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11 Feb 2014: α =. C. A dx. α =. C. A dx =. B dS. Aϕ =g. 4πr. 1 cos θsin θ. B =g. ... 1 cos θsin θ. B =g. 4πr2̂r. References. [1] Gibbons and Rychenkova, 9608085.

21 Jan 2014: 7. Suppose γ : [a,b] R2 is a smooth curve parametrized by arc length. ... r cos z,r sin z,z).(c) U = (0, 2π) (0, 2π), σ(θ,φ) = ((a b cos φ) cos θ, (a b cos φ) sin θ,b sin φ).

29 Aug 2014: cosh ξ cos η ,. z =c sinh ξ. cosh ξ cos η , (B.1). ... UndVndη. , (B.17). ψ̂. ξ= 3. 2sinh ξ. (cosh ξ cos η)5/2.

21 Jan 2014: 7. Suppose γ : [a,b] R2 is a smooth curve parametrized by arc length. ... r cos z,r sin z,z).(c) U = (0, 2π) (0, 2π), σ(θ,φ) = ((a b cos φ) cos θ, (a b cos φ) sin θ,b sin φ).

18 Jan 2014: L =1. 2m(. ρ̇2 ρ2θ̇2). 12kρ2 q. (. Eρ cos θ B. ... A =B. 2. (. ρ sin θ, ρ cos θ, 0).

4 Feb 2014: φ(s,v) = α(s) r(n(s) cos v b(s) sin v)where n = n(s) and b = b(s) are the normal and binormal vectors of α. ... 9. Consider a surface of revolution parametrized by φ : (0, 2π) (a,b) R3, whereφ(u,v) = (f(v) cos u,f(v) sin u,g(v)).

6 Feb 2014: 1. Find the General Solution with f (t) = 1 (t 0).Complementary function is y = A cos(3t) B sin(3t). ... Particular integral is y = 19General solution is y = 19 A cos(3t) B sin(3t).

4 Nov 2014: as. Θ(0, t) = Θ0 A cos (2πt/tD) B cos (2πt/tY) (4.30). – ... We can solve these as. X(x)T(t) =[a sin(. λx) b cos(.

5 Nov 2014: When small-amplitudewaves propagate along ex, the surface of the cylinder is de-scribed by r̂ = [h(s, t) b cos φ]ey b sin φez sex and the. ... p̃ ηωh̃i cos θAK1(kr)ũr h̃ωi cos θ[αArkK1(kr) BK2(kr) CK0(kr)]ũθ h̃ωi sin θ[BK2(kr)

28 Nov 2014: f ′′cot θf ′6f =A cos θ B. (2.14)Together with the particular solution, the full solution is then ψ = cr3(at bt3 1)for t =cos θ. ... The function H(ν) is plotted for cone angles in figure 1(b).