How Do You Calculate the Volume of a Torus Formed by Revolving a Circle?

In summary: 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 becomes Volume=2\pi\int_a^b p(x)h(x)dx.
  • #1
Jovy
17
2

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.
 
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  • #2
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
Monci said:
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?
 
  • #4
Jovy said:

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?
Jovy said:
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?
Jovy said:
The radius or p(y) is (2-x).
The radius of what, exactly? Again, you need a function of y.
 
  • #5
haruspex said:
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?
 
  • #6
Jovy said:
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?
 
  • #7
haruspex said:
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.
 
  • #8
Jovy said:
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?
 
  • #9
haruspex said:
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?
 
  • #10
Jovy said:
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?
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
haruspex said:
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?

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##.
7.3.jpg
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
You are responding to what @haruspex asked in post #10. So why not answer his question?
 
  • #13
Jovy said:
it shouldn't be p(y) or h(y)
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.
 

Related to How Do You Calculate the Volume of a Torus Formed by Revolving a Circle?

What is a Torus?

A Torus is a three-dimensional geometric shape that resembles a donut or inner tube. It is formed by revolving a circle around an axis that does not intersect the circle.

How do you find the volume of a Torus?

To find the volume of a Torus, you can use the formula V = 2π²Rr², where R is the distance from the center of the circle to the center of the torus and r is the radius of the circle.

Can the volume of a Torus be calculated using different units of measurement?

Yes, the volume of a Torus can be calculated using any unit of measurement as long as all measurements are in the same unit.

Are there any real-life applications of finding the volume of a Torus?

Yes, the volume of a Torus is commonly used in architecture and engineering for designing structures such as pipes, tunnels, and roller coasters.

What is the difference between a Torus and a cylinder?

While both shapes have a circular cross-section, a Torus has a curved surface while a cylinder has a flat surface. The volume formula for a Torus is also different from that of a cylinder.

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