Question

Find the divergence and curl of the given vector field. (x, y, z) = 3xy²ᶦ −2xʸʲ+2/3zk

Answer 1

The divergence of the given vector field is 3 and the curl is 2ᶦ. In this case, the curl of the given vector field is ∇ x (x, y, z) = (∂z/∂y - ∂y/∂z)ᶦ + (∂x/∂z - ∂z/∂x)ʲ + (∂y/∂x - ∂x/∂y)ᵏ = (1 - (-1))ᶦ + (0 - 0)ʲ + (0 - 0)ᵏ = 2ᶦ.

Step-by-step explanation:

The given vector field is (x, y, z) = 3xy²ᶦ −2xʸʲ+2/3zk.

To find the divergence and curl of this vector field, we can use the divergence and curl operators.

The divergence of a vector field F = (F₁, F₂, F₃) is given by ∇ · F = ∂F₁/∂x + ∂F₂/∂y + ∂F₃/∂z.

In this case, the divergence of the given vector field is ∇ · (x, y, z) = ∂x/∂x + ∂y/∂y + ∂z/∂z = 1 + 1 + 1 = 3.

The curl of a vector field F = (F₁, F₂, F₃) is given by ∇ x F = (∂F₃/∂y - ∂F₂/∂z)ᶦ + (∂F₁/∂z - ∂F₃/∂x)ʲ + (∂F₂/∂x - ∂F₁/∂y)ᵏ.

In this case, the curl of the given vector field is ∇ x (x, y, z) = (∂z/∂y - ∂y/∂z)ᶦ + (∂x/∂z - ∂z/∂x)ʲ + (∂y/∂x - ∂x/∂y)ᵏ = (1 - (-1))ᶦ + (0 - 0)ʲ + (0 - 0)ᵏ = 2ᶦ.