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

sushifan

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]

cng99

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

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

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

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

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

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

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

Yeah, I don't understand it at all.

And ohhh, so π is pi!

sushifan

So...does anyone have advice for my solution?

cng99

What's the answer given in the book?

sushifan

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

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

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

Bump......

Mark44

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

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...