Questions: Use a triple integral to find the volume of the solid bounded below by the cone z = √(x² + y²) and bounded above by the sphere x² + y² + z² = 338. ∫∫∫ dV = 4π ∫₀^√338 r² (√(338-r²) - r) dr

Solution Steps

The problem involves finding the volume of the solid bounded below by the cone \( z = \sqrt{x^2 + y^2} \) and bounded above by the sphere \( x^2 + y^2 + z^2 = 338 \).

To simplify the integration, convert the given equations to cylindrical coordinates:

• The cone: \( z = r \)
• The sphere: \( r^2 + z^2 = 338 \)

In cylindrical coordinates:

• \( r \) ranges from 0 to the intersection of the cone and the sphere.
• \( \theta \) ranges from 0 to \( 2\pi \).
• \( z \) ranges from the cone \( z = r \) to the sphere \( z = \sqrt{338 - r^2} \).

The volume \( V \) is given by: \[ V = \int_{0}^{2\pi} \int_{0}^{\sqrt{338}} \int_{r}^{\sqrt{338 - r^2}} r \, dz \, dr \, d\theta \]

Integrate the innermost integral with respect to \( z \): \[ \int_{r}^{\sqrt{338 - r^2}} r \, dz = r \left[ z \right]_{r}^{\sqrt{338 - r^2}} = r \left( \sqrt{338 - r^2} - r \right) \]

Now, integrate with respect to \( r \): \[ \int_{0}^{\sqrt{338}} r \left( \sqrt{338 - r^2} - r \right) dr \]

Separate the integral into two parts: \[ \int_{0}^{\sqrt{338}} r \sqrt{338 - r^2} \, dr - \int_{0}^{\sqrt{338}} r^2 \, dr \]

Use substitution for the first integral and direct integration for the second:

1. For \( \int_{0}^{\sqrt{338}} r \sqrt{338 - r^2} \, dr \), let \( u = 338 - r^2 \), then \( du = -2r \, dr \).
2. For \( \int_{0}^{\sqrt{338}} r^2 \, dr \), use the power rule.

Evaluate the integrals and combine the results to find the volume.

Final Answer