What is the volume of the solid generated when the circular region bounded by the graph of x^2+y^2=1 is revolved around the horizontal line y=3 ? (It is known that (-1 to 1)∫√(1-x^2)ⅆx=π/2.)

Answer 1

Note that the center of gravity of the circle is at (0,0) so it travels a distance of 2πr = 6π as the circle revolves. The Theorem of Pappus tells us that the volume of the torus generated is 6π * π = 6π^2

Or you can brute-force the integration, but you have to do the upper and lower halves of the circle separately. Thus, the volume is
v = 2∫[0,1] π(R^2-r^2) dx where R=3 and r = 3-√(1-x^2)
+ 2∫[0,1] π(R^2-r^2) dx where R = 3+√(1-x^2) and r=3