Divergence, Gradient, & Curl - Multivariable Calculus

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Question

Calculate the curl for the following vector field.

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Answer

In order to calculate the curl, we need to recall the formula.

$\curl\ \vec{F}=\triangledown \times \vec{F}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\ $\frac{\partial}{\partial x}$& $\frac{\partial}{\partial y}$ & $\frac{\partial}{\partial z}$ \ P & Q & R \end{vmatrix}\ \ \ =\Big($\frac{\partial R}{\partial y}$-$\frac{\partial Q}{\partial z}$\ \Big)\vec{i}+\Big($\frac{\partial P}{\partial z}$-$\frac{\partial R}{\partial x}$\ \Big)\vec{j}+\Big($\frac{\partial Q}{\partial x}$-$\frac{\partial P}{\partial y}$\ \Big)\vec{k}$

where , , and correspond to the components of a given vector field: $\vec{F}=P\vec{i}+Q\vec{j}+R\vec{k}$

Now lets apply this to out situation.

$curl\ \vec{F}=\triangledown \times \vec{F}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ $\frac{\partial}{\partial x}$& $\frac{\partial}{\partial y}$ & $\frac{\partial}{\partial z}$ \\ $x^3y$^2 & $x^2y$^$3z^4$ & $x^2z$^2 \end{vmatrix}$

$\=\Big($\frac{\partial }{\partial $y}$\Big(x^2z$^2\Big)-$\frac{\partial}{\partial $z}$\Big(x^2y$^$3z^4$\Big)\ \Big)\vec{i}+\Big($\frac{\partial}{\partial $z}$\Big(x^3y$^2\Big)-$\frac{\partial}{\partial $x}$\Big(x^2z$^2\Big)\ \Big)\vec{j}+\Big($\frac{\partial}{\partial $x}$\Big(x^2y$^$3z^4$\Big)-$\frac{\partial}{\partial $y}$\Big(x^3y$^2\Big)\ \Big)\vec{k}$

Thus the curl is

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Question

Compute $div\vec{F}$ , where .

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Answer

All we need to do is calculate the partial derivatives and add them together.

$div\vec{F}=\frac{\partial}{\partial $x}$(10x^2$\ln(yz))+\frac{\partial}{\partial y}$(xyz)+\frac{\partial}{\partial z}$(xz\ln(y))$

$div\vec{F}=20x\ln(yz)+xz+x\ln(y)$

← Didn't Know|Knew It →

Question

Calculate the curl for the following vector field.

Tap to reveal answer

Answer

In order to calculate the curl, we need to recall the formula.

$\curl\ \vec{F}=\triangledown \times \vec{F}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\ $\frac{\partial}{\partial x}$& $\frac{\partial}{\partial y}$ & $\frac{\partial}{\partial z}$ \ P & Q & R \end{vmatrix}\ \ \ =\Big($\frac{\partial R}{\partial y}$-$\frac{\partial Q}{\partial z}$\ \Big)\vec{i}+\Big($\frac{\partial P}{\partial z}$-$\frac{\partial R}{\partial x}$\ \Big)\vec{j}+\Big($\frac{\partial Q}{\partial x}$-$\frac{\partial P}{\partial y}$\ \Big)\vec{k}$

where , , and correspond to the components of a given vector field: $\vec{F}=P\vec{i}+Q\vec{j}+R\vec{k}$

Now lets apply this to out situation.

$curl\ \vec{F}=\triangledown \times \vec{F}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ $\frac{\partial}{\partial x}$& $\frac{\partial}{\partial y}$ & $\frac{\partial}{\partial z}$ \\ $x^3y$^2 & $x^2y$^$3z^4$ & $x^2z$^2 \end{vmatrix}$

$\=\Big($\frac{\partial }{\partial $y}$\Big(x^2z$^2\Big)-$\frac{\partial}{\partial $z}$\Big(x^2y$^$3z^4$\Big)\ \Big)\vec{i}+\Big($\frac{\partial}{\partial $z}$\Big(x^3y$^2\Big)-$\frac{\partial}{\partial $x}$\Big(x^2z$^2\Big)\ \Big)\vec{j}+\Big($\frac{\partial}{\partial $x}$\Big(x^2y$^$3z^4$\Big)-$\frac{\partial}{\partial $y}$\Big(x^3y$^2\Big)\ \Big)\vec{k}$

Thus the curl is

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Question

Compute $div\vec{F}$ , where .

Tap to reveal answer

Answer

All we need to do is calculate the partial derivatives and add them together.

$div\vec{F}=\frac{\partial}{\partial $x}$(10x^2$\ln(yz))+\frac{\partial}{\partial y}$(xyz)+\frac{\partial}{\partial z}$(xz\ln(y))$

$div\vec{F}=20x\ln(yz)+xz+x\ln(y)$

← Didn't Know|Knew It →

Question

Calculate the curl for the following vector field.

Tap to reveal answer

Answer

In order to calculate the curl, we need to recall the formula.

$\curl\ \vec{F}=\triangledown \times \vec{F}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\ $\frac{\partial}{\partial x}$& $\frac{\partial}{\partial y}$ & $\frac{\partial}{\partial z}$ \ P & Q & R \end{vmatrix}\ \ \ =\Big($\frac{\partial R}{\partial y}$-$\frac{\partial Q}{\partial z}$\ \Big)\vec{i}+\Big($\frac{\partial P}{\partial z}$-$\frac{\partial R}{\partial x}$\ \Big)\vec{j}+\Big($\frac{\partial Q}{\partial x}$-$\frac{\partial P}{\partial y}$\ \Big)\vec{k}$

where , , and correspond to the components of a given vector field: $\vec{F}=P\vec{i}+Q\vec{j}+R\vec{k}$

Now lets apply this to out situation.

$curl\ \vec{F}=\triangledown \times \vec{F}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ $\frac{\partial}{\partial x}$& $\frac{\partial}{\partial y}$ & $\frac{\partial}{\partial z}$ \\ $x^3y$^2 & $x^2y$^$3z^4$ & $x^2z$^2 \end{vmatrix}$

$\=\Big($\frac{\partial }{\partial $y}$\Big(x^2z$^2\Big)-$\frac{\partial}{\partial $z}$\Big(x^2y$^$3z^4$\Big)\ \Big)\vec{i}+\Big($\frac{\partial}{\partial $z}$\Big(x^3y$^2\Big)-$\frac{\partial}{\partial $x}$\Big(x^2z$^2\Big)\ \Big)\vec{j}+\Big($\frac{\partial}{\partial $x}$\Big(x^2y$^$3z^4$\Big)-$\frac{\partial}{\partial $y}$\Big(x^3y$^2\Big)\ \Big)\vec{k}$

Thus the curl is

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Question

Compute $div\vec{F}$ , where .

Tap to reveal answer

Answer

All we need to do is calculate the partial derivatives and add them together.

$div\vec{F}=\frac{\partial}{\partial $x}$(10x^2$\ln(yz))+\frac{\partial}{\partial y}$(xyz)+\frac{\partial}{\partial z}$(xz\ln(y))$

$div\vec{F}=20x\ln(yz)+xz+x\ln(y)$

← Didn't Know|Knew It →

Question

Calculate the curl for the following vector field.

Tap to reveal answer

Answer

In order to calculate the curl, we need to recall the formula.

$\curl\ \vec{F}=\triangledown \times \vec{F}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\ $\frac{\partial}{\partial x}$& $\frac{\partial}{\partial y}$ & $\frac{\partial}{\partial z}$ \ P & Q & R \end{vmatrix}\ \ \ =\Big($\frac{\partial R}{\partial y}$-$\frac{\partial Q}{\partial z}$\ \Big)\vec{i}+\Big($\frac{\partial P}{\partial z}$-$\frac{\partial R}{\partial x}$\ \Big)\vec{j}+\Big($\frac{\partial Q}{\partial x}$-$\frac{\partial P}{\partial y}$\ \Big)\vec{k}$

where , , and correspond to the components of a given vector field: $\vec{F}=P\vec{i}+Q\vec{j}+R\vec{k}$

Now lets apply this to out situation.

$curl\ \vec{F}=\triangledown \times \vec{F}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ $\frac{\partial}{\partial x}$& $\frac{\partial}{\partial y}$ & $\frac{\partial}{\partial z}$ \\ $x^3y$^2 & $x^2y$^$3z^4$ & $x^2z$^2 \end{vmatrix}$

$\=\Big($\frac{\partial }{\partial $y}$\Big(x^2z$^2\Big)-$\frac{\partial}{\partial $z}$\Big(x^2y$^$3z^4$\Big)\ \Big)\vec{i}+\Big($\frac{\partial}{\partial $z}$\Big(x^3y$^2\Big)-$\frac{\partial}{\partial $x}$\Big(x^2z$^2\Big)\ \Big)\vec{j}+\Big($\frac{\partial}{\partial $x}$\Big(x^2y$^$3z^4$\Big)-$\frac{\partial}{\partial $y}$\Big(x^3y$^2\Big)\ \Big)\vec{k}$

Thus the curl is

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Question

Compute $div\vec{F}$ , where .

Tap to reveal answer

Answer

All we need to do is calculate the partial derivatives and add them together.

$div\vec{F}=\frac{\partial}{\partial $x}$(10x^2$\ln(yz))+\frac{\partial}{\partial y}$(xyz)+\frac{\partial}{\partial z}$(xz\ln(y))$

$div\vec{F}=20x\ln(yz)+xz+x\ln(y)$

← Didn't Know|Knew It →

Question

Calculate the curl for the following vector field.

Tap to reveal answer

Answer

In order to calculate the curl, we need to recall the formula.

$\curl\ \vec{F}=\triangledown \times \vec{F}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\ $\frac{\partial}{\partial x}$& $\frac{\partial}{\partial y}$ & $\frac{\partial}{\partial z}$ \ P & Q & R \end{vmatrix}\ \ \ =\Big($\frac{\partial R}{\partial y}$-$\frac{\partial Q}{\partial z}$\ \Big)\vec{i}+\Big($\frac{\partial P}{\partial z}$-$\frac{\partial R}{\partial x}$\ \Big)\vec{j}+\Big($\frac{\partial Q}{\partial x}$-$\frac{\partial P}{\partial y}$\ \Big)\vec{k}$

where , , and correspond to the components of a given vector field: $\vec{F}=P\vec{i}+Q\vec{j}+R\vec{k}$

Now lets apply this to out situation.

$curl\ \vec{F}=\triangledown \times \vec{F}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ $\frac{\partial}{\partial x}$& $\frac{\partial}{\partial y}$ & $\frac{\partial}{\partial z}$ \\ $x^3y$^2 & $x^2y$^$3z^4$ & $x^2z$^2 \end{vmatrix}$

$\=\Big($\frac{\partial }{\partial $y}$\Big(x^2z$^2\Big)-$\frac{\partial}{\partial $z}$\Big(x^2y$^$3z^4$\Big)\ \Big)\vec{i}+\Big($\frac{\partial}{\partial $z}$\Big(x^3y$^2\Big)-$\frac{\partial}{\partial $x}$\Big(x^2z$^2\Big)\ \Big)\vec{j}+\Big($\frac{\partial}{\partial $x}$\Big(x^2y$^$3z^4$\Big)-$\frac{\partial}{\partial $y}$\Big(x^3y$^2\Big)\ \Big)\vec{k}$

Thus the curl is

← Didn't Know|Knew It →

Question

Compute $div\vec{F}$ , where .

Tap to reveal answer

Answer

All we need to do is calculate the partial derivatives and add them together.

$div\vec{F}=\frac{\partial}{\partial $x}$(10x^2$\ln(yz))+\frac{\partial}{\partial y}$(xyz)+\frac{\partial}{\partial z}$(xz\ln(y))$

$div\vec{F}=20x\ln(yz)+xz+x\ln(y)$

← Didn't Know|Knew It →

Question

Calculate the curl for the following vector field.

Tap to reveal answer

Answer

In order to calculate the curl, we need to recall the formula.

$\curl\ \vec{F}=\triangledown \times \vec{F}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\ $\frac{\partial}{\partial x}$& $\frac{\partial}{\partial y}$ & $\frac{\partial}{\partial z}$ \ P & Q & R \end{vmatrix}\ \ \ =\Big($\frac{\partial R}{\partial y}$-$\frac{\partial Q}{\partial z}$\ \Big)\vec{i}+\Big($\frac{\partial P}{\partial z}$-$\frac{\partial R}{\partial x}$\ \Big)\vec{j}+\Big($\frac{\partial Q}{\partial x}$-$\frac{\partial P}{\partial y}$\ \Big)\vec{k}$

where , , and correspond to the components of a given vector field: $\vec{F}=P\vec{i}+Q\vec{j}+R\vec{k}$

Now lets apply this to out situation.

$curl\ \vec{F}=\triangledown \times \vec{F}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ $\frac{\partial}{\partial x}$& $\frac{\partial}{\partial y}$ & $\frac{\partial}{\partial z}$ \\ $x^3y$^2 & $x^2y$^$3z^4$ & $x^2z$^2 \end{vmatrix}$

$\=\Big($\frac{\partial }{\partial $y}$\Big(x^2z$^2\Big)-$\frac{\partial}{\partial $z}$\Big(x^2y$^$3z^4$\Big)\ \Big)\vec{i}+\Big($\frac{\partial}{\partial $z}$\Big(x^3y$^2\Big)-$\frac{\partial}{\partial $x}$\Big(x^2z$^2\Big)\ \Big)\vec{j}+\Big($\frac{\partial}{\partial $x}$\Big(x^2y$^$3z^4$\Big)-$\frac{\partial}{\partial $y}$\Big(x^3y$^2\Big)\ \Big)\vec{k}$

Thus the curl is

← Didn't Know|Knew It →

Question

Compute $div\vec{F}$ , where .

Tap to reveal answer

Answer

All we need to do is calculate the partial derivatives and add them together.

$div\vec{F}=\frac{\partial}{\partial $x}$(10x^2$\ln(yz))+\frac{\partial}{\partial y}$(xyz)+\frac{\partial}{\partial z}$(xz\ln(y))$

$div\vec{F}=20x\ln(yz)+xz+x\ln(y)$

← Didn't Know|Knew It →

Question

Calculate the curl for the following vector field.

Tap to reveal answer

Answer

In order to calculate the curl, we need to recall the formula.

$\curl\ \vec{F}=\triangledown \times \vec{F}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\ $\frac{\partial}{\partial x}$& $\frac{\partial}{\partial y}$ & $\frac{\partial}{\partial z}$ \ P & Q & R \end{vmatrix}\ \ \ =\Big($\frac{\partial R}{\partial y}$-$\frac{\partial Q}{\partial z}$\ \Big)\vec{i}+\Big($\frac{\partial P}{\partial z}$-$\frac{\partial R}{\partial x}$\ \Big)\vec{j}+\Big($\frac{\partial Q}{\partial x}$-$\frac{\partial P}{\partial y}$\ \Big)\vec{k}$

where , , and correspond to the components of a given vector field: $\vec{F}=P\vec{i}+Q\vec{j}+R\vec{k}$

Now lets apply this to out situation.

$curl\ \vec{F}=\triangledown \times \vec{F}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ $\frac{\partial}{\partial x}$& $\frac{\partial}{\partial y}$ & $\frac{\partial}{\partial z}$ \\ $x^3y$^2 & $x^2y$^$3z^4$ & $x^2z$^2 \end{vmatrix}$

$\=\Big($\frac{\partial }{\partial $y}$\Big(x^2z$^2\Big)-$\frac{\partial}{\partial $z}$\Big(x^2y$^$3z^4$\Big)\ \Big)\vec{i}+\Big($\frac{\partial}{\partial $z}$\Big(x^3y$^2\Big)-$\frac{\partial}{\partial $x}$\Big(x^2z$^2\Big)\ \Big)\vec{j}+\Big($\frac{\partial}{\partial $x}$\Big(x^2y$^$3z^4$\Big)-$\frac{\partial}{\partial $y}$\Big(x^3y$^2\Big)\ \Big)\vec{k}$

Thus the curl is

← Didn't Know|Knew It →

Question

Compute $div\vec{F}$ , where .

Tap to reveal answer

Answer

All we need to do is calculate the partial derivatives and add them together.

$div\vec{F}=\frac{\partial}{\partial $x}$(10x^2$\ln(yz))+\frac{\partial}{\partial y}$(xyz)+\frac{\partial}{\partial z}$(xz\ln(y))$

$div\vec{F}=20x\ln(yz)+xz+x\ln(y)$

← Didn't Know|Knew It →

Question

Calculate the curl for the following vector field.

Tap to reveal answer

Answer

In order to calculate the curl, we need to recall the formula.

$\curl\ \vec{F}=\triangledown \times \vec{F}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\ $\frac{\partial}{\partial x}$& $\frac{\partial}{\partial y}$ & $\frac{\partial}{\partial z}$ \ P & Q & R \end{vmatrix}\ \ \ =\Big($\frac{\partial R}{\partial y}$-$\frac{\partial Q}{\partial z}$\ \Big)\vec{i}+\Big($\frac{\partial P}{\partial z}$-$\frac{\partial R}{\partial x}$\ \Big)\vec{j}+\Big($\frac{\partial Q}{\partial x}$-$\frac{\partial P}{\partial y}$\ \Big)\vec{k}$

where , , and correspond to the components of a given vector field: $\vec{F}=P\vec{i}+Q\vec{j}+R\vec{k}$

Now lets apply this to out situation.

$curl\ \vec{F}=\triangledown \times \vec{F}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ $\frac{\partial}{\partial x}$& $\frac{\partial}{\partial y}$ & $\frac{\partial}{\partial z}$ \\ $x^3y$^2 & $x^2y$^$3z^4$ & $x^2z$^2 \end{vmatrix}$

$\=\Big($\frac{\partial }{\partial $y}$\Big(x^2z$^2\Big)-$\frac{\partial}{\partial $z}$\Big(x^2y$^$3z^4$\Big)\ \Big)\vec{i}+\Big($\frac{\partial}{\partial $z}$\Big(x^3y$^2\Big)-$\frac{\partial}{\partial $x}$\Big(x^2z$^2\Big)\ \Big)\vec{j}+\Big($\frac{\partial}{\partial $x}$\Big(x^2y$^$3z^4$\Big)-$\frac{\partial}{\partial $y}$\Big(x^3y$^2\Big)\ \Big)\vec{k}$

Thus the curl is

← Didn't Know|Knew It →

Question

Compute $div\vec{F}$ , where .

Tap to reveal answer

Answer

All we need to do is calculate the partial derivatives and add them together.

$div\vec{F}=\frac{\partial}{\partial $x}$(10x^2$\ln(yz))+\frac{\partial}{\partial y}$(xyz)+\frac{\partial}{\partial z}$(xz\ln(y))$

$div\vec{F}=20x\ln(yz)+xz+x\ln(y)$

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Knew It

Divergence, Gradient, & Curl - Multivariable Calculus

Calculate the curl for the following vector field.

In order to calculate the curl, we need to recall the formula. where , , and correspond to the components of a given vector field: Now lets apply this to out situation. Thus the curl is

Compute , where .

All we need to do is calculate the partial derivatives and add them together.

Calculate the curl for the following vector field.

In order to calculate the curl, we need to recall the formula. where , , and correspond to the components of a given vector field: Now lets apply this to out situation. Thus the curl is

Compute , where .

All we need to do is calculate the partial derivatives and add them together.

Calculate the curl for the following vector field.

In order to calculate the curl, we need to recall the formula. where , , and correspond to the components of a given vector field: Now lets apply this to out situation. Thus the curl is

Compute , where .

All we need to do is calculate the partial derivatives and add them together.

Calculate the curl for the following vector field.

In order to calculate the curl, we need to recall the formula. where , , and correspond to the components of a given vector field: Now lets apply this to out situation. Thus the curl is

Compute , where .

All we need to do is calculate the partial derivatives and add them together.

Calculate the curl for the following vector field.

In order to calculate the curl, we need to recall the formula. where , , and correspond to the components of a given vector field: Now lets apply this to out situation. Thus the curl is

Compute , where .

All we need to do is calculate the partial derivatives and add them together.

Calculate the curl for the following vector field.

In order to calculate the curl, we need to recall the formula. where , , and correspond to the components of a given vector field: Now lets apply this to out situation. Thus the curl is

Compute , where .

All we need to do is calculate the partial derivatives and add them together.

Calculate the curl for the following vector field.

In order to calculate the curl, we need to recall the formula. where , , and correspond to the components of a given vector field: Now lets apply this to out situation. Thus the curl is

Compute , where .

All we need to do is calculate the partial derivatives and add them together.

Calculate the curl for the following vector field.

In order to calculate the curl, we need to recall the formula. where , , and correspond to the components of a given vector field: Now lets apply this to out situation. Thus the curl is

Compute , where .

All we need to do is calculate the partial derivatives and add them together.

Compute $div\vec{F}$ , where .

All we need to do is calculate the partial derivatives and add them together.

$div\vec{F}=\frac{\partial}{\partial $x}$(10x^2$\ln(yz))+\frac{\partial}{\partial y}$(xyz)+\frac{\partial}{\partial z}$(xz\ln(y))$

$div\vec{F}=20x\ln(yz)+xz+x\ln(y)$

Compute $div\vec{F}$ , where .

All we need to do is calculate the partial derivatives and add them together.

$div\vec{F}=\frac{\partial}{\partial $x}$(10x^2$\ln(yz))+\frac{\partial}{\partial y}$(xyz)+\frac{\partial}{\partial z}$(xz\ln(y))$

$div\vec{F}=20x\ln(yz)+xz+x\ln(y)$

Compute $div\vec{F}$ , where .

All we need to do is calculate the partial derivatives and add them together.

$div\vec{F}=\frac{\partial}{\partial $x}$(10x^2$\ln(yz))+\frac{\partial}{\partial y}$(xyz)+\frac{\partial}{\partial z}$(xz\ln(y))$

$div\vec{F}=20x\ln(yz)+xz+x\ln(y)$

Compute $div\vec{F}$ , where .

All we need to do is calculate the partial derivatives and add them together.

$div\vec{F}=\frac{\partial}{\partial $x}$(10x^2$\ln(yz))+\frac{\partial}{\partial y}$(xyz)+\frac{\partial}{\partial z}$(xz\ln(y))$

$div\vec{F}=20x\ln(yz)+xz+x\ln(y)$

Compute $div\vec{F}$ , where .

All we need to do is calculate the partial derivatives and add them together.

$div\vec{F}=\frac{\partial}{\partial $x}$(10x^2$\ln(yz))+\frac{\partial}{\partial y}$(xyz)+\frac{\partial}{\partial z}$(xz\ln(y))$

$div\vec{F}=20x\ln(yz)+xz+x\ln(y)$

Compute $div\vec{F}$ , where .

All we need to do is calculate the partial derivatives and add them together.

$div\vec{F}=\frac{\partial}{\partial $x}$(10x^2$\ln(yz))+\frac{\partial}{\partial y}$(xyz)+\frac{\partial}{\partial z}$(xz\ln(y))$

$div\vec{F}=20x\ln(yz)+xz+x\ln(y)$

Compute $div\vec{F}$ , where .

All we need to do is calculate the partial derivatives and add them together.

$div\vec{F}=\frac{\partial}{\partial $x}$(10x^2$\ln(yz))+\frac{\partial}{\partial y}$(xyz)+\frac{\partial}{\partial z}$(xz\ln(y))$

$div\vec{F}=20x\ln(yz)+xz+x\ln(y)$

Compute $div\vec{F}$ , where .

All we need to do is calculate the partial derivatives and add them together.

$div\vec{F}=\frac{\partial}{\partial $x}$(10x^2$\ln(yz))+\frac{\partial}{\partial y}$(xyz)+\frac{\partial}{\partial z}$(xz\ln(y))$

$div\vec{F}=20x\ln(yz)+xz+x\ln(y)$