Question

Find the divergence of v = [2x^2,−3y^2, 8z^2] and its value at P : (3, 1/2 , 0).

Answer 1

Find the divergence of v = [2x^2,−3y^2, 8z^2] and its value at P : (3, 1/2 , 0).

V = Vx.i + Vy.j + Vz.k, where

Vx = 2x^2, Vy = -3y^2, Vz = 8z^2

 

Div (V) = δVx/δx + δVy/δy + δVz/δz

Div(V) = 4x – 6y + 16z

At P(3, ½. 0), x = 3, y = ½, z = 0, giving

Div(V)(3,1/2,0) = 4*3 – 6(1/2) + 0 = 12 – 3 = 9

Div(V) at P(3,1/2,0) = 9