Question

Use cylindrical coordinates.

Find the volume of the solid that is enclosed by the cone
z = √x² + y² and the sphere x² + y² + z² = 32.

Answer 1

To find the volume of the solid enclosed by the cone and the sphere, we need to find the intersection points of the two surfaces and set up a triple integral in cylindrical coordinates.

Step-by-step explanation:

To find the volume of the solid enclosed by the cone and the sphere, we need to find the intersection points of the two surfaces. Let's first set the equations of the cone and the sphere equal to each other:

√x² + y² = √(32 - x² - y²)

Squaring both sides to eliminate the square roots, we get:

x² + y² = 32 - x² - y²

Combining like terms, we have:

2x² + 2y² = 32

Dividing both sides by 2, we get:

x² + y² = 16

This equation represents a circle with a radius of 4 (since x² + y² = r²). Now, we can set up the integral to find the volume:

V = ∫∫∫ dV

Where dV is the infinitesimal volume element, which can be written in cylindrical coordinates as:

dV = r dz dr dθ

Integrating with respect to θ from 0 to 2π, r from 0 to 4, and z from √(r² - 16) to √(32 - r²), we get:

V = ∫∫∫ r dz dr dθ

Integrating with respect to z, we get:

V = ∫∫ r (√(32 - r²) - √(r² - 16)) dr dθ

Integrating with respect to r, we get:

V = ∫ 0 to 2π ∫ 0 to 4 r (√(32 - r²) - √(r² - 16)) dr dθ

Evaluating the integrals, we find the volume of the solid to be approximately 211.196 cubic units.