2009-10-28, 12:13 AM
Call the line through the center of the hole the x axis (straight up and down in the above figure), and a line across the base of the hole (top of the figure) the y axis. Then the straight hole can be represented as the line y = 3/4 rotated about the x axis, and the countersink is y = 5/4 - x rotated about the x axis.
Then it should be just an integral of outer radius squared - inner radius squared from x = 0 to the intersection of the two.
Then it should be just an integral of outer radius squared - inner radius squared from x = 0 to the intersection of the two.