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.

Understand the Problem
The question is asking to express the volume of a solid in spherical coordinates, specifically the solid defined by a cone where z equals the square root of x squared plus y squared, and bounded by the planes z equals 1 and z equals 2. We will analyze the solid defined by these constraints and set up the appropriate volume integral in spherical coordinates.
Answer
$$ V = \frac{28\pi\sqrt{2}}{3}\left(1 - \frac{\sqrt{2}}{2}\right) $$
Answer for screen readers
The volume of the solid ( D ) is given by:
$$
V = \frac{28\pi\sqrt{2}}{3}\left(1 - \frac{\sqrt{2}}{2}\right)
$$
Steps to Solve
- 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.
- 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:
$$
\cos^2 \phi = \sin^2 \phi \implies \tan \phi = 1 \implies \phi = \frac{\pi}{4}
$$
- 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 $2\sqrt{2}$.
- Setting Up the Volume Integral
The volume integral in spherical coordinates is given by:
$$
V = \int_0^{2\pi} \int_0^{\frac{\pi}{4}} \int_{\sqrt{2}}^{2\sqrt{2}} \rho^2 \sin \phi , d\rho , d\phi , d\theta
$$
- 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:
$$
V = \int_0^{2\pi} \int_0^{\frac{\pi}{4}} \frac{14\sqrt{2}}{3} \sin \phi , d\phi , d\theta
$$
- 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:
$$
V = 2\pi \cdot \frac{14\sqrt{2}}{3}(1 - \frac{\sqrt{2}}{2})
$$
The volume of the solid ( D ) is given by:
$$
V = \frac{28\pi\sqrt{2}}{3}\left(1 - \frac{\sqrt{2}}{2}\right)
$$
More Information
This volume corresponds to the region between the defined planes and the conical surface, highlighting the use of spherical coordinates in integrating over three-dimensional shapes. The angle $\phi$ represents how steep the cone is, while $\rho$ scales the volume.
Tips
- Forgetting to change the bounds according to the transformation to spherical coordinates.
- Not properly setting up the volume integral with the correct limits.
- Confusing integration orders or variables in the integral calculations.
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