Question

F(x, y, z) = (x + yz)i + (y + xz)j + (z + xy)k, Find the divergence of the vector field.

Answer 1

The divergence of F is

div(F ) = ∂(x + yz)/∂x + ∂(y + xz)/∂y + ∂(z + xy)/∂z

div(F ) = 1 + 1 + 1

div(F ) = 3

Answer 2

The divergence of the vector field is equal to 3

Data;

• (x+yz)i
• (y + xz)j
• (z+xy)k

Divergence of Vector Field

To find the divergence of the vector field, we have to differentiate the i, j and k component of the vector.

$div F = \frac{\delta}{\delta x} (x+yz) + \frac{\delta}{\delta y} (y + xz) + \frac{\delta}{\delta z} (z + xy)\\div F = (1+0)+(1+0)+(1+0)\\div F = 1 + 1 + 1 \\div F = 3$

The divergence of the vector field is equal to 3

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