Finding Volume with Washer Method for y=x^2+1, x=0, y=1, and y=2

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Homework Help Overview

The discussion revolves around finding the volume of revolution about the x-axis for the region bounded by the curve y=x^2+1, the line x=0, and the horizontal lines y=1 and y=2 using the washer method.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the setup of the integral using the washer method, questioning the inner and outer radii based on the given boundaries.
  • Some participants discuss the need to split the integral into two parts due to the nature of the curves and intersections involved.
  • There is a consideration of whether the inner radius can be zero when integrating over the specified region.

Discussion Status

The discussion is active, with participants providing insights into how to approach the problem. Some guidance has been offered regarding the potential need to break the problem into two integrals, but no consensus has been reached on the exact setup.

Contextual Notes

Participants note that the problem does not require the final volume but rather an understanding of the method and setup for the integral.

madah12
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Homework Statement



is it possible to find the of revolution about x-axis volume of x=y^2+1 ,x=0 between y=1 and y=2
using washer method?

Homework Equations


V=pi integral from a to b R^2-r^2 thickness

The Attempt at a Solution



the problem is when I made the integral i had two inner radii first the line x =1 till the point x=1 because the parabola only starts there so I am confused do I just sum two integrals?
Edit I meant the first radius is line y =1 not x =1 sorry
 
Last edited:
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I don't remember the washer method exactly, but since this area goes right up to the y-axis, isn't the inner radius zero? In other words, can't you just integrate over *cylinders* of infinitesimal thickness dy and radius x(y) from y=1 to y=2?
 
but I am rotating around the x-axis not the y one and it is from y=1 to y=2
 
also the question doesn't want the volume itself it just wants to know which method is possible to use and to setup the integral for them.
 
The "washer" method is exactly the same doing the disc method, using the two boundaries as radii and then subtracting. It looks to me like it would be simplest to break this into to parts.

First, y= 1 and x= y^2+ 1 intersect at x= 2 so for x from 0 to 2, you just have the difference of two cylinders, of raidii 1 and 2, and height 2. That would be the same as
\pi\int_{x=0}^2 (4-1)dx

y= 2 and x= y^2+ 1 intersect at x= 5 so from x= 2 to x= 5, you have an upper bound of y= 2 and a lower bound of y= \sqrt{x- 1}. The volume of that, rotated around the x-axis will be
\pi\int_{x=2}^5 (4- (x-1)) dx= \pi\int_{x=2}^5 4 dx- \pi\int_{x=2}^5 (x-1)dx[/itex]
 
Oh. Okay, so maybe you do have to split it up into two integrals. From x = 0 to the value of x for which y = 1 intersects x(y) (i.e. x(1)) , the inner radius is constant at 1, and the outer radius is constant at 2. From x = x(1) to x = x(2), the inner radius is given by x(y) and the outer radius is still constant and equal to 2. Do you agree?

EDIT: Beaten by HallsofIvy!
 
oh I see thanks I never got a problem when we had to add volumes before so it kind of confused me .
 

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