The divergence of the vector field F⃗ =(x+3y)i^+(y−3z)j^+(x−2z)k^𝐹→=(𝑥+3𝑦)𝑖^+(𝑦−3𝑧)𝑗^+(𝑥−2𝑧)𝑘^ isa.None of theseb.0c.2d.-2

The divergence of a vector field F = P(x,y,z)i + Q(x,y,z)j + R(x,y,z)k is given by the formula:

div F = ∂P/∂x + ∂Q/∂y + ∂R/∂z

Here, P(x,y,z) = x + 3y, Q(x,y,z) = y - 3z, and R(x,y,z) = x - 2z.

So, we have:

∂P/∂x = ∂/∂x (x + 3y) = 1 (since the derivative of x with respect to x is 1 and the derivative of a constant times y with respect to x is 0)

∂Q/∂y = ∂/∂y (y - 3z) = 1 (since the derivative of y with respect to y is 1 and the derivative of a constant times z with respect to y is 0)

∂R/∂z = ∂/∂z (x - 2z) = -2 (since the derivative of a constant times z with respect to z is the constant and the derivative of x with respect to z is 0)

Adding these up, we get:

div F = ∂P/∂x + ∂Q/∂y + ∂R/∂z = 1 + 1 - 2 = 0

So, the divergence of the given vector field is 0. Therefore, the correct answer is (b) 0.