# Finding the volume of a Torus

Jovy

## Homework Statement

A torus is formed by revolving the region bounded by the circle ##x^2+ y^2= 1## about the line x =2 Find the volume of this “doughnut-shaped” solid.
(Hint: The integral ##{\int_{-1}^1} \sqrt{1 -x^2} dx## represents the area of a semicircle.)

2. Homework Equations

$$Volume=2\pi\int_a^b p(y)h(y)dy$$

## The Attempt at a Solution

[/B]
I know that h(y) or the width is ##\sqrt{1 -x^2}##
The radius or p(y) is (2-x).
However, when I plug that into the equation and integrate, I get 0 and I know that's wrong.

Monci
Which limits of integration are you using? Aren't h(y) and p(y) greater than zero if |x| < 1? The integral should be positive.

Jovy
Which limits of integration are you using? Aren't h(y) and p(y) greater than zero if |x| < 1? The integral should be positive.

How do you determine which limits of integration to use?

Homework Helper
Gold Member
2022 Award

## Homework Equations

$$Volume=2\pi\int_a^b p(y)h(y)dy$$
Equations mean nothing out of context. How are y, p, h defined here in relation to the volume to be found?
I know that h(y) or the width is ##\sqrt{1 -x^2}##
h(y) is a function of y. What is the width as a function of y?
The radius or p(y) is (2-x).
The radius of what, exactly? Again, you need a function of y.

Jovy
Equations mean nothing out of context. How are y, p, h defined here in relation to the volume to be found?

h(y) is a function of y. What is the width as a function of y?

The radius of what, exactly? Again, you need a function of y.
I'm not sure I understand. The radius is a function of the circle, ##x^2+y^2=1##, the same goes for the width?

Homework Helper
Gold Member
2022 Award
I'm not sure I understand. The radius is a function of the circle, ##x^2+y^2=1##, the same goes for the width?
A torus has two radii, one for the circular cross-section and one for the circle formed by the locus of centres of the circular cross-sections. Which one does p() represent in that formula?

Jovy
A torus has two radii, one for the circular cross-section and one for the circle formed by the locus of centres of the circular cross-sections. Which one does p() represent in that formula?

I think p(y) represents the circular cross section.

Homework Helper
Gold Member
2022 Award
I think p(y) represents the circular cross section.
Then you have not understood the formula you quoted.
Consider some function f(x) over some range a to b, 0<a<b, with f(a)=f(b)=0. Taken together with the x-axis from a to b this encloses an area.
We can form a volume by rotating this area about the y axis.
Consider a vertical strip of the area from x to x+dx. This has area f(x).dx. Rotated about the y axis, it travels along a path length 2πx, so carves out a volume 2πf(x)x.dx. Integrating that over the range a to b calculates the whole volume.
In your formula, h is the (vertical) width, so corresponds to my f(x). The other factor in my integrand is x, the radius of the rotation. What is the radius of rotation in your problem?

Jovy
Then you have not understood the formula you quoted.
Consider some function f(x) over some range a to b, 0<a<b, with f(a)=f(b)=0. Taken together with the x-axis from a to b this encloses an area.
We can form a volume by rotating this area about the y axis.
Consider a vertical strip of the area from x to x+dx. This has area f(x).dx. Rotated about the y axis, it travels along a path length 2πx, so carves out a volume 2πf(x)x.dx. Integrating that over the range a to b calculates the whole volume.
In your formula, h is the (vertical) width, so corresponds to my f(x). The other factor in my integrand is x, the radius of the rotation. What is the radius of rotation in your problem?

well the circle is from -1 to 1 along the x-axis and -1 to 1 along the y-axis. so the radius would be 1?

Homework Helper
Gold Member
2022 Award
well the circle is from -1 to 1 along the x-axis and -1 to 1 along the y-axis. so the radius would be 1?
Consider the centre of that circle, i.e. the origin. As the circle revolves about the line x=2, what path is taken by its centre?

Jovy
Consider the centre of that circle, i.e. the origin. As the circle revolves about the line x=2, what path is taken by its centre?

I think I made a mistake, it shouldn't be p(y) or h(y), but p(x) and h(x). Also, I think I will understand it with a visual so I tried drawing the graph. The graph at the very top of the image is the torus. The graph below shows a semicircle rotating about the x-axis. A torus has two radii so I would need to insert another integral. The width of a semicircle is ##\sqrt{1-x^2}##. So the equation would look something like this, ##Volume=(something~\pi)\int_{-1}^1 p(x)\sqrt{1-x^2}dx-(some~number~\pi)\int_{-1}^1 p(x)\sqrt{1-x^2}dx##. Because it is a torus, the equation wouldn't have ##2\pi##, I don't necessarily understand why, I just know it's not ##2\pi##.
I do apologize, I bet this is becoming quite tedious, I just still don't understand how to find the two radii. When looking for the radius, do I just look at the part of the graph that is rotated. For example, in my image, would I look at the semicircle on the right of x=2?

Thank you for helping.​

Homework Helper
Gold Member
You are responding to what @haruspex asked in post #10. So why not answer his question?