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```LINE, SURFACE AND VOLUME INTEGRALS
11.1
11.3
11.5
11.7
11.9
11.11
11.13
11.15
11.17
11.19
11.21
11.23
11.25
11.27
Show that ∇ × F = 0. The potential φF (r) = x2 z + y 2 z 2 − z.
(a) c3 ln 2 i + 2 j + (3c/2)k; (b) (−3c4 /8)i − c j − (c2 ln 2)k; (c) c4 ln 2 − c.
For P , x = y = ab/(a2 + b2 )1/2 . The relevant limits are 0 ≤ θ1 ≤ tan−1 (b/a) and
tan−1 (a/b) ≤ θ2 ≤ π/2. The total common area is 4ab tan−1 (b/a).
2
3
Show that,
in the notation of section 11.3, ∂Q/∂x − ∂P /∂y = 2x ; I = πa b/2.
M = I C r × (dr × B). Show that the horizontal sides in the ﬁrst term and the
whole of the second term contribute nothing to the couple.
Note that, if n̂ is the outward normal to the surface, n̂z · n̂ dl is equal to −dρ.
(b) φ = c + z/r.
(a) Yes, F0 (x − y) exp(−r 2 /a2 ); (b) yes, −F0 [(x2 + y 2 )/(2a)] exp(−r2 /a2 );
(c) no, ∇ × F = 0.
A spiral of radius c with its axis parallel to the z-direction and passing through
(a, b). The pitch of the spiral is 2πc2 . No, because (i) γ is not a closed loop and
(ii) the line integral must be zero for every closed loop, not just for a particular
one. In fact ∇ × f = −2k = 0 shows that f is not conservative.
(a) dS = (2a3 cos θ sin2 θ cos φ i + 2a3 cos θ sin2 θ sin φ j + a2 cos θ sin θ k) dθ dφ.
(b) ∇ · r = 3; over the plane z = 0, r · dS = 0.
The necessarily common value is 3πa4 /2.
Write r as ∇( 12 r2 ).
√
The answer is 3 3πα/2 in each case.
Identify the expression for ∇ · (E × B) and use the divergence theorem.
(a) The successive contributions to the integral are:
1 − 2e−1 , 0, 2 + 12 e, − 73 , −1 + 2e−1 , − 21 .
(b) ∇ × F = 2xyz 2 i − y 2 z 2 j + yex k. Show that the contour is equivalent to the
sum of two plane square contours in the planes z = 0 and x = 1, the latter being
traversed in the negative sense. Integral = 16 (3e − 5).
414
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