se a triple integral to find the volume of the solid bounded below by the cone 

zequals=StartRoot x squared plus y squared EndRootx2+y2 

and bounded above by the sphere 

xsquared2plus+ysquared2plus+zsquared2equals=450450


To find the volume of the solid bounded below by the cone \( z = \sqrt{x^2 + y^2} \) and above by the sphere \( x^2 + y^2 + z^2 = 450 \), we can set up a triple integral in spherical coordinates, as this coordinate system simplifies integrals over regions with radial symmetry.

### Step 1: Set Up in Spherical Coordinates
In spherical coordinates:
\[
x = \rho \sin \phi \cos \theta, \quad y = \rho \sin \phi \sin \theta, \quad z = \rho \cos \phi
\]
where:
- \( \rho \) is the radial distance,
- \( \theta \) is the azimuthal angle in the \( xy \)-plane (from \( 0 \) to \( 2\pi \)),
- \( \phi \) is the polar angle (from \( 0 \) to \( \pi \)).

The volume element \( dV \) in spherical coordinates is given by:
\[
dV = \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta
\]

### Step 2: Determine the Bounds
1. **Sphere Bound**: The sphere has the equation \( x^2 + y^2 + z^2 = 450 \), so in spherical coordinates, \( \rho = \sqrt{450} \).
2. **Cone Bound**: The cone equation \( z = \sqrt{x^2 + y^2} \) in spherical coordinates translates to \( \rho \cos \phi = \rho \sin \phi \), leading to \( \tan \phi = 1 \). Therefore, \( \phi = \frac{\pi}{4} \) is the lower bound for \( \phi \).

Thus, the bounds for each variable are:
- \( 0 \leq \theta \leq 2\pi \)
- \( \frac{\pi}{4} \leq \phi \leq \frac{\pi}{2} \)
- \( 0 \leq \rho \leq \sqrt{450} \)

### Step 3: Set Up the Integral
Now, we can express the volume \( V \) as:
\[
V = \int_0^{2\pi} \int_{\pi/4}^{\pi/2} \int_0^{\sqrt{450}} \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta
\]

### Step 4: Evaluate the Integral
Let's evaluate each part of this integral:

1. **Integrate with respect to \( \rho \):**
   \[
   \int_0^{\sqrt{450}} \rho^2 \, d\rho = \left[ \frac{\rho^3}{3} \right]_0^{\sqrt{450}} = \frac{(450)^{3/2}}{3}
   \]

2. **Integrate with respect to \( \phi \):**
   \[
   \int_{\pi/4}^{\pi/2} \sin \phi \, d\phi = \left[ -\cos \phi \right]_{\pi/4}^{\pi/2} = 1 - \frac{\sqrt{2}}{2}
   \]

3. **Integrate with respect to \( \theta \):**
   \[
   \int_0^{2\pi} d\theta = 2\pi
   \]

### Step 5: Combine Results
The volume \( V \) becomes:
\[
V = \frac{(450)^{3/2}}{3} \cdot \left( 1 - \frac{\sqrt{2}}{2} \right) \cdot 2\pi
\]
Simplifying gives the exact volume of the region.

Would you like more details on any step, or have further questions?

---

1. How is the cone's equation expressed in cylindrical coordinates?
2. What is the difference between cylindrical and spherical coordinate systems?
3. How does the integral setup change if we change the solid bounds?
4. What techniques apply for non-symmetric regions?
5. How would a Cartesian triple integral look for this problem?

**Tip**: When working with bounded regions, always identify symmetry to choose the best coordinate system.

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 = 450.

Learn how to use a triple integral in spherical coordinates to find the volume of a solid bounded by a cone and a sphere. This step-by-step solution covers key concepts in calculus, including volume calculation, coordinate transformation, and integration techniques. Suitable for college-level students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation