Find the volume of the solid enclosed by the paraboloids z = 4*( x^{2} + y^{2} ) and z = 8 - 4*( x^{2} + y^{2} ).

To find the volume of the solid enclosed by the paraboloids, we can use double integrals in cylindrical coordinates.

First, let's rewrite the equations of the paraboloids in cylindrical coordinates. In cylindrical coordinates, x = r*cos(theta) and y = r*sin(theta). Substituting these into the equations, we have:

z = 4*( r^{2} ) (equation 1)
z = 8 - 4*( r^{2} ) (equation 2)

To find the limits of integration, we need to determine the region in the xy-plane that corresponds to the volume. Since the paraboloids are symmetric with respect to the z-axis, we only need to find the volume in the positive z region.

From equation 1, we can see that z is always positive or zero. So, the lower bound for z is 0.

From equation 2, we can find the upper bound for z. Setting equation 2 to zero and solving for r:
8 - 4*( r^{2} ) = 0
4*( r^{2} ) = 8
r^{2} = 2
r = sqrt(2)

Now, let's set up the double integral using cylindrical coordinates:

V = ∫∫R (z_upper - z_lower) r dr dθ

Where R is the region in the xy-plane bounded by the intersection of the paraboloids.

The bounds for r and θ are:
0 ≤ r ≤ sqrt(2)
0 ≤ θ ≤ 2π

Now, let's integrate:

V = ∫[0,2π] ∫[0,sqrt(2)] ((8 - 4*( r^{2} )) - (4*( r^{2} ))) r dr dθ

V = ∫[0,2π] ∫[0,sqrt(2)] (8 - 8*( r^{2} )) r dr dθ

V = ∫[0,2π] ( 4*r^4 - 4/3*r^6 ) | [0,sqrt(2)] dθ

Now evaluate this integral:

V = ∫[0,2π] ( 4*2^2 - 4/3*2^3 ) - ( 4*0^2 - 4/3*0^3 ) dθ

V = ∫[0,2π] ( 16 - 32/3 ) dθ

V = ( 16 - 32/3 ) * (2π)

V = ( 48/3 - 32/3 ) * (2π)

V = 16/3 * (2π)

V = (32/3)π

Therefore, the volume of the solid enclosed by the paraboloids is (32/3)π cubic units.