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

  • Thread starter Thread starter ssk13809
  • Start date Start date
  • Tags Tags
    Rotation
Click For Summary
SUMMARY

The volume of the solid obtained by rotating the region R, bounded by the curves y = sqrt(x), y = sqrt(2x-1), and y = 0 about the x-axis, is calculated using integration techniques. The curves intersect at the point (1, 1), creating a closed region. The volume is determined by integrating the area of the region from 0 to 1 for y = sqrt(x) and from 1/2 to 1 for y = sqrt(2x-1), ultimately yielding a result of π/4. An alternative method using cylindrical shells is also suggested for further exploration.

PREREQUISITES
  • Understanding of integral calculus, specifically volume of revolution
  • Familiarity with the equations of curves and their intersections
  • Knowledge of the disk and washer methods for calculating volumes
  • Basic understanding of cylindrical shells method for volume calculation
NEXT STEPS
  • Learn the disk and washer methods for volume of revolution
  • Explore the cylindrical shells method for calculating volumes
  • Study the properties of the functions y = sqrt(x) and y = sqrt(2x-1)
  • Practice finding volumes of solids of revolution using different integration techniques
USEFUL FOR

Students studying calculus, particularly those focusing on volume calculations, educators teaching integral calculus, and anyone interested in solid geometry and applications of integration.

ssk13809
Messages
2
Reaction score
0

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?
 
Physics news on Phys.org
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:
 

Similar threads

Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K