Integral to find the volume of a torus

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Homework Help Overview

The discussion revolves around finding the volume of a torus formed by revolving a circle of radius 2 centered at (3,0) about the y-axis. Participants are exploring the geometric interpretation and integral setup for this problem.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to set up the integral for the volume calculation, with some expressing uncertainty about their equations. There are discussions about integrating from -2 to 2 and considering both sides of the circle. One participant suggests using trigonometric substitution, while another mentions Pappus' theorem as a potential approach.

Discussion Status

The discussion is ongoing, with various approaches being considered. Some participants are questioning their integral setups and exploring different methods, while others are suggesting alternative theorems to evaluate the volume.

Contextual Notes

Participants are working under the constraints of using geometry to evaluate the integral and are expressing confusion about the correct setup for the volume calculation.

togame
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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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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...
 
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.
 
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.
 

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