Find the divergence of the vector field = < 6z cos(z), 2z sin(a), 7z >. -div F-
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Problem Statement:**
Find the divergence of the vector field \(\vec{F} = \langle 6z \cos(x), 2z \sin(x), 7z \rangle\).
**Solution:**
The divergence of a vector field \(\vec{F} = \langle F_1, F_2, F_3 \rangle\) is given by:
\[
\text{div } \vec{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z}
\]
For the given vector field \(\vec{F} = \langle 6z \cos(x), 2z \sin(x), 7z \rangle\), we calculate:
1. \(\frac{\partial}{\partial x}(6z \cos(x)) = -6z \sin(x)\)
2. \(\frac{\partial}{\partial y}(2z \sin(x)) = 0\) (since there is no \(y\) term)
3. \(\frac{\partial}{\partial z}(7z) = 7\)
Substitute these into the divergence formula:
\[
\text{div } \vec{F} = -6z \sin(x) + 0 + 7 = -6z \sin(x) + 7
\]
Therefore, the divergence of the vector field is:
\[
\text{div } \vec{F} = -6z \sin(x) + 7
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F07dc3289-c262-41f1-9fc5-a8d18500de1b%2F4cd7734a-4e06-456c-b80c-1499c34a5e3e%2F0k3xgk7_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find the divergence of the vector field \(\vec{F} = \langle 6z \cos(x), 2z \sin(x), 7z \rangle\).
**Solution:**
The divergence of a vector field \(\vec{F} = \langle F_1, F_2, F_3 \rangle\) is given by:
\[
\text{div } \vec{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z}
\]
For the given vector field \(\vec{F} = \langle 6z \cos(x), 2z \sin(x), 7z \rangle\), we calculate:
1. \(\frac{\partial}{\partial x}(6z \cos(x)) = -6z \sin(x)\)
2. \(\frac{\partial}{\partial y}(2z \sin(x)) = 0\) (since there is no \(y\) term)
3. \(\frac{\partial}{\partial z}(7z) = 7\)
Substitute these into the divergence formula:
\[
\text{div } \vec{F} = -6z \sin(x) + 0 + 7 = -6z \sin(x) + 7
\]
Therefore, the divergence of the vector field is:
\[
\text{div } \vec{F} = -6z \sin(x) + 7
\]
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