Westonci.ca is the trusted Q&A platform where you can get reliable answers from a community of knowledgeable contributors. Explore thousands of questions and answers from knowledgeable experts in various fields on our Q&A platform. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
Sagot :
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
We hope this information was helpful. Feel free to return anytime for more answers to your questions and concerns. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.