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Rotation Question

  • Thread starter ssk13809
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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?
 

Answers and Replies

  • #2
tiny-tim
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Welcome to PF!

Hi ssk13809! Welcome to PF! :smile:

(have a pi: π and a square-root: √ :wink:)
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:)
 
  • #3
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Thanks for the feedback!

I never learned the shell method or the cylindrical method, so I would be curious to see how that works.
 
  • #4
tiny-tim
Science Advisor
Homework Helper
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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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