# Finding the volume of a Torus

1. Feb 22, 2017

### Jovy

1. The problem statement, all variables and given/known data

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. Relevant equations

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

3. The attempt at a solution

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.

2. Feb 22, 2017

### 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.

3. Feb 22, 2017

### Jovy

How do you determine which limits of integration to use?

4. Feb 22, 2017

### haruspex

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.

5. Feb 24, 2017

### Jovy

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?

6. Feb 24, 2017

### haruspex

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?

7. Feb 24, 2017

### Jovy

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

8. Feb 24, 2017

### haruspex

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?

9. Feb 24, 2017

### Jovy

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?

10. Feb 24, 2017

### haruspex

No, still the wrong radius.
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?

11. Feb 25, 2017

### Jovy

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.​

12. Feb 25, 2017

### LCKurtz

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

13. Feb 25, 2017

### haruspex

No, it does need to be as functions of y. They represent two different widths. One is the width of the lamina being rotated (in this case a circle), measured at height y from the x axis. I believe this is what h(y) represents in your formula.
The other is a bit harder to describe.
Consider a thin horizontal slice at height y through the lamina. This has a (horizontal) midpoint. When the lamina is rotated about the axis, that midpoint moves around in a horizontal circle about that axis. That circle has a radius, represented in your formula by p(y).

Now, rather than just learning how to use the formula correctly, it would help you greatly if you were to understand why the formula is correct.
Suppose the horizontal slice has thickness dy. Rotating it about the axis produces a horizontal annulus. The circular hole in the centre has radius p(y)-h(y)/2, while the outer radius of the annulus is p(y)+h(y)/2. (Draw yourself a diagram to check this.). The area of the annulus is therefore π((p+h/2)2-(p-h/2)2)=2πph.
Since the thickness of the annulus is dy, its volume is 2πph.dy. Integrating that gives the volume of rotation.

Note that this works for any lamina rotated around an axis outside of itself and parallel to the y axis. (If the axis of rotation were parallel to the x axis then you would want p(x)h(x).) In the case of the torus, p(y) turns out to be constant.