How to use the divergence theorem to solve this question

Click For Summary

Homework Help Overview

The discussion revolves around the application of the divergence theorem in a problem related to calculating the volume of a cylinder. The original poster attempts to use the theorem to derive a volume expression but encounters a discrepancy in their results.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants question whether the surface integral is over a closed surface and discuss the implications of this on the application of the divergence theorem. There are attempts to clarify the volume being considered in relation to the surface integral.

Discussion Status

There is ongoing clarification regarding the setup of the problem, particularly concerning the volume and surface involved. Some participants provide guidance on recognizing the need to account for the surfaces that contribute to the integral, indicating a productive direction in the discussion.

Contextual Notes

Participants note that the original poster may have misinterpreted the surface area involved, as it pertains to only half of the cylinder rather than the whole. This detail is crucial for the correct application of the divergence theorem.

Leo Liu
Messages
353
Reaction score
156
Homework Statement
.
Relevant Equations
.
1626969702959.png

The correct answer is ##\frac{\pi a^2 h} 2## by using the standard approach. However when I tried using the divergence theorem to solve this problem, I got a different answer. My work is as follows:
$$\iint_S \vec F\cdot\hat n\, dS = \iiint_D \nabla\cdot\vec F\,dV$$
$$= \iiint_D \frac{\partial y}{\partial y}\, dV$$
$$=\iiint_D \, dV=V_{cylinder}=\pi a^2h$$
Where did I go wrong?
 
Physics news on Phys.org
Is the surface integral over a closed surface?
 
  • Like
Likes   Reactions: Leo Liu
Your volume is only half the cylinder.
(After you add the surfaces needed to close the surface where the integral contribution is zero.)
 
  • Like
Likes   Reactions: PeroK and Leo Liu
PeroK said:
Is the surface integral over a closed surface?
Nope. But IMO the flux passing through the led of the cylinder and the bottom of the cylinder is 0.
 
Orodruin said:
Your volume is only half the cylinder.
(After you add the surfaces needed to close the surface where the integral contribution is zero.)
I am sorry. I don't get it.
 
Leo Liu said:
I am sorry. I don't get it.
You've calculated the volume of the whole cylinder. The surface is only half the cylinder.
 
PeroK said:
You've calculated the volume of the whole cylinder. The surface is only half the cylinder.
1626972009451.png

Okay I see -- I didn't see the condition that the surface is to the right of the xz plane. Thank you! :smile:
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
8
Views
3K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
6
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K