One of those cases where using integrals rather than geometry is much simpler. \pi \int_-3^3 (R^2 - x^2) dx = \pi (6 R^2 - 18) is the volume of the rotati

One of those cases where using integrals rather than geometry is much simpler. \pi \int_-3^3 (R^2 - x^2) dx = \pi (6 R^2 - 18) is the volume of the rotational solid without removing the cylinder. While the volume of the cylinder is given by: \pi \int_-3^3 (R^2 - 3^2) dx = \pi 6 (R^2 - 9) As you can see the difference between the two volumes is 36 \pi. You can actually show the solution does not depend on \pi without evaluating the integral, and then compute the 'cheat' case, which would not be a cheat once you've made this observation.