Find the volume of the region bounded by x^2+(y-1)^2 = 1 rotated about the x-axis.


by sushifan
Tags: volume circle
sushifan
sushifan is offline
#1
Mar14-12, 11:40 PM
P: 27
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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cng99
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#2
Mar15-12, 12:20 AM
P: 44
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.
Mark44
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#3
Mar15-12, 12:40 AM
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Quote Quote by sushifan View Post
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.
Quote Quote by sushifan View Post

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

sushifan
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#4
Mar15-12, 12:44 AM
P: 27

Find the volume of the region bounded by x^2+(y-1)^2 = 1 rotated about the x-axis.


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.
Mark44
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#5
Mar15-12, 12:47 AM
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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.
sushifan
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#6
Mar15-12, 12:53 AM
P: 27
Quote Quote by Mark44 View Post
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.
Mark44
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#7
Mar15-12, 01:02 AM
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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.
sushifan
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#8
Mar15-12, 01:10 AM
P: 27
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....
sushifan
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#9
Mar15-12, 01:15 AM
P: 27
Quote Quote by Mark44 View Post
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!
sushifan
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#10
Mar15-12, 01:28 AM
P: 27
So...does anyone have advice for my solution?
cng99
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#11
Mar15-12, 01:59 AM
P: 44
What's the answer given in the book?
sushifan
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#12
Mar15-12, 02:19 AM
P: 27
Quote Quote by cng99 View Post
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.
cng99
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#13
Mar15-12, 02:36 AM
P: 44
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 wanna do integration?
sushifan
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#14
Mar15-12, 02:48 AM
P: 27
Quote Quote by cng99 View Post
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 wanna 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.
sushifan
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#15
Mar15-12, 03:54 AM
P: 27
Bump......
Mark44
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#16
Mar15-12, 08:40 AM
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Quote Quote by sushifan View Post
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.
USN2ENG
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#17
Mar15-12, 09:16 AM
P: 112
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...
sushifan
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#18
Mar15-12, 11:05 AM
P: 27
Quote Quote by USN2ENG View Post
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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