Integral to find the volume of a torus

In summary, to find the volume of a torus formed by rotating a circle around the y-axis, you can use Pappus' theorem by multiplying the distance traveled by the center of the circle (2πr) by the area of the circle (πr^2). In this case, the volume is 16π^2.
  • #1
togame
18
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Homework Statement


Find the volume of the torus formed when the circle of radius 2 centered at (3,0) is revolved about the y-axis. Use geometry to evaluate the integral.


Homework Equations


Formula for the semi-circle: [itex]y=\sqrt{4-x^2}[/itex]
Solving for x give [itex]x=\pm\sqrt{4-y^2}[/itex]


The Attempt at a Solution


I was thinking of solving the integral from -2 to 2 using both the positive and negative sides, then evaluate the integral and then multiply by 2, but I'm not so sure about my equation.

[tex]2\int_{-2}^2\pi\big( (3-\sqrt{4-y^2})^2 - (3+\sqrt{4-y^2})^2 \big) \mathrm d y[/tex]

I just seem lost on this one :(
 
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  • #2
togame said:

Homework Statement


Find the volume of the torus formed when the circle of radius 2 centered at (3,0) is revolved about the y-axis. Use geometry to evaluate the integral.

Homework Equations


Formula for the semi-circle: [itex]y=\sqrt{4-x^2}[/itex]
Solving for x give [itex]x=\pm\sqrt{4-y^2}[/itex]

The Attempt at a Solution


I was thinking of solving the integral from -2 to 2 using both the positive and negative sides, then evaluate the integral and then multiply by 2, but I'm not so sure about my equation.
[tex]2\int_{-2}^2\pi\big( (3-\sqrt{4-y^2})^2 - (3+\sqrt{4-y^2})^2 \big) \mathrm d y[/tex]
I just seem lost on this one :(
I believe the equation is:

[tex]V=\int_{-2}^2\pi\big( (3+\sqrt{4-y^2})^2 - (3-\sqrt{4-y^2})^2 \big) \mathrm d y[/tex]

[tex]V=\int_{-2}^2\pi\cdot 12(\sqrt{4-y^2}) \mathrm d y[/tex]

then trigonometric substitution...
 
  • #3
togame said:

Homework Statement


Find the volume of the torus formed when the circle of radius 2 centered at (3,0) is revolved about the y-axis. Use geometry to evaluate the integral.

Homework Equations


Formula for the semi-circle: [itex]y=\sqrt{4-x^2}[/itex]
Solving for x give [itex]x=\pm\sqrt{4-y^2}[/itex]

The Attempt at a Solution


I was thinking of solving the integral from -2 to 2 using both the positive and negative sides, then evaluate the integral and then multiply by 2, but I'm not so sure about my equation.

[tex]2\int_{-2}^2\pi\big( (3-\sqrt{4-y^2})^2 - (3+\sqrt{4-y^2})^2 \big) \mathrm d y[/tex]

I just seem lost on this one :(

Edit : Got beat to it.
 
  • #4
the way to do this is to use pappus' theorem. the product of the distance traveled by the center of the circle by the area of the circle.
 

FAQ: Integral to find the volume of a torus

1. What is a torus?

A torus is a geometric shape that resembles a donut or a ring. It is a three-dimensional object with a circular cross-section and a hole in the center.

2. How is the volume of a torus calculated using integral?

The volume of a torus can be calculated by using the integral of a function that represents the cross-sectional area of the torus. This integral can be evaluated using the formula V = 2π²Rr², where R is the radius of the torus and r is the radius of the circular cross-section.

3. What is the significance of using integral to find the volume of a torus?

Using integral to find the volume of a torus allows for a more accurate estimation of the volume, as it takes into account the varying cross-sectional areas of the torus. It also allows for the calculation of volumes for more complex torus shapes.

4. Can the volume of a torus be calculated without using integral?

Yes, the volume of a torus can also be calculated using other methods such as the formula V = (π²Rr)(2R-r), or by approximating the torus as a series of smaller cylinders. However, using integral provides a more precise and generalizable solution.

5. Are there any real-world applications of using integral to find the volume of a torus?

Yes, integral is commonly used in engineering and physics to calculate the volume of objects with complex shapes, such as torus-shaped pipes or containers. It is also used in computer graphics to generate 3D models of torus-shaped objects.

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