Calculate Volume of Region Rotated About x-axis: x^2 + (y-1)^2 = 1

In summary, the conversation is about finding the volume of a solid formed by rotating a region bounded by given curves around a specific axis. The methods discussed include using discs or shells, and there is a debate over whether to use the Theorem of Pappus or integration. The final answer is determined to be 2π^2.
  • #1
sushifan
26
0
1. The region bounded by the given curves is rotated about the specific axis. Find the volume of the resulting solid by any method (disc or shell).



2. x^2 + (y-1)^2 = 1



3. This is my first time posting up a homework question, so I apologize if I didn't get the notation down correctly or something.

I tried shells and ended up with ∏ as my answer. (I hope that's a pi symbol.)

V= 2(2∏radius)(width) I doubled the circumference to get the whole circle rather than just having the semi-circle.

V = 4∏ ∫y sqrt[ 1- (y-1)^2] dy on [0,2]
 
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  • #2


The two dimensional figure is definitely a circle which passes through the origin. The rotation about x-axis would make it a horn torus (that's a funny name actually).

Using the formula that I remember (it's derived using integration, plus looks obvious if you see)
V = (π(r^2))(2πR)

Now putting r=1 and R=(1/2)

V=(π(1))(π)

So my answer would be π^2

I might me miserably wrong though.
 
  • #3


sushifan said:
1. The region bounded by the given curves is rotated about the specific axis. Find the volume of the resulting solid by any method (disc or shell).



2. x^2 + (y-1)^2 = 1



3. This is my first time posting up a homework question, so I apologize if I didn't get the notation down correctly or something.

I tried shells and ended up with ∏ as my answer. (I hope that's a pi symbol.)

V= 2(2∏radius)(width) I doubled the circumference to get the whole circle rather than just having the semi-circle.
You doubled the width from x = 0 to x = +√(1 - (y - 1)2).

Also, your volume formula needs something for the thickness of the shell, which in this case is Δy.
sushifan said:
V = 4∏ ∫y sqrt[ 1- (y-1)^2] dy on [0,2]
 
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  • #4


I got [0,2] in regards to the y-axis, not the x-axis, which is [0,1]. My mistake, sorry!

I'll retry what I did but on [0,1] this time.
 
  • #5


No, [0, 2] is the right interval along the y-axis. I took that comment out right after I posted it, but I was a bit too slow.
 
  • #6


Mark44 said:
No, [0, 2] is the right interval along the y-axis. I took that comment out right after I posted it, but I was a bit too slow.

Oh, okay. I'm not sure what action I should take now. But was my mistake the doubling part?

Actually, when I integrate there's one part in which I do substitution and the upper limit and lower limit became the same, so that part of my integral equaled 0. I thought that maybe that accounted for the "empty space". Maybe I should type out my whole solution...I'll do that now.
 
  • #7


Yeah, show what you did. The integral is not amenable to an ordinary substitution, I don't believe.

BTW, wolfram alpha says the value is 2π2, so I'm not sure I trust cng99's formula.
 
  • #8


V = 4∏ ∫ y sqrt[1-(y-1)^2] dy on [0,2]

Substitution: u = y - 1 → du = dy

V = 4∏ ∫ (u+1) sqrt(1-u^2) du on [-1,1]

V = 4∏ ∫u sqrt(1-u^2) du + 4∏ ∫ sqrt(1-u^2) du on [-1,1]

The left integral:

Substitution: t = 1-u^2 → dt = -2du

V = -2∏ ∫ sqrt(t) dt on [0,0] so that just equals 0

The right integral:

Substitution: u = sin∅ → du = cos∅ d∅

V = 4∏ ∫ sqrt[1-sin^2∅] cos∅ d∅ on [arcsin(-1),arcsin(1)]

V = 4∏ ∫ cos^2∅ d∅ on [arcsin(-1),arcsin(1)]

V = 4∏ ∫ 1/2 [1+cos2∅] d∅ on [arcsin(-1),arcsin(1)]

V = 4∏ [ 1/2∅ + 1/4 sin2∅] on [arcsin(-1),arcsin(1)]

V = 4∏ [ 1/2 * ∏/2 + 1/4 * 0] - 4∏ [1/2 * 3∏/2 + 1/4 * 0 ]

V = 4∏ [ ∏/4 - 3∏/4] (Whoops, I just realized my arithmetic mistakes!)

V = -2∏

Okay...so sorry about that. ∏ was not my original answer! And this is definitely wrong since it's negative...
 
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  • #9


Mark44 said:
Yeah, show what you did. The integral is not amenable to an ordinary substitution, I don't believe.

BTW, wolfram alpha says the value is 2π2, so I'm not sure I trust cng99's formula.

Yeah, I don't understand it at all.

And ohhh, so π is pi!
 
  • #10


So...does anyone have advice for my solution?
 
  • #11


What's the answer given in the book?
 
  • #12


cng99 said:
What's the answer given in the book?

There's no answer provided in the book since this is an even numbered problem. But someone on a different forum told me the answer should be 2π^2. He used Theorem of Pappus, though, and I'm limited to discs and shells.
 
  • #14


cng99 said:
Of course!
It's 2π^2

Volume = (2πR)(πr^2)

http://whistleralley.com/torus/torus.htm

And in this case R=1, r=1.

Why do you want to do integration?

The problem's instructions says that I can only use discs or shells, so I have to use integration. Pappus is off limits, especially since I never learned it.
 
  • #15


Bump...
 
  • #16


sushifan said:
V = 4∏ ∫ y sqrt[1-(y-1)^2] dy on [0,2]

Substitution: u = y - 1 → du = dy

V = 4∏ ∫ (u+1) sqrt(1-u^2) du on [-1,1]

V = 4∏ ∫u sqrt(1-u^2) du + 4∏ ∫ sqrt(1-u^2) du on [-1,1]

The left integral:

Substitution: t = 1-u^2 → dt = -2du

V = -2∏ ∫ sqrt(t) dt on [0,0] so that just equals 0

The right integral:

Substitution: u = sin∅ → du = cos∅ d∅

V = 4∏ ∫ sqrt[1-sin^2∅] cos∅ d∅ on [arcsin(-1),arcsin(1)]

V = 4∏ ∫ cos^2∅ d∅ on [arcsin(-1),arcsin(1)]

V = 4∏ ∫ 1/2 [1+cos2∅] d∅ on [arcsin(-1),arcsin(1)]

V = 4∏ [ 1/2∅ + 1/4 sin2∅] on [arcsin(-1),arcsin(1)]

V = 4∏ [ 1/2 * ∏/2 + 1/4 * 0] - 4∏ [1/2 * 3∏/2 + 1/4 * 0 ]

V = 4∏ [ ∏/4 - 3∏/4] (Whoops, I just realized my arithmetic mistakes!)

V = -2∏

Okay...so sorry about that. ∏ was not my original answer! And this is definitely wrong since it's negative...
I don't have time to go through this right now, but I don't see anything obviously wrong in your approach. One suggestion is to just work with indefinite integrals (no limits of integration), and then evaluate your final antiderivative at the original limits of integration, 0 and 2. It could be that a small error in switching several times to different limits of integration might have caused your error.

BTW, bumping a thread before 24 hours have passed is not permitted here at PF.
 
  • #17


Why don't you just use the disk method (well washer I guess) on the top portion of the arc.
For example:

X2+(y-1)2=1 Becomes:

y=+/-sqrt(1-x2) + 1

We just revolve the top arc around y=1 and multiply by 2.

So:

2[itex]\pi[/itex]∫[itex]^{1}_{-1}[/itex] R(x)2 - r(x)2dx

R(x) = sqrt(1-x2) + 1
r(x) = 1 (revolving about y=1)

R(x)2-r(x)2 = (1-x2) + 2 sqrt(1-x2)
So now we have:

2[itex]\pi[/itex]∫[itex]^{1}_{-1}[/itex] (1-x2) + 2sqrt(1-x2)dx

Splitting that into two integrals you get 8[itex]\pi[/itex]/3 for the first and
the next integral 2[itex]\pi[/itex]∫[itex]^{1}_{-1}[/itex] 2sqrt(1-x2)dx needs to be done with trig substitution...
 
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  • #18


USN2ENG said:
Why don't you just use the disk method (well washer I guess) on the top portion of the arc.
For example:

X2+(y-1)2=1 Becomes:

y=+/-sqrt(1-x2) + 1

We just revolve the top arc around y=1 and multiply by 2.

So:

2[itex]\pi[/itex]∫[itex]^{1}_{-1}[/itex] R(x)2 - r(x)2dx

R(x) = sqrt(1-x2) + 1
r(x) = 1 (revolving about y=1)

R(x)2-r(x)2 = (1-x2) + 2 sqrt(1-x2)
So now we have:

2[itex]\pi[/itex]∫[itex]^{1}_{-1}[/itex] (1-x2) + 2sqrt(1-x2)dx

Splitting that into two integrals you get 8[itex]\pi[/itex]/3 for the first and
the next integral 2[itex]\pi[/itex]∫[itex]^{1}_{-1}[/itex] 2sqrt(1-x2)dx needs to be done with trig substitution...

For the second integral, I get 2π^2 for my answer...but then I have 8π/3 for the first integral...

Oh, and sorry about the bump.
 
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  • #19


I took your advice and solved the indefinite integral and then I computed the original limits of integration. It came out to 2pi^2! Thanks a lot!
 
  • #20


There's nothing wrong with doing all the integrals as definite integrals, and transforming the limits of integration according to the substitutions being used. It just seems more prone to getting fouled up by arithmetic errors is all.
 
  • #21


sushifan said:
V = 4∏ ∫ y sqrt[1-(y-1)^2] dy on [0,2]

Substitution: u = y - 1 → du = dy

V = 4∏ ∫ (u+1) sqrt(1-u^2) du on [-1,1]

V = 4∏ ∫u sqrt(1-u^2) du + 4∏ ∫ sqrt(1-u^2) du on [-1,1]

The left integral:

Substitution: t = 1-u^2 → dt = -2du
...
If t = 1-u2, then dt = -2u du .

That may still give a zero when you do the integration.
 
  • #22


SammyS said:
If t = 1-u2, then dt = -2u du .

That may still give a zero when you do the integration.

Yeah, that was just a typo. I already solved the problem, though.
 

Related to Calculate Volume of Region Rotated About x-axis: x^2 + (y-1)^2 = 1

1. What is the formula for calculating the volume of a region rotated about the x-axis?

The formula for calculating the volume of a region rotated about the x-axis is ∫[a,b] π[f(x)]^2 dx, where a and b are the limits of the region and f(x) is the function defining the boundary of the region.

2. How do I determine the limits of integration for this problem?

The limits of integration can be determined by finding the x-values where the boundary of the region intersects the x-axis. In this case, the circle x^2 + (y-1)^2 = 1 intersects the x-axis at x = -1 and x = 1. Therefore, the limits of integration are -1 and 1.

3. Is it necessary to use calculus to solve this problem?

Yes, calculus is necessary to solve this problem as it involves finding the volume of a three-dimensional shape by using integration.

4. Can you provide an example of calculating the volume of a region rotated about the x-axis?

For the region defined by the equation x^2 + (y-1)^2 = 1, the volume can be calculated by first rewriting the equation in terms of y to get y = 1 ± √(1-x^2). Then, using the formula ∫[a,b] π[f(x)]^2 dx, we can set the limits of integration to -1 and 1 and plug in the function (1 ± √(1-x^2))^2. This results in a volume of 4π/3 units^3.

5. Can this formula be applied to any shape rotated about the x-axis?

Yes, this formula can be applied to any shape rotated about the x-axis. However, the shape must be defined in terms of x and y, and the limits of integration must be determined based on the points of intersection with the x-axis.

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