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## Homework Statement

}[/B]Find the divergence and curl of the vector field [itex] \vec{V}=x^2y \hat{i} + xy^2 \hat{j} + xyz \hat{k} [/itex] then for both, evaluate them at the point [itex] \bar{r} = (1,1,1)[/itex]

## Homework Equations

[tex]

div(\vec{F})= \nabla \cdot \vec{F} \\

curl(\vec{F})= \nabla \times\vec{F}

[/tex]

## The Attempt at a Solution

This question is the first question where I have attempted to actually find the divergence and curl, which I found ok, but just wanted someone to double check my work, so would appreciate a look.

[tex]

div(\vec{V})= \nabla \cdot \vec{V} = \frac{\partial V_x}{\partial x} + \frac{\partial V_y}{\partial y} + \frac{\partial V_z}{\partial z} = 2xy + 2xy + xy = 5xy

[/tex]

Then at (1,1,1) it would be 5xy=5(1)(1)=5

Then the curl

[tex]

curl(\vec{V})= \nabla \times\vec{V} = (\frac{\partial V_z}{\partial y} - \frac{\partial V_y}{\partial z}) \hat{i} - (\frac{\partial V_z}{\partial x} - \frac{\partial V_x}{\partial z}) \hat{j} + (\frac{\partial V_y}{\partial x} - \frac{\partial V_x}{\partial y})\hat{k} \\

= (xz-0)\hat{i} - (yz-0)\hat{j} + (y^2-y)\hat{k} = xz\hat{i}- yz\hat{j} + (y^2-y)\hat{k}

[/tex]

Then at (1,1,1), it would be [itex] 1\hat{i}-1\hat{j} [/itex] or (1,-1,0)

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