Question

Se cylindrical coordinates to evaluate the triple integral ∫∫∫e√(x²+y²)¹/²dv, where e is the solid bounded by the circular paraboloid z=1−16(x²+y²) and the xy -plane.

Answer 1

The triple integral of the solid bounded by the circular paraboloid and the xy-plane in cylindrical coordinates is set up and solved step-by-step, starting with the substitution of the integral's function into cylindrical form and determining the bounds of integration.

Step-by-step explanation:

To solve the triple integral ∫∫∫ e√(x²+y²)1/2dv, where e is the solid bounded by the circular paraboloid z=1−16(x²+y²) and the xy-plane using cylindrical coordinates, we start by expressing the bounds of the integration and the function to be integrated in terms of cylindrical coordinates. In cylindrical coordinates, an arbitrary point (x, y, z) is represented as (r cosθ, r sinθ, z), where r is the radius, θ is the angle, and z is the height. The given function transforms into er in these coordinates because √(x²+y²) is simply r.

The bounds for r are from 0 to 1/4 because the paraboloid z = 1 - 16(x² + y²) intersects the xy-plane at r = 1/4. The angle θ varies from 0 to 2π, completing the circular symmetry around the z-axis. The z bounds go from the xy-plane (z=0) up to the paraboloid surface (z=1−16r²). Thus, the integral becomes:

∫02π ∫01/4 ∫01−16r² er r dz dr dθ

Simplifying this integral step-by-step,

4. Integrate with respect to z from 0 to 1-16r², which will yield (1-16r²)er.
8. Next, integrate the resulting expression with respect to r, considering the r term and the expression 1-16r².
12. Finally, integrate the resulting function in terms of θ over the interval from 0 to 2π.

This sequence will determine the total volume of the solid bounded by the paraboloid and the xy-plane.