Solids of Revolution around y = x

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It is possible to revolve a function around the line y = x by changing to new coordinates p = x + y and q = x - y. This transformation allows for the rotation about the q axis, simplifying the integration process. For example, when rotating y = x^2 around y = x, one can substitute x and y in terms of p and q to find the volume. The distance from the axis, originally x = y, is represented as r, which relates to the new coordinates. After solving the problem in p and q, the final solution can be converted back to the original x and y coordinates.
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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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Hi TheAbsoluTurk! :smile:

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. :wink:
 
tiny-tim said:
Hi TheAbsoluTurk! :smile:

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. :wink:

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 ?
 
Last edited:
Easier is to substitute x = (p+q)/2, y = (p-q)/2 :wink:
 
tiny-tim said:
Easier is to substitute x = (p+q)/2, y = (p-q)/2 :wink:

Do you know of any YouTube videos or articles on the internet which show how to do this?
 
uhh? :confused:

just do it … substitute those formulas into y = x2 !​
 
tiny-tim said:
uhh? :confused:

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?
 
TheAbsoluTurk said:
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?)
 
tiny-tim said:
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?
 
  • #10
TheAbsoluTurk said:
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?
 
  • #11
(just got up :zzz:)

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 :smile:
 
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