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

  • Thread starter FaraDazed
  • Start date
  • #1
347
2

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)
 
Last edited:

Answers and Replies

  • #2
11,896
5,551
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:
  • #3
347
2
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 [Broken])
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. [itex]\nabla P [/itex] 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:
  • #4
Dick
Science Advisor
Homework Helper
26,258
619
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. [itex]\nabla P [/itex] 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.
 

Related Threads on Finding Div and Curl of a Vector Field then evaluating a point

Replies
1
Views
627
Replies
4
Views
567
Replies
3
Views
2K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
11
Views
3K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
3
Views
6K
Replies
1
Views
1K
  • Last Post
Replies
2
Views
9K
Top