Solution Manual Energy Principles and Variational Methods in Applied Mechanics 3rd Edition by J. N. Reddy

1. Email: smtb98@gmail.com Telegram: https://t.me/solutionmanual Contact me in order to access the whole complete document. WhatsApp: https://wa.me/message/2H3BV2L5TTSUF1 smtb98@gmail.com smtb98@gmail.com 1 Solutions Manual to ENERGY PRINCIPLES AND VARIATIONAL METHODS IN APPLIED MECHANICS Third Edition J. N. Reddy Department of Mechanical Engineering Texas A&M University College Station, Texas, 77843-3123

2. Email: smtb98@gmail.com Telegram: https://t.me/solutionmanual Contact me in order to access the whole complete document. WhatsApp: https://wa.me/message/2H3BV2L5TTSUF1 Solutions Manual to ENERGY PRINCIPLES AND VARIATIONAL METHODS IN APPLIED MECHANICS Third Edition complete document is available on https://solumanu.com/ *** contact me if site not loaded

3. smtb98@gmail.com Email: smtb98@gmail.com Telegram: https://t.me/solutionmanual Contact me in order to access the whole complete document. WhatsApp: https://wa.me/message/2H3BV2L5TTSUF1 Introduction and Mathematical smtb98@gmail.com 1 Preliminaries 1.1 Let C be a vector along the line passing through the terminal point of vector A and parallel to vector B. Let r be the position vector to an arbitrary point on the line parallel to vector B and passing through the terminal point of vector A. Then the desired equation of the line is (see Fig. P1.1) B |B|, r = A + C = A + β ˆ eB, ˆ eB= FigP2-1 where β is a real number. ˆ e = + β = + α r A A B FigP2-3 B  B A B • Line parallel to vector B  z   r A A B r y B •  ˆ e = β C  C A B r •  A O C A x Fig. P1.1 Fig. P1.2 1.2 Let r denote the position vector. The vectors connecting the terminal points of vectors A, B, C, and r should be in the plane. Thus, for example, the scalar triple product of the vectors B−A, C−A, and r−A should be zero in order that they are co-planar, as shown in Fig. P1.2: (C − A) × (B − A) · (r − A) = 0 [or For example, if A = ˆ e1, B = ˆ e2, and C = ˆ e3, then the equation of the plane is −x−y − z + 1 = 0 or x + y + z = 1. εijk(Ci− Ai)(Bj− Aj)(xk− Ak) = 0] 1.3 Consider the parallelogram formed by points O, A, C, and B, as shown in Fig. P1.3. Let us denote the line segment connecting O to A as vector A, O to B as vector B, O to C as vector C, and B to A as vector D. Suppose that vectors C and D intersect and cross at distances αC and β D. Then we have the following relations among the four vectors: A = αC + (1 − β)D; A = β D + (1 − α)C (1) 1 complete document is available on https://solumanu.com/ *** contact me if site not loaded

4. 2 REDDY: ENERGY PRINCIPLES AND VARIATIONAL METHODS B = αC − β D; B = (1 − α)C − (1 − β)D (2) FigP2-6 From each pair of equations, we obtain the same result, namely, α = β = 0.5, implying that vectors C and D bisect each other.   (1 ) C C       A D (1 ) C  D BA  D OC  C B   (1 ) D   C  A B      A C (1 ) D O Fig. P1.3 1.4 We begin with the left side of the equality and arrive at the right side: A × (B × C) = Aiˆ ei× (ˆ e‘εjk‘BjCk) = AiBjCkεjk‘εpi‘ˆ ep = (δjpδki− δjiδkp)AiBjCkˆ ep= AiBjCiˆ ej− AiBiCkˆ ek = (A · C)B − (A · B)C. 1.5 An arbitrary vector A can be decomposed into a (vector) sum of two vectors, one along any arbitrary line (A1) and the other (A2) perpendicular to it (with magnitudes |A1| = |A|cosθ and |A2| = |A|sinθ, respectively, where θ is the angle between vectors A1and A). Because the unit vector along the line is ˆ e, we have A1= (A·ˆ e)ˆ e. To determine the vector A2perpendicular to the line along ˆ e, consider the vector product A × ˆ e, which is perpendicular to the plane of both A and ˆ e (that is, out of the page). Then, vector ˆ e×(A׈ e) lies in the plane of vectors A and ˆ e, and it is perpendicular to vector A and ˆ e, with the magnitude |A × ˆ e| = |A|sinθ. Hence, we have ˆ ep=ˆ e × (A × ˆ e) |A|sinθ A = |A| cosθ ˆ e + |A| sinθ ˆ ep= (A ·ˆ e)ˆ e + ˆ e × (A ׈ e), . 1.6 (a) δii= δ11+ δ22+ δ33= 1 + 1 + 1 = 3. (b) Fijδij = Fi1δi1+ Fi2δi2+ Fi3δi3. Clearly, the first term is zero unless i = 1, the second is zero unless i = 2, and the third is zero unless i = 3. Thus, we have Fijδij= F11+F22+F33= Fii. The given identity follows as a special case, Fij= δij. (c) Fijδjk= Fi1δ1k+ Fi2δ2k+ Fi3δ3k. Clearly, the first term is zero unless k = 1. If k = 1, the remaining two terms are zero giving Fi1. Similarly, the second term is zero unless k = 2, which makes the first and third terms zero so that the result is Fi2. Finally, the third term is zero unless k = 3, which makes the first two terms zero, complete document is available on https://solumanu.com/ *** contact me if site not loaded

5. smtb98@gmail.com smtb98@gmail.com 3 SOLUTIONS MANUAL (CHAPTER 1) giving Fi3. Thus, Fijδjkis equal to Fik. Alternatively, we have Fijδjk= 0 unless j = k because δjk= 0 unless j = k. (Caution: replace j by k only in the coefficient of δjkand then remove δjkfrom the expression; do not write Fijδjk= Fikδkk, which has no meaning as no subscript can appear more than twice!). The given identity follows as a special case, Fij= δij. (d) εmjkεnjk= δmnδjj− δmjδjn= 3δmn− δmn= 2δmn, where we have used the ε − δ identity (expanded about the common subscript k) and parts (a) and (c) of this problem in arriving at the result. (e) Follows from Parts (a) and (d). (f) Fijεijk= −Fijεikj= −Fjiεjki. In the first step subscripts j and k are interchanged, which results in the change of sign. In the second step, the subscript i is renamed as j and subscript j as i (in the whole expression). Since εjki= εijk, we have Fijεijk= −Fjiεjki → (Fij+ Fji)εijk= 0. Now replace Fijwith AiAjand find that (AiAj+ AjAi)εijk= 2AiAjεijk= 0, which was to be proved. Thus, if Fijis symmetric, it follows that Fijεijk= 0. Con- versely, if Fijεijk= 0, then Fij= Fji, i.e., the second-order tensor F is symmetric. 1.7 We have (A × B) · (B × C) × (C × A) = (εijkAiBjˆ ek) · [(εrstBrCsˆ et) × (εmnpCmAnˆ ep)] = εijkεrstεmnpAiBjBrCsCmAnεtpqδkq = (δitδjp− δipδjt)εrstεmnpAiBjBrCsCmAn = (εrsiεmnj− εrsjεmni)AiBjBrCsCmAn = (B × C · A)(C × A · B) − (B × C · B)(C × A · A) = (B × C · A)(C × A · B) = (B × C · A)2 Note that B × C · A = A × B · C. 1.8 (A × B) × (C × D) = (εijkAiBjˆ ek) × (εmnpCmDnˆ ep) = εijkεkpqεmnpAiBjCmDnˆ eq = εmnp(δipδjq− δiqδjp)AiBjCmDnˆ eq = εmniAiBjCmDnˆ ej− εmnjAiBjCmDnˆ ei = (C × D · A)B − (C × D · B)A. 1.9 Let A = (1,1,0,0), B(0,1,0,1),C = (0,0,1,1) and D = (1,1,1,1). Then the relation αA + βB + γC + µD = 0 complete document is available on https://solumanu.com/ *** contact me if site not loaded

6. 4 REDDY: ENERGY PRINCIPLES AND VARIATIONAL METHODS implies α + µ = 0, α + β + µ = 0, γ + µ = 0, β + γ + µ = 0 whose solution is α = γ = −µ. Hence, the vectors are linearly dependent. 1.10 The linear relation αA + βB + γC = 0 implies 2α − γ = 0, −α + β + γ = 0, α − β = 0 whose solution is trivial. Hence, the vectors are linearly independent. 1.11 (a) First check for linear independence. Set αA + βB + γC = 0 and find that α − 4β + 2γ = 0, −α − 5β + γ = 0. 3α + 3β + γ = 0, The solution of these equations is α = −2β and γ = 3β. Hence, the set is linearly dependent and does not span <3. (b) Set αA + βB + γC = 0 and find that −2β − γ = 0. α + β + γ = 0, α + β = 0, The solution of these equations is trivial α = β = γ = 0. Hence, the set is linearly independent. Next express an arbitrary vector x = x1ˆ e1+ x2ˆ e2+ x3ˆ e3in <3as a linear combination of the given vectors x = c1A + c2B + c3C and show that ciwill not imply a relation among the components xiof the vector (because they are arbitrary). We have −2c1− c3= x3, c1+ c2+ c3= x1, c1+ c2= x2, whose solution is c1=1 c2=1 c3= x1− x2, 2(x2− x1− x3), 2(x1+ x2+ x3). In particular, the vectors ˆ e1, ˆ e2, and ˆ e3are represented by the vectors A, B, and C as ˆ e1= −1 2A +1 ˆ e2=1 2A +1 ˆ e3= −1 2A +1 2B − C, 2B + C, 2B. Hence, the set (A, B, C) spans <3.

7. Email: smtb98@gmail.com Telegram: https://t.me/solutionmanual Contact me in order to access the whole complete document. WhatsApp: https://wa.me/message/2H3BV2L5TTSUF1 smtb98@gmail.com 1.12 Follows as outlined in the problem statement. smtb98@gmail.com 5 SOLUTIONS MANUAL (CHAPTER 1) 1.13 (a) Let ˆ eibe the unit base vectors in the current orthogonal system and ˆ e0 base vectors in the new coordinate system. The vector ˆ e0 vector ˆ e1− ˆ e2+ ˆ e3but its magnitude must be unity ˆ e0 1= |ˆ e1− ˆ e2+ ˆ e3|= The vector ˆ e0 the unit normal to the plane, which is given by ibe the unit 1has the same direction as the ˆ e1− ˆ e2+ ˆ e3 1 √3(ˆ e1− ˆ e2+ ˆ e3). 2is along the normal to the plane 2x1+ 3x2+ x3− 5 = 0. Hence, ˆ e0 2= ˆ n, ∇(2x1+ 3x2+ x3− 5) |∇(2x1+ 3x2+ x3− 5)|= 1 √14(2ˆ e1+ 3ˆ e2+ ˆ e3). 2ˆ e1+ 3ˆ e2+ ˆ e3 p(2)2+ (3)2+ (1)2 ˆ e0 2= = The third basis vector in an orthonormal system is related to the other two vectors by ???????? = ˆ e1 − 1 √42(−4ˆ e1+ ˆ e2+ 5ˆ e3). ???????? ˆ e1 1 √3−1 2 √14 3 √42 ˆ e2 ˆ e3 1 √3 1 √14 ˆ e0 3= ˆ e0 1× ˆ e0 √3 3 √14 ? 2= ? ? ? ? ? 1 1 2 3 2 √42− √42− √42 √42+ √42 − ˆ e2 + ˆ e3 = Thus, the two coordinate systems are related by (note the matrix of direction cosines)   (b) We have         −1 √3 3 √14 1 √42 1 1 √3 2 √14 −4 √42 √3 1 √14 5 √42 ˆ e0 ˆ e0 ˆ e0 ˆ e1 ˆ e2 ˆ e3       1 = . 2 3 1=vector connecting point (1,−1,3) to point(2,−2,4) vector magnitude =(2ˆ e1− 2ˆ e2+ 4ˆ e3) − (ˆ e1− ˆ e2+ 3ˆ e3) magnitude 1 √6(−ˆ e1+ ˆ e2+ 2ˆ e3) ???????? = ˆ e1 √18+ 1 √18(3ˆ e1+ 3ˆ e2+ 0.ˆ e3) = ˆ e0 1 √3(ˆ e1− ˆ e2+ ˆ e3) = ˆ e0 3= (given) ???????? − ˆ e1 −1 √6 1 √3 ˆ e2 1 √6 −1 √3 ? ˆ e3 2 √6 1 √3 ˆ e0 2= ˆ e0 3× ˆ e0 1= ? ? ? ? ? 1 2 1 2 1 1 √18 √18− 1 √2(ˆ e1+ ˆ e2). √18 √18− √18 − ˆ e2 + ˆ e3 =

8. 6 REDDY: ENERGY PRINCIPLES AND VARIATIONAL METHODS The transformation matrix relating (ˆ e0 1,ˆ e0 2,ˆ e0 3) to (ˆ e1,ˆ e2,ˆ e3) is given by  (aij= ˆ e0   √3−1 1 √2 −1 1 1 √3 1 √2 1 √6 √3   0 i· ˆ ej) [A] = 2 √6 √6 (c) We have ˆ e0 1= ˆ e1cos30◦+ ˆ e2sin30◦, ˆ e0 2= ˆ e1(−sin30◦) + ˆ e2cos30◦, ˆ e0   3= ˆ e3( given)   √3 2 −1 1 2 0   √3 2 [A] = 0 2 0 0 1 1.14 Follows from the definition         √3 2 1 2 0   √3 2 [A] = 1 2 0 − 0 0 1 1.15 Follows from the definition 1 1 0 √2 √2 −1   1 2 1 √2 1 √2 [A] = 2 1 2 −1 2 1.16 The vector is given by A =1 2(6ˆ e1− 6ˆ e2+ 4ˆ e3) − (ˆ e1− ˆ e2+ 3ˆ e3) = 2ˆ e1− 2ˆ e2− ˆ e3. The direction cosines are (2/3,−2/3,−1/3). B |B|=6−4 =2 1.17 (a) The orthogonal projection of A onto B is A · (b) The angle between A and B is 3. 3 ?A · B ? ? ? 2 θ = cos−1 = cos−1 → θ = 82.34◦. |A| |B| 5 × 3 1.18 (a) This is obvious. (b) We have ? ? ?dA ? ? ? ? ? ? d2A dt2 d2A dt2 d3A dt3 Ad2A dt2 d2A dt2 d3A dt3 d dt AdA dA dt AdA = + + dt dt AdA dt ? = dt

9. smtb98@gmail.com smtb98@gmail.com 7 SOLUTIONS MANUAL (CHAPTER 1) 1.19 (a) We have ∂ 2=r 1 2=1 1 2−12xi= ˆ eixi(xjxj)−1 ∇(r) = ˆ ei 2ˆ ei(xjxj) ∂xi(xjxj) r. (b) We have ∂ n 2= ˆ ein n−2 2 n 2−12xi= nˆ eixi(xjxj) = nrn−2r. ∇(rn) = ˆ ei ∂xi(xjxj) 2(xjxj) Alternatively, ∂ ∂r ∂xi ∂xi(rn) = nrn−1ˆ ei = nrn−2xiˆ ei= nrn−2r. ∇(rn) = ˆ ei (c) We have, in view of the result in (b) ∂2 ?nrn−2xi ?= n(n − 2)rn−3∂r ∂ ∂xixi+ nrn−2δii ∇2(rn) = ∂xi∂xi(rn) = = n(n − 2)rn−3xi ∂xi rxi+ 3nrn−2= [n(n − 2) + 3n]rn−2= n(n + 1)rn−2. (d) We have (note that A is a constant vector) ? ? δijAj+ xj∂Aj ∂ ∇(r · A) = ˆ ei ∂xi(xjAj) = ˆ ei = ˆ ei(Ai+ 0) = A. ∂xi (e) ?∂xj ? ∂ ∂xiAk+ xj∂Ak ∇ · (r × A) = ˆ ei· ∂xi(εjk‘xjAkˆ e‘) = εjk‘δi‘ = (0 + 0) = 0. ∂xi (f) Carrying out the indicated operation, we obtain ? ? ∂ δirAs+ xr∂As ∇ × (r × A) = ˆ ei× ∂xi(εrstxrAsˆ et) = εrstˆ ei× ˆ et = εrstεjitˆ ej(δirAs+ 0) = εistεjitˆ ejAs= −2ˆ ejδsjAs= −2A. ∂xi (g) ?∂r ? ∂ =xi rAi=1 ∇ · (rA) = ˆ ei· ∂xi(rAjˆ ej) = ˆ ei· ˆ ej rr · A. ∂xiAj (h) Carrying out the indicated operation, we obtain ?xi r(r × A). ? ?√xjxjAkˆ ek rAk ?= ˆ ei× ˆ ek rεiktxiAkˆ et=1 ∂ ∇ × (rA) = ˆ ei× rAk+ 0 ∂xi ?xi ? =1 = εiktˆ et

10. 8 REDDY: ENERGY PRINCIPLES AND VARIATIONAL METHODS ∂2F ∂xi∂xjis symmetric in i and j, we obtain ? 1.20 (a) Since ? ? ? ∂2F ∂xi∂xj ∂ ˆ ej∂F ∂xj ∇ × (∇F) = × ˆ ei = εijkˆ ek = 0. ∂xi (b) ? ? ? ? ∂2Ak ∂xi∂xj ∂2Ak ∂xi∂xj ∂ εjk‘∂Ak ∇ · (∇ × A) = · ˆ ei ˆ e‘ = εjk‘δi‘ = εijk = 0, ∂xi ∂xj because of the symmetry of Ak,ijin i and j. ∂2F ∂xi∂xjis symmetric in i and j and ∂2G ∂xi∂xkis symmetric in i and k, we obtain ? ? ∂2F ∂xi∂xj ∂xk ∂xj ∂xi∂xk (c) Since ? ? ? ∂ ˆ ej∂F ∂xj ∂2F ∂xi∂xj ∂G × ˆ ek∂G ∂G ∂xk +∂F ∇ · (∇F × ∇G) = · ˆ ei ∂xi ∂xk +∂F ? ∂2G ∂xi∂xk ? = εjk‘(ˆ ei· ˆ e‘) ? ∂xj ∂2G = εijk = 0. (d) ?∂F ? ∂ ∂xiG + F∂G ∇(FG) = ˆ ei = ∇F G + F ∇G. ∂xi(FG) = ˆ ei ∂xi (e) ? ∂xiF + Ai∂F ? ?∂Aj ? ∂ ∂xiF + Aj∂F ∇ · (FA) = · (ˆ ejAjF) = δij ˆ ei ∂xi ∂xi =∂Ai = ∇ · AF + A · ∇F. ∂xi (f) ? ? ?∂Aj = F∇ × A − A × ∇F. ? ∂ ∂xiF + Aj∂F ∇ × (FA) = × (ˆ ejAjF) = εijkˆ ek ∂xiF − εjikˆ ekAj∂F ˆ ei ∂xi ∂xi = εijkˆ ek∂Aj ∂xi (g) First, consider ?∂Aj ? ∂xiBj+ Aj∂Bj ∂ ∇(A · B) = ˆ ei = ∇A · B + ∇B · A. ∂xi(AjBj) = ˆ ei (1) ∂xi Next consider ? ? εjkpˆ ep∂Bk = ˆ eqεqipεjkpAi∂Bk A × ∇ × B = (ˆ eiAi) × ∂xj ∂xj complete document is available on https://solumanu.com/ *** contact me if site not loaded

11. smtb98@gmail.com smtb98@gmail.com 9 SOLUTIONS MANUAL (CHAPTER 1) = ˆ eq(δqjδik− δqkδij)Ai∂Bk = ∇B · A − A · ∇B. = ˆ ejAi∂Bi − ˆ ekAi∂Bk ∂xj ∂xj ∂xi (2) Therefore, A × ∇ × B + B × ∇ × A = ∇B · A − A · ∇B + ∇A · B − B · ∇A = ∇A · B + ∇B · A − (A · ∇B + B · ∇A). From Eqs. (1) and (2) the required vector identity follows. (h) ?∂Aj ? ∂ ∂xiBk+ Aj∂Bk ∇ · (A × B) = ˆ ei· ∂xi(εjk‘AjBkˆ e‘) = εijk = ∇ × A · B − ∇ × B · A. ∂xi (i) ∂ ∇ × (A × B) = ˆ ei × (AjBkεjkpˆ ep) ?∂Aj ∂xi ? ∂xiBk+ Aj∂Bk ?∂Aj ∂xiBi+ Aj∂Bi = B · ∇A + A∇ · B − B∇ · A − A · ∇B. = εjkpεqipˆ eq ∂xi ? ∂xiBk+ Aj∂Bk ? = (δjqδki− δjiδkq)ˆ eq ?∂Aj ∂xi ?∂Ai ? ∂xiBk+ Ai∂Bk − ˆ ek = ˆ ej ∂xi ∂xi (j) Since ∇2= ∇ · ∇, we have ? ? ∂2F ∂xi∂xjG + 2ˆ ej∂F ? ∂ ˆ ej∂F ∂xjG + Fˆ ej∂G ∂2F ∂xi∂xjG +∂F ∇2(FG) = ˆ ei · ∂xi ? ∂xj ∂G ∂xi · ˆ ei∂G ? ?∂F ?? ∂2G ∂xj∂xi ∂G ∂xj = ˆ ei· ˆ ej + ˆ ej + F ∂xj ∂xi ∂2G ∂xj∂xi = ˆ ei· ˆ ej + Fˆ ei· ˆ ej ∂xj ∂xi ∂2G ∂xi∂xi ∂2F ∂xi∂xiG + 2∇F · ∇G + F = ∇2FG + 2∇F · ∇G + F∇2G = 1.21 (a) We begin with (see Table 1.2.2) ∂ ∂r+1 ∂ ∂θ+ ˆ ez ∂ ∂z, ∇ = ˆ er rˆ eθ A = Arˆ er+ Aθˆ eθ+ Azˆ ez.

12. 10 REDDY: ENERGY PRINCIPLES AND VARIATIONAL METHODS Then we have ? ? ∂ ∂r+ ˆ eθ1 ∂ ∂θ+ ˆ ez ∂ ∂z + ˆ erˆ ez∂Az ∇A = ˆ er (Arˆ er+ Aθˆ eθ+ Azˆ ez) r = ˆ erˆ er∂Ar + ˆ erˆ eθ∂Aθ +1 rˆ eθˆ er∂Ar +1 ∂r ∂r ∂r rˆ eθ∂ˆ eθ ∂θ rˆ eθ∂ˆ er + ˆ ezˆ er∂Ar +Ar rˆ eθˆ eθ∂Aθ + ˆ ezˆ eθ∂Aθ +1 +Aθ rˆ eθˆ ez∂Az ∂θ ∂θ ∂θ ∂θ + ˆ ezˆ ez∂Az ∂z ∂z ∂z ?∂Ar ? ? = ˆ erˆ er∂Ar + ˆ erˆ eθ∂Aθ + ˆ eθˆ er1 − Aθ ∂r ∂r r ∂θ Ar+∂Aθ ? + ˆ erˆ ez∂Az + ˆ ezˆ er∂Ar + ˆ eθˆ eθ1 ∂r ∂z r ∂θ +1 rˆ eθˆ ez∂Az + ˆ ezˆ eθ∂Aθ + ˆ ezˆ ez∂Az , ∂θ ∂z ∂z where the following derivatives of the base vectors are used: ∂ˆ er ∂θ ∂ˆ eθ ∂θ = −ˆ er. = ˆ eθ, (b) From Table 1.2.2, we have ? ? ∂R+1 ∂ ∂ 1 ∂ ∂θ ∇ = ˆ eR Rˆ eφ Rsinφˆ eθ ∂φ+ and A = ARˆ eR+ Aφˆ eφ+ Aθˆ eθ. Hence, we have ? = ˆ eR ? ∂R+1 ∂ ∂R(ARˆ eR+ Aφˆ eφ+ Aθˆ eθ) +1 1 Rsinφˆ eθ ?∂AR +1 Rˆ eφ 1 Rsinφˆ eθ ∂θ + Aφcosφˆ eθ+∂Aθ ∂θˆ eθ− Aθ(sinφˆ eR+ cosφˆ eφ) =∂AR ∂R ∂ ∂ 1 ∂ ∂θ ˆ eR Rˆ eφ Rsinφˆ eθ (ARˆ eR+ Aφˆ eφ+ Aθˆ eθ) ∂φ+ ∂ ∂φ(ARˆ eR+ Aφˆ eφ+ Aθˆ eθ) Rˆ eφ ∂ ∂θ(ARˆ eR+ Aφˆ eφ+ Aθˆ eθ) ∂Rˆ eR+∂Aφ ?∂AR h∂AR + ? ∂Rˆ eφ+∂Aθ = ˆ eR ∂Rˆ eθ ∂φˆ eφ− Aφˆ eR+∂Aθ ˆ eR+ ARsinφˆ eθ+∂Aφ ? ∂φˆ eR+ ARˆ eφ+∂Aφ ∂φˆ eθ ∂θˆ eφ + i ?∂AR ? ˆ eRˆ eR+∂Aφ ∂Rˆ eRˆ eφ+1 ˆ eφˆ eR+∂Aθ − Aφ ∂Rˆ eRˆ eθ R ∂φ

13. smtb98@gmail.com where the following derivatives of the base vectors are used: smtb98@gmail.com 11 SOLUTIONS MANUAL (CHAPTER 1) ?∂AR ?∂Aφ ? ? ? 1 ? AR+∂Aφ ? 1 ˆ eθˆ eR+1 ˆ eφˆ eφ+1 ∂Aθ ∂φˆ eφˆ eθ − Aθsinφ + Rsinφ 1 Rsinφ ∂θ R ∂φ R ? ARsinφ + Aφcosφ +∂Aθ − Aθcosφ ˆ eθˆ eφ+ ˆ eθˆ eθ, + ∂θ Rsinφ ∂θ ∂ˆ eφ ∂φ ∂ˆ eφ ∂θ ∂ˆ eR ∂φ ∂ˆ eR ∂θ ∂ˆ eθ ∂θ = −ˆ eR, = −(sinφˆ eR+ cosφˆ eφ). = ˆ eφ, = sinφˆ eθ, = cosφˆ eθ, 1.22 Follows from the gradient theorem, Eq. (1.2.55), by setting φ = 1. 1.23 Note that (∂r ∂xi= xi/r) ∂r ∂xi ∇(r2) = 2rˆ ei = 2ˆ eixi= 2r. Using the divergence theorem, we can write I Z ∇ · rdΩ = δiiV = 3V, r · ˆ ndΓ = Γ Ω where V is the volume of the region Ω. 1.24 Using the divergence theorem, we can write Z I I ∂φ ∂ndΓ. ∇ · ∇φdΩ = ˆ n · ∇φdΓ = Ω Γ Γ 1.25 The integral relations are obvious. For example, the identity in (1) is obtained by sub- stituting A = φ∇ψ for A into Eq. (1.2.57) Z The left side of the equality, as per Problem 1.20(e) with A replaced with ∇ψ, is equal to Z The identity in (2) follows directly from (1) by interchanging φ and ψ and subtracting the resulting identity from the one in (1). Finally, identity (3) follows from (2) by replacing ψ with ∇2ψ. I I φ∂ψ ∂ndΓ. ∇ · (φ ∇ψ)dΩ = ˆ n · (φ ∇ψ)dΓ = Ω Γ Γ Z Z ?∇φ · ∇ψ + φ∇2ψ?dΩ. ∇ · (φ ∇ψ)dΩ = (∇φ · ∇ψ + φ∇ · ∇ψ)dΩ = Ω Ω Ω 1.26 See Problem 1.13(a) for the basis vectors of the barred coordinate system in terms of the unbarred system; the matrix of direction cosines [A] is given there. Then the components of the dyad in the barred coordinate system are   0 −1 3 −2 1 0 [¯T] = [A][T][A]T= [A]  [A]T. −1 −2 0 complete document is available on https://solumanu.com/ *** contact me if site not loaded