Problem:
Let R be the region consisting of the set of points in the coordinate plane that satisfy both ∣8−x∣+y≤10 and 3y−x≥15. When R is revolved around the line whose equation is 3y−x=15, the volume of the resulting solid is np​mπ​, where m,n, and p are positive integers, m and n are relatively prime, and p is not divisible by the square of any prime. Find m+n+p.
Solution:
The region R is a triangular region bounded by the lines 3y−x=15, y=x+2, and y=−x+18. The vertices of this triangle are A=(29​,213​),B=(439​,433​), and C=(8,10). Let D be the foot of the perpendicular from C to line AB. It can be verified that the coordinates of point D are (8.7,7.9), and hence D is between A and B. Thus the solid of revolution consists of two right circular cones with heights AD and BD, each having a base radius of CD. The desired volume is therefore 31​π⋅CD2⋅AD+31​π⋅CD2⋅BD=31​π⋅CD2⋅AB. Note that
​AB=(421​)2+(47​)2​=47​32+12​=4710​​ and CD=(8−8.7)2+(10−7.9)2​=10​7​​
Thus the desired volume is 31​π⋅1049​⋅4710​​=1210​343π​, and m+n+p=343+12+10=365​.