**Calculating the Divergence of a Vector Field**To find the divergence of the vector field $\vec{F}(x, y, z) = \langle yx^6, xx^5, xy^2 \rangle $, follow these steps:1. **Define the Vector Field:** \[\vec{F}(x, y, z) = \langle yx^6, xx^5, xy^2 \rangle\]2. **Calculate the Partial Derivatives:** - The divergence operator (div) is defined as the dot product of the del operator ($\nabla$) with the vector field $\vec{F}$. - The del operator in Cartesian coordinates is: \[\nabla = \left( \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right)\] Apply this to each component of the vector field: - For the $x$-component ($yx^6$): \[\frac{\partial}{\partial x} (yx^6) = y \cdot 6x^5\] - For the $y$-component ($xx^5$): \[\frac{\partial}{\partial y} (xx^5) = x^5\] - For the $z$-component ($xy^2$): \[\frac{\partial}{\partial z} (xy^2) = 0\]3. **Combine the Partial Derivatives:** The divergence of $\vec{F}$ is given by the sum of these partial derivatives: \[ \text{div}( \vec{F} ) = 6yx^5 + x^5 + 0 \] \[ \text{div}( \vec{F} ) = 6yx^5 + x^5 \]4. **Final Expression:** \[ \text{div}( \vec{F} ) = x^5(6y + 1) \]Conclusion: The divergence of the vector field $\vec{F}(x, y, z) = \langle yx^6, xx^5, xy^2 \rangle $ is $ x^5(6y + 1) $.

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The divergence of the vector field F(z, y, z) = (ya°, az", zy?) is: %3D

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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Question

**Calculating the Divergence of a Vector Field**

To find the divergence of the vector field $\vec{F}(x, y, z) = \langle yx^6, xx^5, xy^2 \rangle $, follow these steps:

1. **Define the Vector Field:**

\[\vec{F}(x, y, z) = \langle yx^6, xx^5, xy^2 \rangle\]

2. **Calculate the Partial Derivatives:**

- The divergence operator (div) is defined as the dot product of the del operator ($\nabla$) with the vector field $\vec{F}$.
- The del operator in Cartesian coordinates is: \[\nabla = \left( \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right)\]

Apply this to each component of the vector field:

- For the $x$-component ($yx^6$):
\[\frac{\partial}{\partial x} (yx^6) = y \cdot 6x^5\]
- For the $y$-component ($xx^5$):
\[\frac{\partial}{\partial y} (xx^5) = x^5\]
- For the $z$-component ($xy^2$):
\[\frac{\partial}{\partial z} (xy^2) = 0\]

3. **Combine the Partial Derivatives:**

The divergence of $\vec{F}$ is given by the sum of these partial derivatives:

\[ \text{div}( \vec{F} ) = 6yx^5 + x^5 + 0 \]
\[ \text{div}( \vec{F} ) = 6yx^5 + x^5 \]

4. **Final Expression:**

\[ \text{div}( \vec{F} ) = x^5(6y + 1) \]

Conclusion: The divergence of the vector field $\vec{F}(x, y, z) = \langle yx^6, xx^5, xy^2 \rangle $ is $ x^5(6y + 1) $.

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