Solids of Revolution around y = x


by TheAbsoluTurk
Tags: revolution, solids
TheAbsoluTurk
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#1
Feb5-13, 04:10 PM
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Is it possible to revolve a function around y = x? If so how would you do it?

I suppose the main difficulty is in finding the radius for the area of a disk or cylinder. Is there any method that works will all or most functions?
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tiny-tim
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Feb5-13, 04:52 PM
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Hi TheAbsoluTurk!

Easiest way is to change to new coordinates p = x + y, q = x - y (or the same but divided by √2, if you prefer).

Then x = y is the q axis, so that's just a rotation about the q axis.
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#3
Feb5-13, 05:25 PM
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Quote Quote by tiny-tim View Post
Hi TheAbsoluTurk!

Easiest way is to change to new coordinates p = x + y, q = x - y (or the same but divided by √2, if you prefer).

Then x = y is the q axis, so that's just a rotation about the q axis.
Ok, let's say that I'm trying to rotate y = x^2 around y = x.

p = x + y

q = x - y

So do I have to insert (q + y) into x to make y = (q + y)^2 ?

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Feb5-13, 05:35 PM
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Solids of Revolution around y = x


Easier is to substitute x = (p+q)/2, y = (p-q)/2
TheAbsoluTurk
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#5
Feb5-13, 05:44 PM
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Quote Quote by tiny-tim View Post
Easier is to substitute x = (p+q)/2, y = (p-q)/2
Do you know of any YouTube videos or articles on the internet which show how to do this?
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#6
Feb5-13, 05:49 PM
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uhh?
just do it substitute those formulas into y = x2 !
TheAbsoluTurk
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#7
Feb5-13, 05:53 PM
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Quote Quote by tiny-tim View Post
uhh?
just do it substitute those formulas into y = x2 !
I understand that but I don't know what to do after that. Does r in ∏r^2 equal (p-q)/2? How do you integrate that?
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Feb5-13, 06:01 PM
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Quote Quote by TheAbsoluTurk View Post
I understand that but I don't know what to do after that. Does r in ∏r^2 equal (p-q)/2? How do you integrate that?
no, the r is the distance from your axis

your axis (originally called x=y) is the q axis, so r is the distance from the q axis, which is p (or is it p/2?)
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#9
Feb5-13, 06:26 PM
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Quote Quote by tiny-tim View Post
no, the r is the distance from your axis

your axis (originally called x=y) is the q axis, so r is the distance from the q axis, which is p (or is it p/2?)
Let me get this straight:

What is the volume of y = x^2 rotated about y = x?

Define p = x +y

Define q = x - y

I don't understand why you chose to insert x = (p+q)/2 and y = (p-q)/2 ? How did you get these?
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#10
Feb5-13, 08:17 PM
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Quote Quote by TheAbsoluTurk View Post
Let me get this straight:

What is the volume of y = x^2 rotated about y = x?

Define p = x +y

Define q = x - y

I don't understand why you chose to insert x = (p+q)/2 and y = (p-q)/2 ? How did you get these?
Ok, I understand how you got those expressions. But what's to do next? Do you solve for p?
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#11
Feb6-13, 02:04 AM
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(just got up )

first you convert everything into p and q

then you solve the problem, in p and q (you've said you know how to do this)
finally you convert your solution back to x and y


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