\( \operatorname { div } F = c [ \frac { \partial } { \partial x } ( \frac { x } { r ^ { 3 } } ) + \frac { \partial } { \partial y } ( \frac { y } { r ^ { 3 } } ) + \frac { \partial } { \partial z } ( \frac { z } { r ^ { 3 } } ) ] \)

To compute the divergence of the vector field \( F \) given by \[ F = \left( \frac{x}{r^3}, \frac{y}{r^3}, \frac{z}{r^3} \right) \] where \( r = \sqrt{x^2 + y^2 + z^2} \), we need to evaluate the expression: \[ \operatorname{div} F = \frac{\partial}{\partial x} \left( \frac{x}{r^3} \right) + \frac{\partial}{\partial y} \left( \frac{y}{r^3} \right) + \frac{\partial}{\partial z} \left( \frac{z}{r^3} \right) \] Let's compute each term separately. ### Step 1: Compute \( \frac{\partial}{\partial x} \left( \frac{x}{r^3} \right) \) Using the quotient rule: \[ \frac{\partial}{\partial x} \left( \frac{x}{r^3} \right) = \frac{(r^3)(1) - x \frac{\partial}{\partial x}(r^3)}{(r^3)^2} \] Now, we need to compute \( \frac{\partial}{\partial x}(r^3) \): \[ r^3 = (x^2 + y^2 + z^2)^{3/2} \] Using the chain rule: \[ \frac{\partial}{\partial x}(r^3) = 3r^2 \frac{\partial r}{\partial x} = 3r^2 \cdot \frac{x}{r} = 3xr \] Substituting this back into our expression: \[ \frac{\partial}{\partial x} \left( \frac{x}{r^3} \right) = \frac{r^3 - x(3xr)}{r^6} = \frac{r^3 - 3x^2r}{r^6} = \frac{r^3 - 3x^2r}{r^6} = \frac{r^2 - 3x^2}{r^4} \] ### Step 2: Compute \( \frac{\partial}{\partial y} \left( \frac{y}{r^3} \right) \) By symmetry, we can write: \[ \frac{\partial}{\partial y} \left( \frac{y}{r^3} \right) = \frac{r^2 - 3y^2}{r^4} \] ### Step 3: Compute \( \frac{\partial}{\partial z} \left( \frac{z}{r^3} \right) \) Similarly, we have: \[ \frac{\partial}{\partial z} \left( \frac{z}{r^3} \right) = \frac{r^2 - 3z^2}{r^4} \] ### Step 4: Combine the results Now we can combine all three terms: \[ \operatorname{div} F = \frac{r^2 - 3x^2}{r^4} + \frac{r^2 - 3y^2}{r^4} + \frac{r^2 - 3z^2}{r^4} \] Combining these gives: \[ \operatorname{div} F = \frac{(r^2 - 3x^2) + (r^2 - 3y^2) + (r^2 - 3z^2)}{r^4} = \frac{3r^2 - 3(x^2 + y^2 + z^2)}{r^4} \] Since \( x^2 + y^2 + z^2 = r^2 \), we have: \[ \operatorname{div} F = \frac{3r^2 - 3r^2}{r^4} = \frac{0}{r^4} = 0 \] ### Conclusion Thus, the divergence of the vector field \( F \) is: \[ \operatorname{div} F = 0 \]