Use a triple integral to find the volume of the solid bounded below by the cone zequalsStartRoot x squared plus y squared EndRoot and bounded above by the sphere xsquaredplusysquaredpluszsquaredequals450.


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.

Use a triple integral 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.

Learn how to compute the volume of a solid bounded by the cone z = sqrt(x^2 + y^2) and the sphere x^2 + y^2 + z^2 = 450 using a triple integral in cylindrical coordinates. This step-by-step guide explains setting integration limits, simplifying integrals, and applying substitution. Suitable for undergraduate calculus students.

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