Question

Use polar coordinates to find the volume of the given solid.

Below the cone z = √x² + y² and above the ring 1 ≤ x² + y² ≤ 64

Answer 1

To find the volume of the given solid, we can use polar coordinates. Convert the equations of the cone and the ring into polar form. Integrate the height of the cone with respect to r from 1 to 8 and evaluate the integral to find the volume.

Step-by-step explanation:

To find the volume of the given solid, we can use polar coordinates. First, let's convert the equations of the cone and the ring into polar form.

The equation of the cone in polar form is z = √(r²), where r is the distance from the origin to a point in the xy-plane. The equation of the ring in polar form is 1 ≤ r² ≤ 64, which simplifies to 1 ≤ r ≤ 8.

The volume of the solid can be found by integrating the height of the cone with respect to r from 1 to 8. The height of the cone is given by z = √(r²), so the integral becomes ∫(1 to 8) (z) dr = ∫(1 to 8) √(r²) dr.

Using the formula for the volume of a cone, the integral becomes ∫(1 to 8) (1/3)πr² dr.

Evaluating the integral, we get the volume of the solid as V = (1/3)π[(8)³ - (1)³] = 168π cubic units.

Answer 2

The volume of a solid below the cone z = √x² + y² and above the ring 1 ≤ x² + y² ≤ 64 is found using polar coordinates by setting up and evaluating the double integral ∫_{0}^{2π} ∫_{1}^{8} r^2 dr dθ.

Step-by-step explanation:

To find the volume of the solid that lies below the cone z = √x² + y² and above the ring 1 ≤ x² + y² ≤ 64 using polar coordinates, we need to set up an integral in polar coordinates.

In polar coordinates, x² + y² becomes r² and the area element dA is given by r dr dθ. So, the volume integral in polar coordinates can be expressed as:

1. Convert the bounds of x² + y² to polar coordinates, which gives 1 ≤ r² ≤ 64.
2. Set up the integral with the appropriate limits for r from 1 to 8 (since r² = 64 means r = 8).
3. The angle θ goes from 0 to 2π to complete the full revolution around the ring.
4. Compute the integral ∫_{0}^{2π} ∫_{1}^{8} r^2 dr dθ which represents the volume under the cone and above the ring.

Notice that the expression for the cone, z = √x² + y², when converted to polar coordinates, becomes z = r. Therefore, we integrate r to get the volume.