To solve this problem, let's break down the requirements and set up the integral in spherical coordinates to find the volume of the solid region described.

Step 1: Understand the Region in Spherical Coordinates

Given:

1. Two spheres:

   • $x^2 + y^2 + z^2 = a^2$
   • $x^2 + y^2 + z^2 = b^2$, where $b > a$.

   These spheres have radii $a$ and $b$, respectively, and define the bounds for $\rho$ in spherical coordinates.

2. Cone:

   • $z^2 = x^2 + y^2$, which is equivalent to $z = \pm \sqrt{x^2 + y^2}$.
   • This cone restricts the angle $\phi$ (the angle between the positive $z$-axis and the position vector) to a fixed value. Since $z = \rho \cos \phi$ and $\sqrt{x^2 + y^2} = \rho \sin \phi$, we find $\tan \phi = 1$, giving $\phi = \frac{\pi}{4}$.

Thus, the region is bounded by:

• $a \leq \rho \leq b$
• $0 \leq \theta < 2\pi$
• $0 \leq \phi \leq \frac{\pi}{4}$

Step 2: Set Up the Volume Integral

In spherical coordinates, the volume element $dV$ is given by: $dV = \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta$

The volume $V$ of the solid is: $V = \int_{\theta=0}^{2\pi} \int_{\phi=0}^{\pi/4} \int_{\rho=a}^{b} \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta$

Step 3: Evaluate the Integral

1. Integrate with respect to $\rho$: $\int_{a}^{b} \rho^2 \, d\rho = \left[ \frac{\rho^3}{3} \right]_{a}^{b} = \frac{b^3}{3} - \frac{a^3}{3}$

2. Integrate with respect to $\phi$: $\int_{0}^{\pi/4} \sin \phi \, d\phi = \left[ -\cos \phi \right]_{0}^{\pi/4} = -\cos \frac{\pi}{4} + \cos 0 = -\frac{\sqrt{2}}{2} + 1 = 1 - \frac{\sqrt{2}}{2}$

3. Integrate with respect to $\theta$: $\int_{0}^{2\pi} d\theta = 2\pi$

Step 4: Combine Results

Now, combining all parts, the volume $V$ is: $V = \left( \frac{b^3}{3} - \frac{a^3}{3} \right) \left( 1 - \frac{\sqrt{2}}{2} \right) \cdot 2\pi$ $V = \frac{2\pi}{3} \left( b^3 - a^3 \right) \left( 1 - \frac{\sqrt{2}}{2} \right)$

This is the volume of the solid region bounded by the two spheres and the cone.

Would you like a detailed step-by-step walkthrough or have any questions about this solution?

Related Questions

1. How would the integral change if the cone angle were different?
2. How does the volume vary if the outer radius $b$ increases while $a$ remains constant?
3. What would be the volume if the region were bounded only by the outer sphere and the cone?
4. How would Cartesian coordinates affect the complexity of solving this integral?
5. Could cylindrical coordinates be useful for solving a similar problem?

Tip

In spherical coordinates, always remember to include the $\rho^2 \sin \phi$ factor in the volume element, as it accounts for the spherical geometry's area and volume scaling.