)ˆy+∂∂z(f /g)ˆz=g∂f∂x−f∂g∂xg2ˆx+g∂f∂y−f∂g∂yg2ˆy+g∂f∂z−f∂g∂zg2ˆz=1g2[g(∂f∂xˆx+∂f∂yˆy+∂f∂zˆz)−f(∂g∂xˆx+∂g∂yˆy+∂g∂zˆz)]=g∇f−f∇gg2.qed∇·(A/g) =∂∂x(Ax/g) +∂∂y(Ay/g) +∂∂z(Az/g)=g∂Ax∂x−Ax ∂g∂xg2+g∂Ay∂y−Ay ∂g∂yg2+g∂Az∂z−Az ∂g∂xg2=1g2[g(∂Ax∂x+∂Ay∂y+∂Az∂z)−(Ax ∂g∂x+Ay ∂g∂y+Az ∂g∂z)]=g∇·A−A·∇gg2.qed[∇×(A/g)]x=∂∂y(Az/g)−∂∂z(Ay/g)=g∂Az∂y−Az ∂g∂yg2−g∂Ay∂z−Ay ∂g∂zg2=1g2[g(∂Az∂y−∂Ay∂z)−(Az ∂g∂y−Ay ∂g∂z)]=g(∇×A)x+(A×∇g)xg2(same foryandz).qedProblem 1.25(a)A×B=∣∣∣∣∣∣∣∣∣ˆxˆyˆzx2y3z3y−2x0∣∣∣∣∣∣∣∣∣=ˆx(6xz) +ˆy(9zy) +ˆz(−2x2−6y2)∇·(A×B) =∂∂x(6xz) +∂∂y(9zy) +∂∂z(−2x2−6y2) = 6z+ 9z+ 0 = 15z∇×A=ˆx(∂∂y(3z)−∂∂z(2y))+ˆy(∂∂z(x)−∂∂x(3z))+ˆz(∂∂x(2y)−∂∂y(x))= 0;B·(∇×A) = 0∇×B=ˆx(∂∂y(0)−∂∂z(−2x))+ˆy(∂∂z(3y)−∂∂x(0))+ˆz(∂∂x(−2x)−∂∂y(3y))=−5ˆz;A·(∇×B) =−15z∇·(A×B)?=B·(∇×A)−A·(∇×B) = 0−(−15z) = 15z.X(b)A·B= 3xy−4xy=−xy;∇(A·B) =∇(−xy) =ˆx∂∂x(−xy) +ˆy∂∂y(−xy) =−yˆx−xˆyA×(∇×B) =∣∣∣∣∣∣∣∣∣ˆxˆyˆzx2y3z00−5∣∣∣∣∣∣∣∣∣=ˆx(−10y) +ˆy(5x);B×(∇×A) =0(A·∇)B=(x∂∂x+ 2y∂∂y+ 3z∂∂z)(3yˆx−2xˆy) =ˆx(6y) +ˆy(−2x)(B·∇)A=(3y∂∂x−2x∂∂y)(xˆx+ 2yˆy+ 3zˆz) =ˆx(3y) +ˆy(−4x)A×(∇×B) +B×(∇×A) + (A·∇)B+ (B·∇)A=−10yˆx+ 5xˆy+ 6yˆx−2xˆy+ 3yˆx−4xˆy=−yˆx−xˆy=∇·(A·B).X(c)∇×(A×B) =ˆx(∂∂y(−2x2−6y2)−∂∂z(9zy))+ˆy(∂∂z(6xz)−∂∂x(−2x2−6y2))+ˆz(∂∂x(9zy)−∂∂y(6xz))=ˆx(−12y−9y) +ˆy(6x+ 4x) +ˆz(0) =−21yˆx+ 10xˆy∇·A=∂∂x(x) +∂∂y(2y) +∂∂z(3z) = 1 + 2 + 3 = 6;∇·B=∂∂x(3y) +∂∂y(−2x) = 0c

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CHAPTER 1. VECTOR ANALYSIS11(B·∇)A−(A·∇)B+A(∇·B)−B(∇·A) = 3yˆx−4xˆy−6yˆx+ 2xˆy−18yˆx+ 12xˆy=−21yˆx+ 10xˆy=∇×(A×B).XProblem 1.26(a)∂2Ta∂x2= 2;∂2Ta∂y2=∂2Ta∂z2= 0⇒∇2Ta= 2.(b)∂2Tb∂x2=∂2Tb∂y2=∂2Tb∂z2=−Tb⇒∇2Tb=−3Tb=−3 sinxsinysinz.(c)∂2Tc∂x2= 25Tc;∂2Tc∂y2=−16Tc;∂2Tc∂z2=−9Tc⇒∇2Tc= 0.(d)∂2vx∂x2= 2 ;∂2vx∂y2=∂2vx∂z2= 0⇒∇2vx= 2∂2vy∂x2=∂2vy∂y2= 0 ;∂2vy∂z2= 6x⇒∇2vy= 6x∂2vz∂x2=∂2vz∂y2=∂2vz∂z2= 0⇒∇2vz= 0∇2v= 2ˆx+ 6xˆy.Problem 1.27∇·(∇×v) =∂∂x(∂vz∂y−∂vy∂z)+∂∂y(∂vx∂z−∂vz∂x)+∂∂z(∂vy∂x−∂vx∂y)=(∂2vz∂x ∂y−∂2vz∂y ∂x)+(∂2vx∂y ∂z−∂2vx∂z ∂y)+(∂2vy∂z ∂x−∂2vy∂x ∂z)= 0, by equality of cross-derivatives.From Prob. 1.18:∇×va=−6xzˆx+2zˆy+3z2ˆz⇒∇·(∇×va) =∂∂x(−6xz)+∂∂y(2z)+∂∂z(3z2) =−6z+6z= 0.Problem 1.28∇×(∇t) =∣∣∣∣∣∣∣∣∣ˆxˆyˆz∂∂x∂∂y∂∂z∂t∂x∂t∂y∂t∂z∣∣∣∣∣∣∣∣∣=ˆx(∂2t∂y ∂z−∂2t∂z ∂y)+ˆy(∂2t∂z ∂x−∂2t∂x ∂z)+ˆz(∂2t∂x ∂y−∂2t∂y ∂x)= 0, by equality of cross-derivatives.In Prob. 1.11(b),∇f= 2xy3z4ˆx+ 3x2y2z4ˆy+ 4x2y3z3ˆz, so∇×(∇f) =∣∣∣∣∣∣∣∣∣ˆxˆyˆz∂∂x∂∂y∂∂z2xy3z43x2y2z44x2y3z3∣∣∣∣∣∣∣∣∣=ˆx(3·4x2y2z3−4·3x2y2z3) +ˆy(4·2xy3z3−2·4xy3z3) +ˆz(2·3xy2z4−3·2xy2z4) = 0.XProblem 1.29(a) (0,0,0)−→(1,0,0). x: 0→1, y=z= 0;dl=dxˆx;v·dl=x2dx;∫v·dl=∫10x2dx= (x3/3)|10= 1/3.(1,0,0)−→(1,1,0). x= 1, y: 0→1, z= 0;dl=dyˆy;v·dl= 2yz dy= 0;∫v·dl= 0.(1,1,0)−→(1,1,1). x=y= 1, z: 0→1;dl=dzˆz;v·dl=y2dz=dz;∫v·dl=∫10dz=z|10= 1.Total:∫v·dl= (1/3) + 0 + 1 =4/3.(b) (0,0,0)−→(0,0,1). x=y= 0, z: 0→1;dl=dzˆz;v·dl=y2dz= 0;∫v·dl= 0.(0,0,1)−→(0,1,1). x= 0, y: 0→1, z= 1;dl=dyˆy;v·dl= 2yz dy= 2y dy;∫v·dl=∫102y dy=y2|10= 1.(0,1,1)−→(1,1,1). x: 0→1, y=z= 1;dl=dxˆx;v·dl=x2dx;∫v·dl=∫10x2dx= (x3/3)|10= 1/3.Total:∫v·dl= 0 + 1 + (1/3) =4/3.(c)x=y=z: 0→1;dx=dy=dz;v·dl=x2dx+ 2yz dy+y2dz=x2dx+ 2x2dx+x2dx= 4x2dx;∫v·dl=∫104x2dx= (4x3/3)|10=4/3.(d)∮v·dl= (4/3)−(4/3) =0.c

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12CHAPTER 1. VECTOR ANALYSISProblem 1.30x, y: 0→1, z= 0;da=dx dyˆz;v·da=y(z2−3)dx dy=−3y dx dy;∫v·da=−3∫20dx∫20y dy=−3(x|20)(y22|20) =−3(2)(2) =−12.In Ex. 1.7 we got 20, for the same boundary line (the square in thexy-plane), so the answer isno:the surface integral doesnotdepend only on the boundary line. Thetotalfluxfor the cube is 20 + 12 =32.Problem 1.31∫T dτ=∫z2dx dy dz.You can do the integrals in any order—here it is simplest to savezfor last:∫z2[∫(∫dx)dy]dz.The sloping surface isx+y+z= 1, so thexintegral is∫(1−y−z)0dx= 1−y−z.For a givenz,yranges from 0 to1−z, so theyintegral is∫(1−z)0(1−y−z)dy= [(1−z)y−(y2/2)]|(1−z)0= (1−z)2−[(1−z)2/2] = (1−z)2/2 =(1/2)−z+ (z2/2).Finally, thezintegral is∫10z2(12−z+z22)dz=∫10(z22−z3+z42)dz= (z36−z44+z510)|10=16−14+110=1/60.ˆProblem 1.32T(b) = 1 + 4 + 2 = 7;T(a) = 0.⇒T(b)−T(a) = 7.∇T= (2x+ 4y)ˆx+ (4x+ 2z3)y+ (6yz2)ˆz;∇T·dl= (2x+ 4y)dx+ (4x+ 2z3)dy+ (6yz2)dz(a) Segment 1:x: 0→1, y=z=dy=dz= 0.∫∇T·dl=∫10(2x)dx=x2∣∣∣10= 1.Segment 2:y: 0→1, x= 1, z= 0, dx=dz= 0.∫∇T·dl=∫10(4)dy= 4y|10= 4.Segment 3:z: 0→1, x=y= 1, dx=dy= 0.∫∇T·dl=∫10(6z2)dz= 2z3∣∣∣10= 2.∫ba∇T·dl= 7.X(b) Segment 1:z: 0→1, x=y=dx=dy= 0.∫∇T·dl=∫10(0)dz= 0.Segment 2:y: 0→1, x= 0, z= 1, dx=dz= 0.∫∇T·dl=∫10(2)dy= 2y|10= 2.Segment 3:x: 0→1, y=z= 1, dy=dz= 0.∫∇T·dl=∫10(2x+ 4)dx= (x2+ 4x)∣∣∣10= 1 + 4 = 5.∫ba∇T·dl= 7.X(c)x: 0→1, y=x, z=x2, dy=dx, dz= 2x dx.∇T·dl= (2x+ 4x)dx+ (4x+ 2x6)dx+ (6xx4)2x dx= (10x+ 14x6)dx.∫ba∇T·dl=∫10(10x+ 14x6)dx= (5x2+ 2x7)∣∣∣10= 5 + 2 = 7.XProblem 1.33∇·v=y+ 2z+ 3x∫(∇·v)dτ=∫(y+ 2z+ 3x)dx dy dz=∫∫ {∫20(y+ 2z+ 3x)dx}dy dz↪→[(y+ 2z)x+32x2]20= 2(y+ 2z) + 6=∫ {∫20(2y+ 4z+ 6)dy}dz↪→[y2+ (4z+ 6)y]20= 4 + 2(4z+ 6) = 8z+ 16=∫20(8z+ 16)dz= (4z2+ 16z)∣∣∣20= 16 + 32 =48.Numbering the surfaces as in Fig. 1.29:c

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CHAPTER 1. VECTOR ANALYSIS13(i)da=dy dzˆx, x= 2.v·da= 2y dy dz.∫v·da=∫∫2y dy dz= 2y2∣∣∣20= 8.(ii)da=−dy dzˆx, x= 0.v·da= 0.∫v·da= 0.(iii)da=dx dzˆy, y= 2.v·da= 4z dx dz.∫v·da=∫∫4z dx dz= 16.(iv)da=−dx dzˆy, y= 0.v·da= 0.∫v·da= 0.(v)da=dx dyˆz, z= 2.v·da= 6x dx dy.∫v·da= 24.(vi)da=−dx dyˆz, z= 0.v·da= 0.∫v·da= 0.⇒∫v·da= 8 + 16 + 24 = 48XProblem 1.34∇×v=ˆx(0−2y) +ˆy(0−3z) +ˆz(0−x) =−2yˆx−3zˆy−xˆz.da=dy dzˆx, if we agree that the path integral shall run counterclockwise. So(∇×v)·da=−2y dy dz.∫(∇×v)·da=∫ {∫2−z0(−2y)dy}dz↪→y2∣∣∣2−z0=−(2−z)2=−∫20(4−4z+z2)dz=−(4z−2z2+z33)∣∣∣∣20=−(8−8 +83)=−83-6zy@@@@@@y= 2−zMeanwhile,v·dl= (xy)dx+ (2yz)dy+ (3zx)dz.There are three segments.-6zy@@@@@@-(1)@@I(2)?(3)(1)x=z= 0;dx=dz= 0. y: 0→2.∫v·dl= 0.(2)x= 0;z= 2−y;dx= 0, dz=−dy, y: 2→0.v·dl= 2yz dy.∫v·dl=∫022y(2−y)dy=−∫20(4y−2y2)dy=−(2y2−23y3)∣∣∣20=−(8−23·8)=−83.(3)x=y= 0;dx=dy= 0;z: 2→0.v·dl= 0.∫v·dl= 0.So∮v·dl=−83.XProblem 1.35By Corollary 1,∫(∇×v)·dashould equal43.∇×v= (4z2−2x)ˆx+ 2zˆz.(i)da=dy dzˆx, x= 1;y, z: 0→1.(∇×v)·da= (4z2−2)dy dz;∫(∇×v)·da=∫10(4z2−2)dz= (43z3−2z)∣∣∣10=43−2 =−23.(ii)da=−dx dyˆz, z= 0;x, y: 0→1.(∇×v)·da= 0;∫(∇×v)·da= 0.(iii)da=dx dzˆy, y= 1;x, z: 0→1.(∇×v)·da= 0;∫(∇×v)·da= 0.(iv)da=−dx dzˆy, y= 0;x, z: 0→1.(∇×v)·da= 0;∫(∇×v)·da= 0.(v)da=dx dyˆz, z= 1;x, y: 0→1.(∇×v)·da= 2dx dy;∫(∇×v)·da= 2.⇒∫(∇×v)·da=−23+ 2 =43.Xc

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14CHAPTER 1. VECTOR ANALYSISProblem 1.36(a) Use the product rule∇×(fA) =f(∇×A)−A×(∇f) :∫Sf(∇×A)·da=∫S∇×(fA)·da+∫S[A×(∇f)]·da=∮PfA·dl+∫S[A×(∇f)]·da.qed(I used Stokes’ theorem in the last step.)(b) Use the product rule∇·(A×B) =B·(∇×A)−A·(∇×B) :∫VB·(∇×A)dτ=∫V∇·(A×B)dτ+∫VA·(∇×B)dτ=∮S(A×B)·da+∫VA·(∇×B)dτ.qed(I used the divergence theorem in the last step.)Problem 1.37r=√x2+y2+z2;θ= cos−1(z√x2+y2+z2);φ= tan−1(yx).Problem 1.38There are many ways to do this one—probably the most illuminating way is to work it out by trigonometryfrom Fig. 1.36. The most systematic approach is to study the expression:r=xˆx+yˆy+zˆz=rsinθcosφˆx+rsinθsinφˆy+rcosθˆz.If I only varyrslightly, thendr=∂∂r(r)dris a short vector pointing in the direction of increase inr. To makeit a unit vector, I must divide by its length. Thus:ˆr=∂r∂r∣∣∣∂r∂r∣∣∣;ˆθ=∂r∂θ∣∣∣∂r∂θ∣∣∣;ˆφ=∂r∂φ∣∣∣∂r∂φ∣∣∣.∂r∂r= sinθcosφˆx+ sinθsinφˆy+ cosθˆz;∣∣∣∂r∂r∣∣∣2= sin2θcos2φ+ sin2θsin2φ+ cos2θ= 1.∂r∂θ=rcosθcosφˆx+rcosθsinφˆy−rsinθˆz;∣∣∣∂r∂θ∣∣∣2=r2cos2θcos2φ+r2cos2θsin2φ+r2sin2θ=r2.∂r∂φ=−rsinθsinφˆx+rsinθcosφˆy;∣∣∣∂r∂φ∣∣∣2=r2sin2θsin2φ+r2sin2θcos2φ=r2sin2θ.⇒ˆr= sinθcosφˆx+ sinθsinφˆy+ cosθˆz.ˆθ= cosθcosφˆx+ cosθsinφˆy−sinθˆz.ˆφ=−sinφˆx+ cosφˆy.Check:ˆr·ˆr= sin2θ(cos2φ+ sin2φ) + cos2θ= sin2θ+ cos2θ= 1,Xˆθ·ˆφ=−cosθsinφcosφ+ cosθsinφcosφ= 0,Xetc.sinθˆr= sin2θcosφˆx+ sin2θsinφˆy+ sinθcosθˆz.cosθˆθ= cos2θcosφˆx+ cos2θsinφˆy−sinθcosθˆz.Add these:(1) sinθˆr+ cosθˆθ= + cosφˆx+ sinφˆy;(2)ˆφ=−sinφˆx+ cosφˆy.Multiply (1) by cosφ, (2) by sinφ, and subtract:ˆx= sinθcosφˆr+ cosθcosφˆθ−sinφˆφ.c

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CHAPTER 1. VECTOR ANALYSIS15ˆMultiply (1) by sinφ, (2) by cosφ, and add:y= sinθsinφˆr+ cosθsinφˆθ+ cosφˆφ.cosθˆr= sinθcosθcosφˆx+ sinθcosθsinφˆy+ cos2θˆz.sinθˆθ= sinθcosθcosφˆx+ sinθcosθsinφˆy−sin2θˆz.Subtract these:ˆz= cosθˆr−sinθˆθ.Problem 1.39(a)∇·v1=1r2∂∂r(r2r2) =1r24r3= 4r∫(∇·v1)dτ=∫(4r)(r2sinθ dr dθ dφ) = (4)∫R0r3dr∫π0sinθ dθ∫2π0dφ= (4)(R44)(2)(2π) = 4πR4∫v1·da=∫(r2ˆr)·(r2sinθ dθ dφˆr) =r4∫π0sinθ dθ∫2π0dφ= 4πR4X(Note:at surface of spherer=R.)(b)∇·v2=1r2∂∂r(r2 1r2)= 0⇒∫(∇·v2)dτ= 0∫v2·da=∫ (1r2ˆr)(r2sinθ dθ dφˆr) =∫sinθ dθ dφ=4π.Theydon’tagree!The point is that this divergence is zeroexcept at the origin, where it blows up, so ourcalculation of∫(∇·v2) isincorrect. The right answer is 4π.Problem 1.40∇·v=1r2∂∂r(r2rcosθ) +1rsinθ∂∂θ(sinθ rsinθ) +1rsinθ∂∂φ(rsinθcosφ)=1r23r2cosθ+1rsinθr2 sinθcosθ+1rsinθrsinθ(−sinφ)= 3 cosθ+ 2 cosθ−sinφ= 5 cosθ−sinφ∫(∇·v)dτ=∫(5 cosθ−sinφ)r2sinθ dr dθ dφ=∫R0r2dr∫θ20[∫2π0(5 cosθ−sinφ)dφ]dθsinθ↪→2π(5 cosθ)=(R33)(10π)∫π20sinθcosθ dθ↪→sin2θ2∣∣∣∣π20=12=5π3R3.Two surfaces—one the hemisphere:da=R2sinθ dθ dφˆr;r=R;φ: 0→2π, θ: 0→π2.∫v·da=∫(rcosθ)R2sinθ dθ dφ=R3∫π20sinθcosθ dθ∫2π0dφ=R3(12)(2π) =πR3.other the flat bottom:da= (dr)(rsinθ dφ)(+ˆθ) =r dr dφˆθ(hereθ=π2).r: 0→R, φ: 0→2π.∫v·da=∫(rsinθ)(r dr dφ) =∫R0r2dr∫2π0dφ= 2πR33.Total:∫v·da=πR3+23πR3=53πR3.XProblem 1.41∇t= (cosθ+ sinθcosφ)ˆr+ (−sinθ+ cosθcosφ)ˆθ+1sinθ/(−sinθ/sinφ)ˆφ∇2t=∇·(∇t)=1r2∂∂r(r2(cosθ+ sinθcosφ))+1rsinθ∂∂θ(sinθ(−sinθ+ cosθcosφ)) +1rsinθ∂∂φ(−sinφ)=1r22r(cosθ+ sinθcosφ) +1rsinθ(−2 sinθcosθ+ cos2θcosφ−sin2θcosφ)−1rsinθcosφ=1rsinθ[2 sinθcosθ+ 2 sin2θcosφ−2 sinθcosθ+ cos2θcosφ−sin2θcosφ−cosφ]=1rsinθ[(sin2θ+ cos2θ) cosφ−cosφ]= 0.c

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16CHAPTER 1. VECTOR ANALYSIS⇒∇2t= 0Check:rcosθ=z, rsinθcosφ=x⇒in Cartesian coordinatest=x+z. Obviously Laplacian is zero.Gradient Theorem:∫ba∇t·dl=t(b)−t(a)Segment 1:θ=π2, φ= 0, r: 0→2. dl=drˆr;∇t·dl= (cosθ+ sinθcosφ)dr= (0 + 1)dr=dr.∫∇t·dl=∫20dr= 2.Segment 2:θ=π2, r= 2, φ: 0→π2.dl=rsinθ dφˆφ= 2dφˆφ.∇t·dl= (−sinφ)(2dφ) =−2 sinφ dφ.∫∇t·dl=−∫π202 sinφ dφ= 2 cosφ|π20=−2.Segment 3:r= 2, φ=π2;θ:π2→0.dl=r dθˆθ= 2dθˆθ;∇t·dl= (−sinθ+ cosθcosφ)(2dθ) =−2 sinθ dθ.∫∇t·dl=−∫0π22 sinθ dθ= 2 cosθ|0π2= 2.Total:∫ba∇t·dl= 2−2 + 2 =2 . Meanwhile,t(b)−t(a) = [2(1 + 0)]−[0( )] = 2.XˆˆˆˆProblem 1.42From Fig. 1.42,ˆs= cosφˆx+ sinφˆy;ˆφ=−sinφˆx+ cosφˆy;ˆz=ˆzMultiply first by cosφ, second by sinφ, and subtract:scosφ−ˆφsinφ= cos2φˆx+ cosφsinφˆy+ sin2φˆx−sinφcosφˆy=ˆx(sin2φ+ cos2φ) =ˆx.Soˆx= cosφs−sinφˆφ.Multiply first by sinφ, second by cosφ, and add:ssinφ+ˆφcosφ= sinφcosφˆx+ sin2φˆy−sinφcosφˆx+ cos2φˆy=ˆy(sin2φ+ cos2φ) =ˆy.Soˆy= sinφs+ cosφˆφ.ˆz=ˆz.Problem 1.43(a)∇·v=1s∂∂s(s s(2 + sin2φ))+1s∂∂φ(ssinφcosφ) +∂∂z(3z)=1s2s(2 + sin2φ) +1ss(cos2φ−sin2φ) + 3= 4 + 2 sin2φ+ cos2φ−sin2φ+ 3= 4 + sin2φ+ cos2φ+ 3 =8.(b)∫(∇·v)dτ=∫(8)s ds dφ dz= 8∫20s ds∫π20dφ∫50dz= 8(2)(π2)(5) =40π.Meanwhile, the surface integral has five parts:top:z= 5, da=s ds dφˆz;v·da= 3z s ds dφ= 15s ds dφ.∫v·da= 15∫20s ds∫π20dφ= 15π.bottom:z= 0, da=−s ds dφˆz;v·da=−3z s ds dφ= 0.∫v·da= 0.back:φ=π2, da=ds dzˆφ;v·da=ssinφcosφ ds dz= 0.∫v·da= 0.left:φ= 0, da=−ds dzˆφ;v·da=−ssinφcosφ ds dz= 0.∫v·da= 0.front:s= 2, da=s dφ dzˆs;v·da=s(2 + sin2φ)s dφ dz= 4(2 + sin2φ)dφ dz.∫v·da= 4∫π20(2 + sin2φ)dφ∫50dz= (4)(π+π4)(5) = 25π.So∮v·da= 15π+ 25π= 40π.X(c)∇×v=(1s∂∂φ(3z)−∂∂z(ssinφcosφ))ˆs+(∂∂z(s(2 + sin2φ))−∂∂s(3z))ˆφ+1s(∂∂s(s2sinφcosφ)−∂∂φ(s(2 + sin2φ)))ˆz=1s(2ssinφcosφ−s2 sinφcosφ)ˆz=0.c

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