The problem asks us to calculate the volume of the region bounded by a cone and a sphere using spherical polar coordinates.

1. Problem Overview

• Cone equation: $z = \sqrt{x^2 + y^2}$, which is equivalent to $z = r$, where $r = \sqrt{x^2 + y^2}$ in cylindrical coordinates.
• Sphere equation: $x^2 + y^2 + z^2 = z^2$, which simplifies to $r^2 + z^2 = 1$ or $r = 1$, corresponding to a unit sphere in spherical coordinates.

2. Convert to Spherical Coordinates

In spherical coordinates, we use the following transformations:

• $x = \rho \sin\theta \cos\phi$
• $y = \rho \sin\theta \sin\phi$
• $z = \rho \cos\theta$ Where:
• $\rho$ is the radial distance,
• $\theta$ is the polar angle,
• $\phi$ is the azimuthal angle.

For the given setup:

• The cone equation $z = r$ becomes $\cos\theta = \sin\theta$, giving $\theta = \frac{\pi}{4}$.
• The sphere equation $x^2 + y^2 + z^2 = 1$ gives $\rho = 1$.

3. Volume Integral Setup

The volume element in spherical coordinates is given by: $dV = \rho^2 \sin\theta \, d\rho \, d\theta \, d\phi$ We now set up the limits:

• $\rho$ ranges from the sphere (which is a unit sphere) to the cone.
• $\theta$ ranges from 0 to $\frac{\pi}{4}$ because the cone intersects at that angle.
• $\phi$ ranges from $0$ to $2\pi$.

4. Integral for Volume

The volume integral becomes: $V = \int_0^{2\pi} \int_0^{\frac{\pi}{4}} \int_0^1 \rho^2 \sin\theta \, d\rho \, d\theta \, d\phi$

Now, let's calculate the integral step by step:

1. Radial integral: $\int_0^1 \rho^2 \, d\rho = \frac{1}{3}$
2. Angular integral (in $\theta$): $\int_0^{\frac{\pi}{4}} \sin\theta \, d\theta = 1 - \frac{1}{\sqrt{2}}$
3. Azimuthal integral: $\int_0^{2\pi} d\phi = 2\pi$

5. Final Volume Calculation

Combining all these results: $V = 2\pi \times \frac{1}{3} \times \left( 1 - \frac{1}{\sqrt{2}} \right)$ This gives the final volume of the region bounded by the cone and the sphere.

Would you like to see the detailed steps in computing the integrals, or have any clarifications on the setup?

5 Related Questions:

1. How would the solution change if the cone had a different equation?
2. What if the sphere had a different radius, say $R$ instead of 1?
3. Can you derive the equation of the cone in spherical coordinates explicitly?
4. How would you compute the surface area instead of the volume for this setup?
5. What if the volume was bounded between two concentric spheres?

Tip:

Always start by identifying the symmetry of the problem to simplify the setup of your integrals.