To find the volume of the solid bounded below by the cone $z = \sqrt{x^2 + y^2}$ and bounded above by the sphere $x^2 + y^2 + z^2 = 450$, we can use cylindrical coordinates. In cylindrical coordinates:

$x = r \cos \theta, \quad y = r \sin \theta, \quad z = z$

In these coordinates:

• The cone $z = \sqrt{x^2 + y^2}$ becomes $z = r$.
• The sphere $x^2 + y^2 + z^2 = 450$ becomes $r^2 + z^2 = 450$.

Step 1: Set up the bounds

We need to determine the region of integration. The solid is bounded below by the cone $z = r$, and above by the sphere $z = \sqrt{450 - r^2}$.

• The limits for $r$ come from the projection of the region onto the $xy$-plane. The radius of the base of the solid is determined by the intersection of the cone and the sphere. From the equation $r^2 + z^2 = 450$, substitute $z = r$ to find the intersection: $r^2 + r^2 = 450 \quad \Rightarrow \quad 2r^2 = 450 \quad \Rightarrow \quad r = \sqrt{225} = 15$ Therefore, $r$ ranges from 0 to 15.

• The limits for $z$ are from $z = r$ (the cone) to $z = \sqrt{450 - r^2}$ (the sphere).

• The limits for $\theta$ go from $0$ to $2\pi$, since the region is symmetric about the $z$-axis.

Step 2: Write the volume integral

In cylindrical coordinates, the volume element $dV$ is given by $r \, dz \, dr \, d\theta$. Thus, the volume of the solid is:

$V = \int_0^{2\pi} \int_0^{15} \int_r^{\sqrt{450 - r^2}} r \, dz \, dr \, d\theta$

Step 3: Compute the integrals

1. Integrating with respect to $z$: $\int_r^{\sqrt{450 - r^2}} r \, dz = r \left( \sqrt{450 - r^2} - r \right)$

2. Integrating with respect to $r$: Now, we compute: $\int_0^{15} r \left( \sqrt{450 - r^2} - r \right) \, dr$ This splits into two integrals: $\int_0^{15} r \sqrt{450 - r^2} \, dr - \int_0^{15} r^2 \, dr$ The second integral is straightforward: $\int_0^{15} r^2 \, dr = \frac{r^3}{3} \Big|_0^{15} = \frac{15^3}{3} = 1125$

   For the first integral, use the substitution $u = 450 - r^2$, hence $du = -2r \, dr$, and the limits change from $r = 0$ to $r = 15$, corresponding to $u = 450$ to $u = 0$: $\int_0^{15} r \sqrt{450 - r^2} \, dr = -\frac{1}{2} \int_{450}^0 \sqrt{u} \, du = \frac{1}{2} \int_0^{450} \sqrt{u} \, du$ Now, integrate: $\frac{1}{2} \int_0^{450} u^{1/2} \, du = \frac{1}{2} \cdot \frac{2}{3} u^{3/2} \Big|_0^{450} = \frac{1}{3} (450)^{3/2}$ Simplifying $450^{3/2}$: $450^{3/2} = 450 \cdot \sqrt{450} = 450 \cdot 21.213 = 9545.85$ Therefore: $\frac{1}{3} \cdot 9545.85 \approx 3181.95$

   So, the total integral is: $3181.95 - 1125 = 2056.95$

3. Integrating with respect to $\theta$: Finally, multiply by $\int_0^{2\pi} d\theta = 2\pi$: $V = 2\pi \times 2056.95 \approx 12924.58$

Thus, the volume of the solid is approximately $12924.58$ cubic units.

Do you have any questions or need more details?

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Here are five related questions:

1. How would you find the surface area of the solid instead of the volume?
2. What changes in the integral if the cone has a different angle?
3. How do you adapt the approach if the cone is open at the bottom (z = 0)?
4. Can you use spherical coordinates instead of cylindrical for this problem?
5. How would you calculate the volume if the sphere had a different radius?

Tip: Substitution methods simplify integrals when functions inside are complex.