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
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:
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top