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Find the volume of the region bounded by x^2+(y1)^2 = 1 rotated about the xaxis.by sushifan
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#1
Mar1412, 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 + (y1)^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 semicircle. V = 4∏ ∫y sqrt[ 1 (y1)^2] dy on [0,2] 


#2
Mar1512, 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. 


#3
Mar1512, 12:40 AM

Mentor
P: 21,305

Also, your volume formula needs something for the thickness of the shell, which in this case is Δy. 


#4
Mar1512, 12:44 AM

P: 27

Find the volume of the region bounded by x^2+(y1)^2 = 1 rotated about the xaxis.
I got [0,2] in regards to the yaxis, not the xaxis, which is [0,1]. My mistake, sorry!
I'll retry what I did but on [0,1] this time. 


#5
Mar1512, 12:47 AM

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P: 21,305

No, [0, 2] is the right interval along the yaxis. I took that comment out right after I posted it, but I was a bit too slow.



#6
Mar1512, 12:53 AM

P: 27

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
Mar1512, 01:02 AM

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P: 21,305

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 2π^{2}, so I'm not sure I trust cng99's formula. 


#8
Mar1512, 01:10 AM

P: 27

V = 4∏ ∫ y sqrt[1(y1)^2] dy on [0,2]
Substitution: u = y  1 → du = dy V = 4∏ ∫ (u+1) sqrt(1u^2) du on [1,1] V = 4∏ ∫u sqrt(1u^2) du + 4∏ ∫ sqrt(1u^2) du on [1,1] The left integral: Substitution: t = 1u^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[1sin^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.... 


#9
Mar1512, 01:15 AM

P: 27

And ohhh, so π is pi! 


#10
Mar1512, 01:28 AM

P: 27

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



#11
Mar1512, 01:59 AM

P: 44

What's the answer given in the book?



#12
Mar1512, 02:19 AM

P: 27




#13
Mar1512, 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? 


#14
Mar1512, 02:48 AM

P: 27




#15
Mar1512, 03:54 AM

P: 27

Bump......



#16
Mar1512, 08:40 AM

Mentor
P: 21,305

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


#17
Mar1512, 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: X^{2}+(y1)^{2}=1 Becomes: y=+/sqrt(1x^{2}) + 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)^{2}dx R(x) = sqrt(1x^{2}) + 1 r(x) = 1 (revolving about y=1) R(x)^{2}r(x)^{2} = (1x^{2}) + 2 sqrt(1x^{2}) So now we have: 2[itex]\pi[/itex]∫[itex]^{1}_{1}[/itex] (1x^{2}) + 2sqrt(1x^{2})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(1x^{2})dx needs to be done with trig substitution... 


#18
Mar1512, 11:05 AM

P: 27

Oh, and sorry about the bump. 


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