Consider the circle D on the plane with the center (2,0) and radius r=1. Revolving D

[booker]

1. The problem statement, all variables and given/known data

Consider the circle D on the plane with the center (2,0) and radius r=1. Revolving D about the y-axis, we obtain a donut (torus). What is the volume of this donut?

2. Relevant equations

Assume:
F(x)=\int^{a}_{-a} sqrt(a^2-u^2)du = (PI * a^2 )/2 where a\geq0


3. The attempt at a solution

I sincerely apologize, I have no idea as to where to even get started.

[HallsofIvy]

Can we at least assume you are taking a Calculus class and that you are dealling with "volumes of rotation" right now? You can do this by using "washers" or "cylinders".
In either case, the first thing you should do is draw a picture- graph the circle.

Washers: Draw a horizontal line across the circle at a fixed "y" value and imagine that being rotated around the y-axis. It forms a "washer"- the region between two circles. Its area is the difference between the areas of those two circles. If you think of it as having height "dy" then its volume is that area times dy. Integrate from the lowest y value in the circle to the highest.

Cylinders: Draw a vertical line through the circle at a specific x and imagine it rotated around the y-axis so that it forms a cylinder. Its surface area is the circumference of the circle, 2\pi x, times the length of that line. If you think of it has having thickness dx, its volume is that area times dx and the volume of the whole thing is the integral of that from the lowest x value in the circle to the highest.