Solved Examples/Numericals - Divergence of a Vector Field.

Q.1 Determine the divergence of the following vector fields and evaluate them at the specified points:
a) A = yz a_x + 4xy a_y + y a_z at point (1, -2, 3)
Ans:

Given A = yz a_x + 4xy a_y + y a_z at point (1, -2, 3)

Divergence of A is given as:

∇ . A = 0 + 4x + 0 = 4x
At Point (1, -2, 3)  Hence, ∇ . A = 4

b) A = ρzsinφ a_ρ + 3ρz² cosφ a_φ at (5, π / 2, 1)
Ans:

Divergence of a vector field A in cylindrical co-ordinate system is given as:

=> ∇ . A = (1/ρ) 2ρz sinφ - (1/ρ) 3ρz² sinφ

= 2zsinφ - 3z² sinφ =  (2 - 3z) zsinφ

At point (5, π / 2, 1)       ∇ . A = (2 – 3) (1) = -1

c) A = 2r cosθ cosφ a_r + r^1/2 a_φ   at (1, π / 6, π / 3)
Ans:

Divergence of a vector field A in Spherical co-ordinate system is given as:

=> ∇ . A = (1/r²) 6r² cosθ cosφ = 6 cosθ cosφ

At point (1, π/6, π/3)

∇ . A= 6 cos π/6 cos π/3 = 2.598.


Q. 2 Determine the flux of D = ρ² cos²φ a_ρ + zsin φ a_φ over the closed surface of the cylinder, 0 < z < 1, ρ = 4. Verify the divergence theorem for this case.
Ans:

We know a closed cylinder has three surface - top(t), bottom(b) and curved surface(s).

Ψ = Ψ_t + Ψ_b + Ψ_s

Ψ_t = Ψ_b = 0, since D has no z component.

It means,
Ψ_s = ∫ ∫ ρ² cos²φ (ρ dφ dz)
= ρ³ ∫ ₀^2πcos²φ dφ ∫ ₀¹dz
= (4)³ (π) (1) = 64 π

Therefore,
Ψ = Ψ_t + Ψ_b + Ψ_s = 64 π

By the divergence theorem,

Divergence of D is given as:

∇ . A = 3ρ cos²φ - (1/ρ) cosφ

= ∫ ∫ ∫(3ρ cos²φ - (1/ρ) cosφ)   ρdρ dφ dz

= 3 ∫₀⁴ ρ²dρ  ∫₀^2π cos²φ dφ  ∫₀¹dz  +  ∫₀⁴dρ  ∫₀^2π cosφdφ  ∫₀¹ dz

= 3 (4³ /3 ) (π ) (1) = 64π



ALSO READ:

- Line , Surface and Volume Intergral.

- Del Operator - Definition and Significance.

- Gradient Of a Scalar (∇ V).

- Numericals / Solved Examples - Gradient Of a Scalar.

- Divergence Of a Vector ( ∇ . A ).

- Numericals / Solved Examples - Divergence Of a Vector.

- Curl Of a Vector ( ∇ x A).

- Laplacian Of a Scalar ( ∇² V).


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