How do I find the volume of the solid formed by revolving the region bounded by x= 0, y= 0, y= (6 - x)^1/2 about the x-axis?
I know the answer is pi something... because that's what all the solutions have in my notes.
The teacher gave me the following possibilities.
2p
9p /2
8p
25p /2
18p
49p/2
32p
50p
2006-12-03 04:30:57 · 3 answers · asked by chris 2 in Science & Mathematics ➔ Mathematics
I don't know I am only in the 8th grade. gosh!!!!!!!!!!
2006-12-03 04:39:15 · answer #1 · answered by Kelsey 2 · 0⤊ 1⤋
The volume of a thin disk thickness dx at position x is pi*r^2*dx, or in the case of the problem, pi*(6-x)*dx
integrate this from zero to the x location of y=0 (x=6) and you have your answer:
integral(pi*(6-x)*dx,0,6) = 18 pi
2006-12-03 04:36:16 · answer #2 · answered by MooseBoys 6 · 0⤊ 0⤋
consult u r teacher??
2006-12-03 05:07:42 · answer #3 · answered by Paassion 3 · 0⤊ 0⤋