Final answer:
The volume of the solid between z=4 and z=\(\sqrt{x^{2}+y^{2}}\) is found using cylindrical coordinates. The integral setup involves radial distance limits from 0 to 4, angular limits from 0 to \(2\pi\), and vertical limits from r to 4. Upon integration, the volume of the solid is obtained.
Step-by-step explanation:
The student is asking for the volume of a three-dimensional solid that is bounded above by the plane z=4 and below by the cone z=\(\sqrt{x^{2}+y^{2}}\). To find this volume, we can use a method known as cylindrical coordinates, which is suitable for dealing with volumes of rotationally symmetric solids. In cylindrical coordinates, the problem is simplified because the cone can be described by z=r, where r represents the radial distance from the z-axis, and the plane z=4 becomes the upper limit of integration.
To compute the volume, we set up an integral in cylindrical coordinates:
- The limits of r are from 0 to 4, since at z=4, the cone z=\(\sqrt{x^{2}+y^{2}}\) intersects the plane at \(r=\sqrt{x^{2}+y^{2}}=4\).
- The limits for \(\theta\) are from 0 to \(2\pi\), encompassing the full circle in the xy-plane.
- The limits for z are from r (the bottom surface of the cone) to 4 (the top plane).
The volume V is then given by the triple integral:
V = \(\int_{0}^{2\pi} \int_{0}^{4} \int_{r}^{4} r dz dr d\theta\)
Carrying out the integration gives us the volume of the solid.