# Solids of Revolution around y = x

• TheAbsoluTurk
In summary, to rotate a function around the line y = x, one can use the substitution p = x + y and q = x - y (or p = x + y/√2 and q = x - y/√2). Then, solve for p and q in terms of x and y, and use these expressions to solve the problem. Finally, convert the solution back to x and y to obtain the final answer.
TheAbsoluTurk
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?

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.

tiny-tim said:
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 ?

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

tiny-tim said:
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?

uhh?

just do it … substitute those formulas into y = x2 !​

tiny-tim said:
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?

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?

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?

(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

## 1. What is a solid of revolution around y = x?

A solid of revolution around y = x is a three-dimensional shape formed by rotating a two-dimensional shape around the line y = x. The resulting shape is symmetrical about the line y = x and has a circular cross-section.

## 2. What types of shapes can be used to create a solid of revolution around y = x?

Any shape that is symmetrical about the line y = x can be used to create a solid of revolution. This includes circles, squares, rectangles, triangles, and more complex shapes such as ellipses or parabolas.

## 3. How is the volume of a solid of revolution around y = x calculated?

The volume of a solid of revolution around y = x can be calculated using the formula V = π∫(x)^2 dy, where the limits of integration are the y-values of the shape's endpoints. This formula can be derived using the disk method or the shell method.

## 4. What is the significance of y = x in solids of revolution?

The line y = x is significant in solids of revolution because it serves as the axis of rotation. This means that when a shape is rotated around this line, the resulting solid will have a consistent cross-section and be symmetrical about the line.

## 5. Can a solid of revolution around y = x have a hole or hollow center?

Yes, a solid of revolution around y = x can have a hole or hollow center. This can occur if the original shape being rotated has a hole or if the shape is hollow to begin with. The resulting solid will still be symmetrical about the line y = x and will have a circular cross-section.

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