To find the mass of the solid bounded by the cone $z = \sqrt{x^2 + y^2}$ and the hemisphere $z = \sqrt{9.25^2 - x^2 - y^2}$, with a density function $\rho(x, y, z) = \sqrt{x^2 + y^2 + z^2}$, we can use spherical coordinates to set up and evaluate the integral.

Step-by-Step Solution

1. Convert to Spherical Coordinates:

   • In spherical coordinates, we have: $x = \rho \sin\phi \cos\theta, \quad y = \rho \sin\phi \sin\theta, \quad z = \rho \cos\phi$
   • The density function becomes $\rho(x, y, z) = \rho$ since $\sqrt{x^2 + y^2 + z^2} = \rho$.
   • The volume element in spherical coordinates is $dV = \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta$.

2. Set the Limits:

   • The cone $z = \sqrt{x^2 + y^2}$ in spherical coordinates corresponds to $\phi = \frac{\pi}{4}$.
   • The hemisphere $z = \sqrt{9.25^2 - x^2 - y^2}$ has radius $9.25$, so $\rho$ goes from $0$ to $9.25$.
   • The angle $\theta$ ranges from $0$ to $2\pi$, and $\phi$ ranges from $0$ to $\frac{\pi}{4}$.

3. Set Up the Integral for Mass: The mass $M$ is given by: $M = \int\int\int_V \rho(x, y, z) \, dV$ Substituting the density $\rho(x, y, z) = \rho$ and the volume element: $M = \int_0^{2\pi} \int_0^{\pi/4} \int_0^{9.25} \rho \cdot \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta$ Simplifying, we get: $M = \int_0^{2\pi} \int_0^{\pi/4} \int_0^{9.25} \rho^3 \sin\phi \, d\rho \, d\phi \, d\theta$

4. Evaluate the Integral:

   • First, integrate with respect to $\rho$: $\int_0^{9.25} \rho^3 \, d\rho = \left[ \frac{\rho^4}{4} \right]_0^{9.25} = \frac{(9.25)^4}{4}$
   • Next, integrate with respect to $\phi$: $\int_0^{\pi/4} \sin\phi \, d\phi = \left[ -\cos\phi \right]_0^{\pi/4} = 1 - \frac{\sqrt{2}}{2}$
   • Finally, integrate with respect to $\theta$: $\int_0^{2\pi} d\theta = 2\pi$

5. Combine Results: $M = 2\pi \cdot \frac{(9.25)^4}{4} \cdot \left(1 - \frac{\sqrt{2}}{2}\right)$

6. Compute the Mass: Substitute $9.25^4 \approx 731.1616$: $M \approx 2\pi \cdot \frac{731.1616}{4} \cdot \left(1 - \frac{\sqrt{2}}{2}\right)$ After simplifying and rounding, you should get the final answer in terms of $\pi$.

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Related Questions

1. What are the advantages of using spherical coordinates in volume integrals for symmetric solids?
2. How would the problem change if the cone had a different equation?
3. What if the density function was constant instead of varying with $\rho$?
4. How can you verify if the limits in spherical coordinates are correctly set for a given solid?
5. How does the volume element change in cylindrical vs. spherical coordinates?

Tip

For symmetrical shapes (like hemispheres or cones), using spherical or cylindrical coordinates simplifies the integral limits and reduces computational complexity.