Finding Div and Curl of a Vector Field then evaluating a point

[FaraDazed]

Homework Statement

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

Homework Equations

<br /> div(\vec{F})= \nabla \cdot \vec{F} \\<br /> curl(\vec{F})= \nabla \times\vec{F}<br />

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.

<br /> 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<br />

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

Then the curl
<br /> 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} \\<br /> = (xz-0)\hat{i} - (yz-0)\hat{j} + (y^2-y)\hat{k} = xz\hat{i}- yz\hat{j} + (y^2-y)\hat{k}<br />

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

 

[jedishrfu]

Looks right to me. The j part of the curl fooled me for a moment as its usually written as + (Vxz - Vzx) j but then I saw the minus sign.

http://en.wikipedia.org/wiki/Curl_(mathematics )

 

[FaraDazed]

Looks right to me. The j part fooled me for a moment as its usually written as + (Vxz - Vzx) j but then I saw the minus sign.

http://en.wikipedia.org/wiki/Curl_(mathematics )

Ah right. Yeah I just did the cross product as I would normally have done with any two vectors, find it easier to remember like that. Thanks for taking a look.

Just wanted to check if I was doing things correctly. there is another question which I am stuck on, not sure whether to make another thread, but it basically asks to find the maximum [pressure] gradient at a certain point that's given; I have found the gradient, i.e. \nabla P but am not sure what to do next. Do I just put the coordinates of the point in, or do I need to first set it to zero and somehow solve it?

 

[Dick]

Ah right. Yeah I just did the cross product as I would normally have done with any two vectors, find it easier to remember like that. Thanks for taking a look.

Just wanted to check if I was doing things correctly. there is another question which I am stuck on, not sure whether to make another thread, but it basically asks to find the maximum [pressure] gradient at a certain point that's given; I have found the gradient, i.e. \nabla P but am not sure what to do next. Do I just put the coordinates of the point in, or do I need to first set it to zero and somehow solve it?

You need to find the maximum of $|\nabla P|$, the point where the gradient vector has maximum length.