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

1. Feb 19, 2015

1. The problem statement, all variables and given/known data}
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 $\bar{r} = (1,1,1)$

2. Relevant equations

$$div(\vec{F})= \nabla \cdot \vec{F} \\ curl(\vec{F})= \nabla \times\vec{F}$$

3. 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.

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

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

Then the curl
$$curl(\vec{V})= \nabla \times\vec{V} = (\frac{\partial V_z}{\partial y} - \frac{\partial V_y}{\partial z}) \hat{i} - (\frac{\partial V_z}{\partial x} - \frac{\partial V_x}{\partial z}) \hat{j} + (\frac{\partial V_y}{\partial x} - \frac{\partial V_x}{\partial y})\hat{k} \\ = (xz-0)\hat{i} - (yz-0)\hat{j} + (y^2-y)\hat{k} = xz\hat{i}- yz\hat{j} + (y^2-y)\hat{k}$$

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

Last edited: Feb 19, 2015
2. Feb 19, 2015

### Staff: Mentor

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 [Broken])

Last edited by a moderator: May 7, 2017
3. Feb 19, 2015

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 thats 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?

Last edited by a moderator: May 7, 2017
4. Feb 19, 2015

### Dick

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