Efficiently Solve a Challenging Rotation Problem | Homework with Shell Method

Click For Summary

Homework Help Overview

The discussion revolves around a problem involving the shell method for calculating volumes of solids of revolution, specifically focusing on the integration limits and the visualization of the radius in relation to the function \( \sqrt{\sin(y)} \).

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the setup of the integral and the limits of integration, with one participant expressing uncertainty about visualizing the radius. There is also a focus on the interpretation of the function being integrated and its graphical representation.

Discussion Status

The conversation is ongoing, with participants providing feedback on the integral setup and questioning the limits of integration. Clarifications regarding the relationship between the variables in the function are being explored, indicating a productive exchange of ideas.

Contextual Notes

There is mention of potential confusion regarding the limits of integration and the correct interpretation of the function \( \sqrt{\sin(y)} \) as it relates to the integration variable. Participants are navigating these assumptions without reaching a definitive conclusion.

iRaid
Messages
558
Reaction score
8

Homework Statement


serocp.png



Homework Equations


Shell method


The Attempt at a Solution


Not sure if this is right, but the integral I set up is:
[tex]2\pi \int_0^1 (4-y)(\sqrt{siny})dy[/tex]


Finding the radius is my big problem with these problems, I can't visualize it very well.

Any help is appreciated.
 
Physics news on Phys.org
iRaid said:

Homework Statement


serocp.png


Homework Equations


Shell method

The Attempt at a Solution


Not sure if this is right, but the integral I set up is:
[tex]2\pi \int_0^1 (4-y)(\sqrt{siny})dy[/tex]
Finding the radius is my big problem with these problems, I can't visualize it very well.

Any help is appreciated.
That looks fine except for the limits of integration.

How did you get those limits?
 
graphing [tex]\sqrt{siny}[/tex] from 0 to pi gives me a curve that goes to 1 and back to 0. Since I'm integrating wrt y, I figured it's 0 to 1?
 
Last edited:
iRaid said:
graphing [tex]\sqrt{siny}[/tex] from 0 to pi gives me a curve that goes to 1 and back to 0. Since I'm integrating wrt y, I figured it's 0 to 1?
You don't have [itex]\ \ y=\sqrt{\sin(x)}\ \[/itex] with x going from 0 to π.

You have [itex]\ \ x=\sqrt{\sin(y)}\ \[/itex] with y going from 0 to π --- and your integration is w.r.t. y.
 
SammyS said:
You don't have [itex]\ \ y=\sqrt{\sin(x)}\ \[/itex] with x going from 0 to π.

You have [itex]\ \ x=\sqrt{\sin(y)}\ \[/itex] with y going from 0 to π --- and your integration is w.r.t. y.


Thanks, probably shouldn't agree with wolfram straight after I graph...
 

Similar threads

  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
4
Views
1K
Replies
3
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K