Answer :
To find the divergence of the vector field \(\overrightarrow{F}\), we start with its expression:
[tex]\[ \overrightarrow{F} = \frac{-2 x y z}{\left(x^2+y^2\right)^2} \hat{i} + \frac{\left(x^2-y^2\right) z}{\left(x^2+y^2\right)^2} \hat{j} + \frac{y}{\left(x^2+y^2\right)} \hat{k} \][/tex]
The divergence \(\operatorname{div} \vec{F}\) is calculated as:
[tex]\[ \operatorname{div} \vec{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z} \][/tex]
where \(F_1\), \(F_2\), and \(F_3\) are the components of \(\overrightarrow{F}\) along the \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\) directions, respectively:
[tex]\[ F_1 = \frac{-2 x y z}{\left(x^2+y^2\right)^2}, \quad F_2 = \frac{\left(x^2-y^2\right) z}{\left(x^2+y^2\right)^2}, \quad F_3 = \frac{y}{\left(x^2+y^2\right)} \][/tex]
### Step 1: Compute \(\frac{\partial F_1}{\partial x}\)
[tex]\[ \frac{\partial F_1}{\partial x} = \frac{\partial}{\partial x} \left( \frac{-2 x y z}{\left(x^2+y^2\right)^2} \right) \][/tex]
### Step 2: Compute \(\frac{\partial F_2}{\partial y}\)
[tex]\[ \frac{\partial F_2}{\partial y} = \frac{\partial}{\partial y} \left( \frac{\left(x^2-y^2\right) z}{\left(x^2+y^2\right)^2} \right) \][/tex]
### Step 3: Compute \(\frac{\partial F_3}{\partial z}\)
[tex]\[ \frac{\partial F_3}{\partial z} = \frac{\partial}{\partial z} \left( \frac{y}{\left(x^2+y^2\right)} \right) = 0 \quad (\text{since } F_3 \text{ does not depend on } z) \][/tex]
### Combine the partial derivatives
[tex]\[ \operatorname{div} \vec{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z} \][/tex]
By performing the differentiation and combining the terms, the divergence \(\operatorname{div} \vec{F}\) results in:
[tex]\[ \operatorname{div} \vec{F} = \frac{8.0 x^2 y z}{(x^2 + y^2)^3} - \frac{4.0 y z (x^2 - y^2)}{(x^2 + y^2)^3} - \frac{4.0 y z}{(x^2 + y^2)^2} \][/tex]
Therefore, substituting the expressions, we achieve the following result:
[tex]\[ \boxed{0} \][/tex]
Thus, the divergence of the vector field \(\overrightarrow{F}\) is:
(d) zero
[tex]\[ \overrightarrow{F} = \frac{-2 x y z}{\left(x^2+y^2\right)^2} \hat{i} + \frac{\left(x^2-y^2\right) z}{\left(x^2+y^2\right)^2} \hat{j} + \frac{y}{\left(x^2+y^2\right)} \hat{k} \][/tex]
The divergence \(\operatorname{div} \vec{F}\) is calculated as:
[tex]\[ \operatorname{div} \vec{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z} \][/tex]
where \(F_1\), \(F_2\), and \(F_3\) are the components of \(\overrightarrow{F}\) along the \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\) directions, respectively:
[tex]\[ F_1 = \frac{-2 x y z}{\left(x^2+y^2\right)^2}, \quad F_2 = \frac{\left(x^2-y^2\right) z}{\left(x^2+y^2\right)^2}, \quad F_3 = \frac{y}{\left(x^2+y^2\right)} \][/tex]
### Step 1: Compute \(\frac{\partial F_1}{\partial x}\)
[tex]\[ \frac{\partial F_1}{\partial x} = \frac{\partial}{\partial x} \left( \frac{-2 x y z}{\left(x^2+y^2\right)^2} \right) \][/tex]
### Step 2: Compute \(\frac{\partial F_2}{\partial y}\)
[tex]\[ \frac{\partial F_2}{\partial y} = \frac{\partial}{\partial y} \left( \frac{\left(x^2-y^2\right) z}{\left(x^2+y^2\right)^2} \right) \][/tex]
### Step 3: Compute \(\frac{\partial F_3}{\partial z}\)
[tex]\[ \frac{\partial F_3}{\partial z} = \frac{\partial}{\partial z} \left( \frac{y}{\left(x^2+y^2\right)} \right) = 0 \quad (\text{since } F_3 \text{ does not depend on } z) \][/tex]
### Combine the partial derivatives
[tex]\[ \operatorname{div} \vec{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z} \][/tex]
By performing the differentiation and combining the terms, the divergence \(\operatorname{div} \vec{F}\) results in:
[tex]\[ \operatorname{div} \vec{F} = \frac{8.0 x^2 y z}{(x^2 + y^2)^3} - \frac{4.0 y z (x^2 - y^2)}{(x^2 + y^2)^3} - \frac{4.0 y z}{(x^2 + y^2)^2} \][/tex]
Therefore, substituting the expressions, we achieve the following result:
[tex]\[ \boxed{0} \][/tex]
Thus, the divergence of the vector field \(\overrightarrow{F}\) is:
(d) zero
