082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

866 Chapter 26 Vector calculus
14 Thevector fieldsAand B are given by
A =x 2 yi+2xzj+xk
B=yzi+x 2 yj+3k
(a) Calculate the vector productA ×B.
(b) Calculate ∇ · (A ×B).
(c) CalculateB·(∇ ×A) −A·(∇ ×B).
(d) What doyou conclude from (b)and(c)?
(e) Can you prove thisresult istrueforany two
vector fields?
15 Given
F = (3x,2y,4z)
φ =x 2 yz
G = (1,−1,3)
find
(a) ∇·F
(b) ∇φ
(c) ∇×F
(d) ∇ ·(φF)
(e) ∇ × (φG)
(f) ∇(F·G)
(g) ∇×(F×G)
(h) ∇·(F+G)
(i) ∇ × (3F +4G)
Solutions
1 (a) 3yzi +3xzj +3xyk
(b) (2xyz +y 2 z +yz 2 )i
+ (x 2 z +2xyz +xz 2 )j
+ (x 2 y +xy 2 +2xyz)k
(c) (y 2 z 2 +2xyz 2 +2xy 2 z)i
+ (2xyz 2 +x 2 z 2 +2x 2 yz)j
+ (2xy 2 z +2x 2 yz +x 2 y 2 )k
4 (a) zsinxi+zcosxj+(−ycosx−xsinx)k
(b) zcosx (c) 0 (d) −cosxk
6 ∇ψ =2xyi+x 2 j
∇×a=0
∇ × (ψa) =x 2 zi −2xyzj + (2xy 2 −x 3 )k
8 (4xy +15x 4 )i +ye xy j +yzk,
(4y +60x 3 )i +y 2 e xy j
9 (a) E = −∇V = −5yzi −5xzj −5xyk
√
(b) 75
10 (a) 12i +8j
(b) 21i +24j
11 (2y −1)k
12 (y +1)k
13 5i +10j
14 (a) (6xz −x 3 y)i + (xyz −3x 2 y)j
+(x 4 y 2 −2xyz 2 )k
(b) −3x 2 (y +1) −xz(4y −1) +6z
(c) −3x 2 (y +1) −xz(4y −1) +6z
(d) ∇·(A×B)=B·(∇×A)−A·(∇×B)
15 (a) 9
(b) 2xyzi +x 2 zj +x 2 yk
(c) 0
(d) 21x 2 yz
(e) x 2 (y +3z)i + (x 2 y −6xyz)j
− (x 2 z +2xyz)k
(f) 3i −2j +12k
(g) −6i +7j −15k
(h) 9
(i) 0