Find Vol of Rotated Region R: y=sqrt x, y=sqrt(2x-1), y=0

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Homework Help Overview

The problem involves finding the volume of a solid obtained by rotating a region bounded by the curves y = sqrt(x), y = sqrt(2x-1), and y = 0 around the x-axis. The curves intersect at a single point, which adds complexity to the problem.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to calculate the volume by considering the rotation of each curve separately and then subtracting the volumes. They express uncertainty about their strategy. Other participants suggest alternative methods, such as using the shell method, and inquire about the original poster's familiarity with these concepts.

Discussion Status

Contextual Notes

Participants note that the original poster has not learned the shell method, which may limit their exploration of alternative solutions. The problem's complexity arises from the unique intersection of the curves.

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Homework Statement



Find the Volume of the solid obtained by rotating the region R, that is bounded by the graphs y = sqrt x, and y = sqrt (2x-1) and y = 0, about the x-axis



Homework Equations



The 3 equations
y= sqrt x
y = sqrt (2x-1)
y = 0


The Attempt at a Solution



Well, the tricky part of this problem is that the 2 curves, x^(1/2) and (2x-1)^(1/2) intersect at just 1 point, which is 1. So it's not your usual problem. Nonetheless, it is still a closed region.

So here is what I decided to do

find the volume of the region sqrt(x) rotated from 0 to 1 around the x-axis. Then find the volume of the region sqrt(2x-1) rotated from 1/2 (because that is when it is equal y=0) and 1, and then subtract the first and bigger volume from the smaller one.

I ended up with pi/4.

Good strategy?
 
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Welcome to PF!

Hi ssk13809! Welcome to PF! :smile:

(have a pi: π and a square-root: √ :wink:)
ssk13809 said:
Find the Volume of the solid obtained by rotating the region R, that is bounded by the graphs y = sqrt x, and y = sqrt (2x-1) and y = 0, about the x-axis

So here is what I decided to do

find the volume of the region sqrt(x) rotated from 0 to 1 around the x-axis. Then find the volume of the region sqrt(2x-1) rotated from 1/2 (because that is when it is equal y=0) and 1, and then subtract the first and bigger volume from the smaller one.

I ended up with pi/4.

Good strategy?

Perfect! :biggrin:

(though i haven't checked the answer)

(an alternative, if you want to have just one integral, but with both limits variable, would be to slice it into horizontal cylindrical shells of thickness dy …

do you want to to see whether that gives the same result? :wink:)
 
Thanks for the feedback!

I never learned the shell method or the cylindrical method, so I would be curious to see how that works.
 
ok, using two circular cookie-cutters (or napkin-rings, if you're posh :wink:), cut a slice of thickness dx … that will be a cylindrical shell.

Its volume will be 2π times its radius times its length times dx. :smile:
 

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