e is the region between the spheres x^2+y^2+z^2=1 and x^2+y^2+z^2=4 in the first octant.
theta range from 0 to 2pi, fee range from 0 to pi/2, roe range from 1 to 2. correct?
using these bounds, i get 7pi/8. it's supposed to be 15pi/16. where did the extra pi/16 go? what could possibly be the problem?
2006-12-04 14:29:03 · 2 answers · asked by nickname 3 in Science & Mathematics ➔ Mathematics
correction, theta goes from 0 to pi/2, not what i said up there.
2006-12-04 14:30:08 · update #1
calculus. sorry, there's a density of z i didn't see. i'll try that. thanks you two!
2006-12-04 14:49:10 · update #2
never mind, i've got it! thanks again!
2006-12-04 14:56:58 · update #3
You are right
theta=(0,pi/2)
phi=(0,pi/2)
rho=(1,2)
do you want calculus or geometry answer?
2006-12-04 14:40:38 · answer #1 · answered by Micah M 1 · 0⤊ 0⤋
Are you sure you typed in the problem right? Using those bounds (which are correct, at least after you fixed theta), working through the triple-integral I get 7pi/6, which isn't even what you got.
However, thats obviously right - just thinking about volumes, the outer sphere has volume 4/3 * pi * 2^3, and the inner sphere has volume 4/3 * pi * 1^3, so the difference is 28pi/3, so divide by 8 to find the volume in the first octant, which is 7pi/6.
2006-12-04 14:44:01 · answer #2 · answered by stephen m 4 · 0⤊ 0⤋