Question

Divergence of f (x,y,z)=(xi+yj+zk)/(x^2+y^2+z^2)^3/2,(x,y,z)≠(0,0,0).(i,j,k are unit vectors.)

Answer 1

Answer:

3

Explanation:

The Divergence theorem states that:

where ∇.D is the divergence of the vector field D.

In Rectangular coordinates, the divergence is defined as:

Analysis:

Answer 2

Answer:

The divergence of f (x,y,z)=(xi+yj+zk)/$(x^2+y^2+z^2)^{3/2$,(x,y,z)≠(0,0,0) is equal to 0

Explanation:

From the above question,

They have given :

The gradient of the function f (x,y,z)=(xi+yj+zk)/$(x^2+y^2+z^2)^{3/2$,(x,y,z)≠(0,0,0) can be found using the following formula for the gradient:

∇f = ∂f/∂x i + ∂f/∂y j + ∂f/∂z k

$\frac{\partial f}{\partial x} = \frac{x \left(x^2+y^2+z^2\right) - 3x \left(xi+yj+zk\right)}{\left(x^2+y^2+z^2\right)^4}$

$\frac{\partial f}{\partial y} = \frac{y \left(x^2+y^2+z^2\right) - 3y \left(xi+yj+zk\right)}{\left(x^2+y^2+z^2\right)^4}$

$\frac{\partial f}{\partial z} = \frac{z \left(x^2+y^2+z^2\right) - 3z \left(xi+yj+zk\right)}{\left(x^2+y^2+z^2\right)^4}$

The divergence of the vector field f (x, y, z) = (xi + yj + zk)/(x2 + y2 + z2)3/2 , (x, y, z) ̸= (0, 0, 0) can be found by computing the partial derivatives of the function with respect to each of the three variables. Using the chain rule, the divergence is given by:

The divergence of the gradient is given by the following formula:

div(∇f) = ∂/∂x (∂f/∂x) + ∂/∂y (∂f/∂y) + ∂/∂z (∂f/∂z)

$\nabla \cdot f = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} + \frac{\partial f}{\partial z}$

The divergence of the gradient is given by the following formula:

div(∇f) = ∂/∂x (∂f/∂x) + ∂/∂y (∂f/∂y) + ∂/∂z (∂f/∂z)

Calculating the partial derivatives and performing the summation, we find that the divergence of f (x,y,z)=(xi+yj+zk)/$(x^2+y^2+z^2)^{3/2$,(x,y,z)≠(0,0,0) is:

div(∇f) = 0

Thus, the divergence of f (x,y,z)=(xi+yj+zk)/$(x^2+y^2+z^2)^{3/2$,(x,y,z)≠(0,0,0) is equal to 0.

Substituting the partial derivatives, we get the divergence of the vector field as:

$\nabla \cdot f = \frac{\left(x^2+y^2+z^2\right)}{\left(x^2+y^2+z^2\right)^3} - \frac{3 \left(xi+yj+zk\right)}{\left(x^2+y^2+z^2\right)^4}$

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