Question

Divergence of \( \vec{f}(x,y,z) = \frac{(x\hat{i}+y\hat{j}+z\hat{k})}{(x^2+y^2+z^2)^{3/2}}, (x, y, z) ≠ (0, 0, 0).\)

(a) 0

(b) 1

(c) 2

(d) 3

This question was addressed to me in an online quiz.

The query is from Divergence and Curl of a Vector Field topic in portion Vector Differential Calculus of Engineering Mathematics

Answer 1

Correct choice is (a) 0

To explain: Here \(\vec{f}(x,y,z) = \frac{x}{r^3}\hat{i} + \frac{y}{r^3}\hat{j} + \frac{z}{r^3}\hat{k},\) where \(r=(x^2+y^2+z^2) \)

Also here we can see that

\( \frac{∂r}{∂x} = \frac{x}{r} , \frac{∂r}{∂y} = \frac{y}{r} , \frac{∂r}{∂z} = \frac{z}{r} \)

∴ div \(\vec{f} = \frac{∂}{∂x} (\frac{x}{r}) + \frac{∂}{∂y} (\frac{y}{r}) + \frac{∂}{∂z} (\frac{z}{r}) \)

\(\frac{∂}{∂x} (\frac{x}{r^3}) = \frac{r^3-x 3r^2 (x/r)}{(r^3)^2}  =  \frac{1}{r^3} -\frac{3x^2}{r^5} \)

\(\frac{∂}{∂y} (\frac{y}{r^3}) = \frac{1}{r^3} – \frac{3y^2}{r^5}   ,  \frac{∂}{∂y} (\frac{y}{r}) = \frac{1}{r^3} – \frac{3z^2}{r^5} \)

∴ we get div \(\vec{f} = \frac{3}{r^3} – \frac{3x^2+3y^2+3z^2}{r^5}   = 0. \)