Home › Calculus › Elements of field theory › Divergence of a vector field at M(1;1;1)

Divergence of a vector field at M(1;1;1)

This problem asks to find the divergence of a vector field at a given point.

Problem

Find the divergence of the vector field:

a → = x 2 y i → − 2 y 3 z j → + 3 z 2 x k → , M ( 1 ; 1 ; 1 )

Solution

Step 1

Write the formula for the divergence of the vector field ā = P ī + Q j̄ + R k̄:

div a → = ∂ P ∂ x + ∂ Q ∂ y + ∂ R ∂ z

Step 2

Write out the components of the vector field:

P = x 2 y , Q = − 2 y 3 z , R = 3 z 2 x

Step 3

Compute the partial derivatives appearing in the divergence formula:

Why is this a scalar quantity?

Unlike the curl, the divergence is a scalar quantity. It is the sum of three partial derivatives, each of which is a scalar. Hence the result is also a scalar, not a vector.

∂ P ∂ x = 2 x y ∂ Q ∂ y = − 6 y 2 z ∂ R ∂ z = 6 z x

Step 4

Substitute the found derivatives into the divergence formula:

div a → = 2 x y − 6 y 2 z + 6 z x

Step 5

Substitute the coordinates of the point M(1; 1; 1) into the expression for the divergence:

div a → ( M ) = 2 − 6 + 6 = 2

Answer

div ā(M) = 2.

  • Antiderivative and integral
  • Definite integral
  • Improper integrals
  • Double integrals
  • Triple integrals
  • Line integrals
  • Surface integrals
  • Elements of field theory
  • Bringing a factor under the differential sign
  • Tangent line equations
  • USE profile level
  • USE basic level
  • BSE