Integral to find the volume of a torus

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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.