Find the volume of a solid torus (donut shape) with r as the radius of the circular?

Question

Find the volume of a solid torus (donut shape) with r as the radius of the circular cross-section of the torus, and R as the distance from the torus center to the center of any cross-section.

Answer 1

It just works out that it's the same as a cylinder of radius r and length equal to the circumference of the circle of radius R.

Here's a simple way to see why this is so: Slice the torus into layers of "racetracks" of uniform thicknesses. The area of any racetrack is just the difference between the areas of the larger and smaller circle, or the width of the racetrack times the circumference of the circle halfway between the two. Ergo, the rest follows.

Answer 2

Integrate the infinitesimally small cross-sections * their depth

The depth will be R*dθ. The cs area is π*r^2

the equation becomes Integral(π*r^2*R*dθ)
Everything except for θ is a constant, so the equation reduces to
π*r^2*R* Integral(dθ)
You want the volume of the entire torus, so you have to integrate from θ= 0 to θ=2π

The final equation is therefore

π*r^2*R*(2π-0)

or

2π^2*r^2*R



edit: your teacher will like mine better 8^)

Answer 3

1. you dunk it in a tub of water and see how much the water rises....there's your volume.

2. melt the torus in a zero-g environment until it's a sphere. Let the sphere cool and measure it's diameter. Divide the diameter in half, then use the 4/3 * pi * r^3 to find your volume.

Answer 4

The question is clever, yet ill posed. Technically, the ambience of any planet is evaporating consistently because of the tail end of the Boltzman distribution extending to infinity. it relatively is in simple terms a query of time.