Problem:
Let R be the region consisting of the set of points in the coordinate plane that satisfy both β£8βxβ£+yβ€10 and 3yβxβ₯15. When R is revolved around the line whose equation is 3yβx=15, the volume of the resulting solid is npβmΟβ, 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 3yβx=15, y=x+2, and y=βx+18. The vertices of this triangle are A=(29β,213β),B=(439β,433β), 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 31βΟβ CD2β AD+31βΟβ CD2β BD=31βΟβ CD2β AB. Note that
βAB=(421β)2+(47β)2β=47β32+12β=4710ββ and CD=(8β8.7)2+(10β7.9)2β=10β7ββ
Thus the desired volume is 31βΟβ 1049ββ 4710ββ=1210β343Οβ, and m+n+p=343+12+10=365β.