Multiple Integrals

(a)
(i) ∫_[-3,3]∫[_[x^2,9]∫_[0, 9-y] f(x,y,z) dz dy dx
The plane y+z=9 intersects the xy-plane at y=9, so that, the base of D is
{ (x, y) | x^2 < = y < = 9 }
(ii)∫_[-2, 2]∫_[-√(4-z^2), √(4-z^2)]∫_[√(x^2+z^2), √(8-x^2-z^2)] f(x,y,z) dydxdz
x^2+y^2+z^2=8 intersects y=√(x^2+z^2) at the circle y=2 and x^2+z^2=4,
so that, the bottom of D is y=√(x^2+z^2) and covered with y=√(8-x^2-z^2).
The projection of D to the zx-plane is the circle x^2+z^2=4.

(b)
The cone z=√(3x^2+3y^2) intersects the sphere x^2+y^2+z^2=9 at
x^2+y^2= 9/4 and z= 3√3 /2, so that z=√(3x^2+3y^2) can be expressed as
φ=π/6 (φ is the angle between z=√(3x^2+3y^2) and the positive z-axis).
Hence, ∫∫∫_D √(x^2+y^2) dV
=∫_[π/6,π] ∫_[0,2π]∫_[0,3] ρsinφ*ρ^2 sinφ dρ dθ dφ
= (81π/4)*(5π/6 + √3 /4)

2011-06-22 13:46:14 補充：
I think the inner parts of the zone z=√(3x^2+3y^2) should be z > = √(3x^3+3y^2).
If you recognize the outside parts of the cone as z > = √(3x^2+3y^2), then φ=0~π/6.