Answer: 6 units
Step-by-step explanation: To find the mass of the lamina, we need to integrate the density function fi(x,y,z) over the volume of the lamina. The lamina is a portion of the cone z=sqrt(x^2+y^2) between the planes z=1 and z=3. We can express this volume using cylindrical coordinates, where rho is the distance from the origin to the point (x,y,z), phi is the angle between the positive x-axis and the line connecting the origin to the point (x,y,z), and z is the height of the point (x,y,z) above the xy-plane.
We have:
1 <= z <= 3
0 <= rho <= 3sin(phi)
0 <= phi <= 2pi
The density function is given by fi(x,y,z) = x^2z. In cylindrical coordinates, we have x = rhocos(phi), y = rhosin(phi), and z = z. Therefore, we can express the density function as:
fi(rho, phi, z) = (rhocos(phi))^2 * z = rho^2cos^2(phi) * z
The mass of the lamina is given by the triple integral of the density function over the volume of the lamina:
M = ∭ fi(rho, phi, z) dV
= ∫∫∫ rho^2cos^2(phi) * z dz dA
= ∫∫[1,3] ∫[0,2pi] ∫[0,3sin(phi)] rho^2cos^2(phi) * z dz d(phi) drho
= ∫[0,2pi] ∫[0,3] ∫[0,rhosin(phi)] rho^2cos^2(phi) * z dz drho d(phi)
= ∫[0,2pi] ∫[0,3] (1/3)rho^3sin(phi)cos^2(phi) d(phi) drho
= ∫[0,2pi] (1/3)[9sin(phi)cos^2(phi)] d(phi)
= (1/3)[9(2/3)]
= 6
Therefore, the mass of the lamina is 6 units.