Volume of the Solid bounds and integral

Click For Summary

Homework Help Overview

The discussion revolves around finding the volume of a solid formed by rotating a region bounded by the curves y = e^(-2x^2), y = 0, x = 0, and x = 1 about the y-axis. Participants are exploring the appropriate bounds and methods for setting up the integral.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are questioning whether to take bounds from the y-axis or the x-axis and discussing the setup of the integral for volume calculation. There is exploration of using discs versus cylinders for the volume calculation, with some uncertainty about the correct formulation of the integral and the definition of radius in this context.

Discussion Status

The discussion is active, with participants raising questions about the setup of the integral and the interpretation of the volume elements. Some guidance has been provided regarding the use of cylinders and the need to clarify the definitions of height, radius, and thickness in the context of the problem.

Contextual Notes

Participants are navigating the complexities of integrating with respect to different axes and are considering the implications of their choices on the volume calculation. There is a lack of consensus on the best approach and the definitions involved in the setup.

alexs2jennisha
Messages
14
Reaction score
0
Find the volume of the solid obtained by rotating about the y-axis the region bounded by the curves y= e-2x^2, y=0, x=0, x=1.

Should the bounds for the problem be taken from the y-axis or the x-axis?

I think that the integral for this problem would be:

∏∫(e-2x^2)dx , is this correct
 
Last edited:
Physics news on Phys.org
alexs2jennisha said:
∏∫(e-2x^2)dx , is this correct
That would be the same as ∏∫ydx, which is clearly wrong.
You need to decide how you want to carve up the volume. You could do it in discs centred on the y axis, but that gets a bit messy because the integral falls into two parts (and it will involve logs). More natural is to carve it into cylinders centred on the y axis.
 
Carving it into cylinders still uses the ∏∫R^2dx formula, right? and then my bounds would be on the y axis?
 
alexs2jennisha said:
Carving it into cylinders still uses the ∏∫R^2dx formula, right?
Yes, but what is R in this case?
then my bounds would be on the y axis?
No. Each cylinder is centred on the y axis. What are its height, radius and thickness?
 
would r be the formula but solved for x?
 
alexs2jennisha said:
would r be the formula but solved for x?

I don't know what you mean by that question. r is not much of a formula.
If you take a slice from y = 0 to y = f(x), and from x to x+dx, in the xy plane, then rotate it around the y axis, what do you get? What is its volume? What integral would you write to add up all these volumes between x = 0 and x = 1?
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
5
Views
2K
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K