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