Question

39. what is the divergence of the vector
field F where F=xzi+yzj+xyk.

​

Answer 1

SOLUTION

TO DETERMINE

The divergence of the vector field $\vec{F}$

Where $\vec{F} = xz \hat{i} + yz \hat{j} + xy \hat{k}$

EVALUATION

Here the given vector is

$\vec{F} = xz \hat{i} + yz \hat{j} + xy \hat{k}$

Now

$div \: \vec{F}$

$= \nabla \: . \vec{F}$

$\displaystyle = \bigg(\frac{ \partial}{ \partial x} \hat{i} + \frac{ \partial}{ \partial y} \hat{j} + \frac{ \partial}{ \partial z} \hat{k} \bigg).(xz \hat{i} + yz \hat{j} + xy \hat{k})$

$\displaystyle = \frac{ \partial}{ \partial x} (xz)+ \frac{ \partial}{ \partial y}(yz) + \frac{ \partial}{ \partial z} (xy)$

$= z + z + 0$

$= 2z$

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Answer 2

Given:-

Function, F= $xz\hat{i}+yz\hat{j}+xy\hat{k}$

To Find:-

Divergence of F.

Explanation:-

Divergence of any function is defined as $\bigtriangledown.\vec{F}=\dfrac{\partial F_{1}}{\partial x}+\dfrac{\partial F_{2}}{\partial y}+\dfrac{\partial F_{3}}{\partial z}\text{ where,}\\\\F_{1}=xz\Rightarrow\dfrac{\partial F_{1}}{\partial x}=z\\\\F_{2}=yz\Rightarrow\dfrac{\partial F_{2}}{\partial y}=z\\\\F_{3}=xy\Rightarrow\dfrac{\partial F_{3}}{\partial z}=0\\\\\Rightarrow \text{Divergence of the function, }\bigtriangledown.\vec{F}=z+z+0=2z$