Final answer:
To evaluate the triple integrals, we define the regions of integration. The first integral is over the volume of a cylinder with radius √4 and height 2. The limits of integration depend on the specific boundaries of the regions, and cylindrical coordinates are often employed.
Step-by-step explanation:
To evaluate the triple integrals, we first need to understand the regions of integration. For the first integral ∫∫∫_D (x^2 + y^2) dv, the region D is defined by z=2 and x^2 + y^2 = 2z. Rewriting the second equation, we have z = \frac{1}{2}(x^2 + y^2), which describes a cone. The region of integration is the volume between this cone and the plane z=2, which is, in fact, a cylinder of radius √4 and height 2.
The integral can be computed in cylindrical coordinates where x^2 + y^2 converts to r^2 and dv becomes r dr d\theta dz. The integration limits for r are from 0 to √4, for \theta from 0 to 2\pi, and for z from z=r^2/2 to z=2. The integral then is ∫_{0}^{2\pi} ∫_{0}^{√4} ∫_{r^2/2}^{2} r^3 dz dr d\theta .
For the second integral ∫∫∫_D z dv, the region D is not fully described in the question prompt. Assuming z=h/k represents a plane, the integral would typically be over the volume beneath this plane within a specified boundary. Knowing that boundary is essential to set the limits of integration. If we suppose the limits describe a simple shape like a cylinder, one would again convert to cylindrical coordinates to perform the integral.