Evaluate the triple integral of xyz over the region bounded by z=x^2+y^2 and z=4-x^2-y^2.
To evaluate the triple integral, we'll first need to find the limits of integration for x, y, and z. We can start by finding where the two surfaces intersect:

x^2 + y^2 = 4 - x^2 - y^2

2x^2 + 2y^2 = 4

x^2 + y^2 = 2

This is a circle with radius √2 in the xy-plane.

Now let's set up the triple integral using cylindrical coordinates. Recall that in cylindrical coordinates, x = rcos(θ), y = rsin(θ), and z = z. The Jacobian for the transformation is r, so our integrand becomes (r^3)cos(θ)sin(θ).

We have the following bounds for our variables:
- θ goes from 0 to 2π (full circle)
- r goes from 0 to √2 (radius of the circle)
- z goes from r^2 to 4-r^2 (from lower surface to upper surface)

Now we can write out the triple integral:

∬∬∬ xyz dz dy dx = ∬∬∬ (r^3)cos(θ)sin(θ) rdz dr dθ

Let's integrate with respect to z first:

∫(r^3)cos(θ)sin(θ) dz = (r^3)cos(θ)sin(θ)z | evaluated from z=r^2 to z=4-r^2

= (r^3)cos(θ)sin(θ)[(4-r^2)-r^2]

= (r^3)cos(θ)sin(θ)(4-2r^2)

Now integrate with respect to r:

∫(r^3)cos(θ)sin(θ)(4-2r^2) dr = cos(θ)sin(θ)∫(4r^3 - 2r^5) dr

= cos(θ)sin(θ)[(r^4 - (1/3)r^6)] | evaluated from r=0 to r=√2

= cos(θ)sin(θ)[(4 - (8/3))]

= (4/3)cos(θ)sin(θ)

Finally, integrate with respect to θ:

∫(4/3)cos(θ)sin(θ) dθ

To evaluate this integral, we can use the double angle identity sin(2θ) = 2cos(θ)sin(θ). Then our integral becomes:

(2/3)∫sin(2θ) dθ = -(1/3)cos(2θ) | evaluated from θ=0 to θ=2π

= -(1/3)[cos(4π) - cos(0)]

= -(1/3)[1 - 1] = 0

So the triple integral evaluates to 0. This makes sense because the integrand xyz is an odd function and the region of integration is symmetric about the origin.