## Solids of Revolution around y = x

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?
 Blog Entries: 27 Recognitions: Gold Member Homework Help Science Advisor 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.

 Quote by tiny-tim 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 ?

Blog Entries: 27
Recognitions:
Gold Member
Homework Help

## Solids of Revolution around y = x

Easier is to substitute x = (p+q)/2, y = (p-q)/2

 Quote by tiny-tim 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?
 Blog Entries: 27 Recognitions: Gold Member Homework Help Science Advisor uhh? just do it … substitute those formulas into y = x2 !

 Quote by tiny-tim 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?

Blog Entries: 27
Recognitions:
Gold Member
Homework Help
 Quote by TheAbsoluTurk 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?)

 Quote by tiny-tim 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?

 Quote by TheAbsoluTurk 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?
 Blog Entries: 27 Recognitions: Gold Member Homework Help Science Advisor (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