Questions: Find the volume of the solid bounded below by the circular cone z=3 sqrt(x^2+y^2) and above by the sphere x^2+y^2+z^2=4 z.

Solution Steps

To find the volume of the solid bounded by the given cone and sphere, we can use triple integrals in spherical coordinates. First, we convert the equations of the cone and sphere into spherical coordinates. The cone equation \( z = 3\sqrt{x^2 + y^2} \) becomes \( \phi = \frac{\pi}{6} \) in spherical coordinates. The sphere equation \( x^2 + y^2 + z^2 = 4z \) simplifies to \( \rho = 4\cos(\phi) \). We then set up the integral with the appropriate limits for \(\rho\), \(\phi\), and \(\theta\) to cover the region between the cone and the sphere.

We are tasked with finding the volume of the solid bounded below by the circular cone given by the equation \( z = 3\sqrt{x^2 + y^2} \) and above by the sphere defined by \( x^2 + y^2 + z^2 = 4z \). In spherical coordinates, the cone corresponds to \( \phi = \frac{\pi}{6} \) and the sphere can be rewritten as \( \rho = 4\cos(\phi) \).

The volume \( V \) can be expressed as a triple integral in spherical coordinates: \[ V = \int_0^{2\pi} \int_0^{\frac{\pi}{6}} \int_0^{4\cos(\phi)} \rho^2 \sin(\phi) \, d\rho \, d\phi \, d\theta \] where:

• \( \rho \) ranges from \( 0 \) to \( 4\cos(\phi) \),
• \( \phi \) ranges from \( 0 \) to \( \frac{\pi}{6} \),
• \( \theta \) ranges from \( 0 \) to \( 2\pi \).

Calculating the integral yields: \[ V = \int_0^{2\pi} d\theta \int_0^{\frac{\pi}{6}} \left( \int_0^{4\cos(\phi)} \rho^2 \sin(\phi) \, d\rho \right) d\phi \] The inner integral evaluates to \( \frac{64}{3} \sin(\phi) \cos^3(\phi) \). Integrating with respect to \( \phi \) and \( \theta \) gives: \[ V = \frac{14\pi}{3} \]

Final Answer