Calculating Line Integrals with Vector Fields on a Bounded Region in 3D Space

Click For Summary

Homework Help Overview

The discussion revolves around evaluating a line integral over a bounded region in 3D space, specifically using a vector field defined by a given function. The context involves applying the Gauss/Divergence theorem to relate surface integrals to volume integrals.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of the Gauss/Divergence theorem and express confusion regarding the notation used for the surface integral. There are questions about the relevance of the given coordinates and the interpretation of the integral notation.

Discussion Status

The conversation is ongoing, with participants seeking clarification on the notation and the mathematical concepts involved. Some guidance has been offered regarding the divergence of the vector field and the need to refer to related theory for better understanding.

Contextual Notes

There is uncertainty about the notation for the surface integral and the specific boundaries of integration. Participants are also questioning the assumptions underlying the problem setup.

ElDavidas
Messages
78
Reaction score
0
Again, I'm stuck on a question:

"Let C be the region in space given by [itex]0 \leq x,y,z \leq 1[/itex] and let [itex]\partial C[/itex] be the boundary of C oriented by the outward pointing unit normal. Suppose that v is the vector field given by

[tex]v = (y^3 -2xy, y^2+3y+2zy, z-z^2)[/tex].

Evaluate [tex]\int_{\partial C} v . dA[/tex]

Stating clearly any result used"

Thanks in advance
 
Last edited:
Physics news on Phys.org
Well if the integral you need to evaluate is a surface integral then just use Gauss/Divergence theorem. But in this context, your notation, more specifically the dA, is unfamiliar to me.
 
Benny said:
more specifically the dA, is unfamiliar to me.

This is what I don't understand. If you have to calculate the area, why are you given x,y, and z? Unless it's the surface of the cube you have to find out. I'm not sure how to go about doing that though
 
If it's the surface integral over the cube then it should be

[tex] \int\limits_{}^{} {\int\limits_{\partial W}^{} {\mathop v\limits^ \to } } .d\mathop S\limits^ \to = \int\limits_{}^{} {\int\limits_{}^{} {\int\limits_W^{} {\nabla \bullet \mathop v\limits^ \to } } } [/tex]

Where the terminals of the triple integral go from -1 to 1 for each of x,y and z.
 
Benny said:
[tex] \int\limits_{}^{} {\int\limits_{\partial W}^{} {\mathop v\limits^ \to } } .d\mathop S\limits^ \to = \int\limits_{}^{} {\int\limits_{}^{} {\int\limits_W^{} {\nabla \bullet \mathop v\limits^ \to } } } [/tex]

Sorry but I don't understand this notation. Are you integrating the gradient of v?
 
It's the divergence of v (there is a dot in between grad and v). If you are unsure about what the problem is asking then you should ask whoever set the question. If it is a textbook problem then surely there should be a related theory section with relevant formulas and explanations.
 

Similar threads

Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
12
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K