Express the volume of D in spherical coordinates, where D is the solid enclosed by z = √(x² + y²) between the planes z = 1 and z = 2.

Steps to Solve

1. Identify the Spherical Coordinates Transformation

In spherical coordinates, we define the variables as follows:
$$ x = \rho \sin \phi \cos \theta $$
$$ y = \rho \sin \phi \sin \theta $$
$$ z = \rho \cos \phi $$
where $\rho$ is the radius, $\phi$ is the angle from the positive z-axis, and $\theta$ is the angle in the xy-plane.

2. Express the Cone Equation in Spherical Coordinates

The cone given by the equation ( z = \sqrt{x^2 + y^2} ) can be expressed in spherical coordinates:
$$ \rho \cos \phi = \sqrt{\rho^2 \sin^2 \phi} $$
Squaring both sides:
$$ \rho^2 \cos^2 \phi = \rho^2 \sin^2 \phi $$
Thus, simplifying gives:
$$ \cos^2 \phi = \sin^2 \phi \implies \tan \phi = 1 \implies \phi = \frac{\pi}{4} $$

3. Determining the Bounds for z

The bounds for the z coordinate are given as ( z = 1 ) and ( z = 2 ). In spherical coordinates, these translate to:
$$ 1 = \rho \cos \phi \quad \text{and} \quad 2 = \rho \cos \phi $$
Thus, using ( \phi = \frac{\pi}{4} ):
$$ 1 = \rho \cos \left( \frac{\pi}{4} \right) \implies 1 = \frac{\rho}{\sqrt{2}} \implies \rho = \sqrt{2} $$
$$ 2 = \rho \cos \left( \frac{\pi}{4} \right) \implies 2 = \frac{\rho}{\sqrt{2}} \implies \rho = 2\sqrt{2} $$
So, the bounds for $\rho$ are from $\sqrt{2}$ to $2\sqrt{2}$.

4. Setting Up the Volume Integral

The volume integral in spherical coordinates is given by:
$$ V = \int_0^{2\pi} \int_0^{\frac{\pi}{4}} \int_{\sqrt{2}}^{2\sqrt{2}} \rho^2 \sin \phi , d\rho , d\phi , d\theta $$

5. Evaluating the Integral

We will evaluate the integral:
First, integrate with respect to $\rho$:
$$ \int_{\sqrt{2}}^{2\sqrt{2}} \rho^2 , d\rho = \left[\frac{\rho^3}{3}\right]_{\sqrt{2}}^{2\sqrt{2}} $$
Calculating:
$$ = \frac{(2\sqrt{2})^3}{3} - \frac{(\sqrt{2})^3}{3} = \frac{16\sqrt{2}}{3} - \frac{2\sqrt{2}}{3} = \frac{14\sqrt{2}}{3} $$

The integral then simplifies to:
$$ V = \int_0^{2\pi} \int_0^{\frac{\pi}{4}} \frac{14\sqrt{2}}{3} \sin \phi , d\phi , d\theta $$

6. Integrating Over phi and theta

First, integrate with respect to $\phi$:
$$ \int_0^{\frac{\pi}{4}} \sin \phi , d\phi = \left[-\cos \phi \right]_0^{\frac{\pi}{4}} = -\left(\frac{\sqrt{2}}{2} - 1\right) = 1 - \frac{\sqrt{2}}{2} $$

Then, integrate with respect to $\theta$:
$$ \int_0^{2\pi} d\theta = 2\pi $$

Putting it all together, the volume is:
$$ V = 2\pi \cdot \frac{14\sqrt{2}}{3}(1 - \frac{\sqrt{2}}{2}) $$