Divergence and surfaces integral, very hard

Click For Summary

Homework Help Overview

The discussion revolves around calculating a surface integral of a vector field in cylindrical polar coordinates and comparing it with the divergence theorem. The problem involves a closed surface bounded by two cylinders and two planes.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to calculate the surface integral over six surfaces but questions the discrepancy between their results and those obtained using the divergence theorem. Some participants question the identification and count of the surfaces involved in the calculation.

Discussion Status

Participants are actively discussing the surfaces involved in the problem, with some suggesting that the original poster may have miscounted the surfaces in cylindrical coordinates. There is an ongoing exploration of the correct interpretation of the problem setup.

Contextual Notes

The original poster expresses confusion regarding their calculations and seeks clarification on the surfaces, indicating a potential misunderstanding of the geometry involved in cylindrical coordinates.

matt222
Messages
130
Reaction score
0

Homework Statement



A vector field h is described in cylindrical polar coordinates by ( h equation attached )
where i, j, and k are the unit vectors along the Cartesian axes and
(er) is the unit vector (x/r) i+(y/r) j
Calculate (1) by surface integral h through the closed surface bounded by the cylinders r=a and r=2a and the planes z=-a*pi/2 to +a*pi/2
and (2) by divergence theorem.


Homework Equations





The Attempt at a Solution


1- I draw the equation and I found it is like hollow tube cylinder with radius of r of the inner surface and 2r for the outer surface
2-I found there are 6 surfaces

a- S1 pointing into -z direction
b- S2 pointing into +z direction
c- S3 pointing into -Y direction
d- S4 pointing into -x direction
e- S5 surface of the outer tube r= 2a
f- S6 surface of the inner tube r=a


I solved the 6 surfaces and the outcome

S1=0

S2=sin(lamda*a*pi/2)*pi*a^2/4

S3=y/lamda*sin(a*pi*lmda/2)

S4=x/lamda*sin(a*pi*lmda/2)

S5= pi/2*lamda[2sin(a*pi*lmda/2)-2acos(a*pi*lmda/2)+2a]

S6=pi/2*lamda[sin(a*pi*lmda/2)-acos(a*pi*lmda/2)+a]

adding them all up and compared them with divergence which I got are not the same, could anyone find out what mistake I did

the divergence I got was

3a*pi/4lamda*sin(a*pi*lmda/2)+7a^2*pi/6*cos(a*pi*lmda/2)+3a*pi/4sin(a*pi*lmda/2)+
7a^2*pi/6

which is not the same with what I got
 

Attachments

  • Untitled.png
    Untitled.png
    2.3 KB · Views: 518
Physics news on Phys.org
It's very hard to see what you may have done wrong when you've only shown your final result, however just taking a look at what you took for your surfaces:

You mentioned you had 6 surfaces? If you're dealing in cylindrical coordinates I see only 4,

S1: The top of the hollow tube (pointing in the positive \hat{e_z} direction.)
S2: The bottom of the hollow tube (pointing in the negative \hat{e_z} direction.)
S3: The outside of the cylinder r=2a (pointing radially outwards in the \hat{e_r} direction.)
S4: The inside of the cylinder r=a (pointing radially inwards in the \hat{e_r} direction.)
 
what about the other two surfaces which is pointing in -y and -x direction
 
All the surfaces are accounted for. You have two planes which have normal vectors oriented with the z-axis.

You have two cylinders, one twice the radius of the other. The surfaces of cylinders have normal vectors which point radially.

What other surfaces do you have?
 
really lost don't know where is my mistake
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
8
Views
3K