Uncertain about volume of bounded region question

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    Bounded Volume
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Homework Help Overview

The problem involves finding the volume of a solid formed by rotating a region bounded by the curves y=lnX, y=1, y=2, and x=0 about the y-axis. The original poster expresses uncertainty about the specific region to consider for the volume calculation and seeks clarification on setting up the integration.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to clarify the region for integration and has provided a graph for reference. Some participants discuss the correct orientation of the disks used for volume calculation, indicating a need to integrate with respect to y.

Discussion Status

The discussion is ongoing, with participants providing guidance on the setup of the integration problem. There is a focus on ensuring the correct interpretation of the region and the method of integration, but no consensus has been reached yet.

Contextual Notes

Participants note the importance of correctly identifying the bounds of the region and the orientation of the disks for the volume calculation. The original poster has indicated some understanding of integration but is seeking further clarification on the setup.

warfreak131
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Homework Statement



The question states:

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line.

The lines are, y=lnX, y = 1, y = 2, and x = 0, rotated about the y-axis.

I know how to integrate it, I just don't exactly know which region I'm taking the volume of, so I am having trouble setting up the integration problem.

I have provided a graph of the curves.

http://img691.imageshack.us/img691/2071/boundregion.png

EDIT:

I think I got it, here's my attempt at the solution:

I'm going to use the area of circles method going vertical. So I find the distance from the y-axis to the lnX curve, which is equal to e^y. From there I do \int_{0}^{2}\pi(e^{2y})dy -\int_{0}^{1}\pi(e^{2y})dy
 
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The region being revolved is bounded below by the line y = 1, above by the line y = 2, on the right by y = ln x, and on the left by the line x = 0. On your graph, this region has a roughly trapezoidal shape, and lies between the purple line and the light brown line. You first integral (it should include dy) gives you the volume of the rotated region.

BTW, you are integrating by using disks, not circles.
 
And the disks are horizontal, not vertical.
 
HallsofIvy said:
And the disks are horizontal, not vertical.

I meant that the disks are stacked on top of each other vertically.
 

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