If D is the solid region in the first octant bounded above by the paraboloid z =

1 − x2 − y2 and below by the xy-plane, the volume of D is:

so i set this up integrating over dz dr dtheta and got pi/2 but the answer i spi/8

To find the volume of the solid region D, bounded above by the paraboloid and below by the xy-plane, you need to set up a triple integral in cylindrical coordinates:

∭dV = ∭r dz dr dθ

The limits of integration for each variable are as follows:

- For z: Since D is bounded above by the paraboloid 1 - x^2 - y^2, the upper limit of integration for z is given by the equation of the paraboloid: z = 1 - x^2 - y^2. The lower limit of integration for z is 0 since D is bounded below by the xy-plane.

- For r: In cylindrical coordinates, r represents the radial distance from the origin. In the first octant, the region D is bound by the coordinate planes x = 0, y = 0, and the quarter-circle x^2 + y^2 = 1. Thus, the lower limit of integration for r is 0, and the upper limit is the radius of the quarter-circle, which is 1.

- For θ: In the first octant, θ varies from 0 to π/2.

Therefore, the triple integral to calculate the volume becomes:

V = ∫[0 to π/2] ∫[0 to 1] ∫[0 to 1 - r^2] r dz dr dθ

Evaluating this integral provides the desired volume of the region D.