66.2k views
0 votes
Find div F for F = (x ^ 2 * y ^ 3 - z ^ 4) * i 4x ^ 5 * y ^ 2 * zj - y ^ 4 * z ^ 6 * k

1 Answer

3 votes

Final answer:

The divergence of the vector field F = (x^2*y^3 - z^4)*i + 4x^5*y^2*z*j - y^4*z^6*k is calculated using partial derivatives of each component with respect to their respective variables and is found to be div F = 2*x*y^3 + 8*x^5*y*z - 6*y^4*z^5.

Step-by-step explanation:

To find the divergence of the vector field F, denoted as div F, we apply the divergence operator, which in Cartesian coordinates is given by:

div F = ∇ ⋅ F = ∂Fx/∂x + ∂Fy/∂y + ∂Fz/∂z

For F = (x^2*y^3 - z^4)*i + 4x^5*y^2*z*j - y^4*z^6*k, the divergence is calculated by taking the partial derivatives:

  • ∂Fx/∂x = ∂/(x^2*y^3 - z^4)/∂x
  • ∂Fy/∂y = ∂/(4x^5*y^2*z)/∂y
  • ∂Fz/∂z = ∂/(-y^4*z^6)/∂z

Substitute the respective terms with their partial derivatives to find the divergence:

div F = 2*x*y^3 + 8*x^5*y*z - 6*y^4*z^5

User TonioElGringo
by
8.1k points