Use polar coordinates to find the volume of the given solid.
above the cone 
z = 

root(x2 + y2)
 and below the sphere 
x2 + y2 + z2 = 9

To find the volume of the solid above the cone \( z = \sqrt{x^2 + y^2} \) and below the sphere \( x^2 + y^2 + z^2 = 9 \) using polar coordinates, we can follow these steps:

1. **Convert the equations to polar coordinates:**
   - In polar coordinates, \( x = r \cos \theta \) and \( y = r \sin \theta \).
   - The cone equation becomes:
     \[
     z = \sqrt{x^2 + y^2} = \sqrt{r^2} = r.
     \]
   - The sphere equation transforms to:
     \[
     r^2 + z^2 = 9 \implies z^2 = 9 - r^2 \implies z = \sqrt{9 - r^2}.
     \]

2. **Determine the bounds for \( r \):**
   - The cone intersects the sphere where \( z = r \) and \( z = \sqrt{9 - r^2} \):
     \[
     r = \sqrt{9 - r^2}.
     \]
   - Squaring both sides, we get:
     \[
     r^2 = 9 - r^2 \implies 2r^2 = 9 \implies r^2 = \frac{9}{2} \implies r = \frac{3}{\sqrt{2}}.
     \]

3. **Set up the volume integral:**
   - The volume \( V \) can be expressed as:
     \[
     V = \int_0^{2\pi} \int_0^{\frac{3}{\sqrt{2}}} (z_{\text{top}} - z_{\text{bottom}}) r \, dr \, d\theta,
     \]
   - where \( z_{\text{top}} = \sqrt{9 - r^2} \) (sphere) and \( z_{\text{bottom}} = r \) (cone).
   - Therefore, the integral becomes:
     \[
     V = \int_0^{2\pi} \int_0^{\frac{3}{\sqrt{2}}} \left( \sqrt{9 - r^2} - r \right) r \, dr \, d\theta.
     \]

4. **Evaluate the integral:**
   - First, calculate the inner integral:
     \[
     \int_0^{\frac{3}{\sqrt{2}}} \left( \sqrt{9 - r^2} - r \right) r \, dr = \int_0^{\frac{3}{\sqrt{2}}} \left( r\sqrt{9 - r^2} - r^2 \right) \, dr.
     \]

   - Let’s break it into two parts:
     \[
     \int_0^{\frac{3}{\sqrt{2}}} r\sqrt{9 - r^2} \, dr - \int_0^{\frac{3}{\sqrt{2}}} r^2 \, dr.
     \]

   - For the first integral, use the substitution \( u = 9 - r^2 \), which leads to:
     \[
     du = -2r \, dr \implies r \, dr = -\frac{1}{2} du.
     \]
     Changing the limits accordingly, when \( r = 0, u = 9 \), and when \( r = \frac{3}{\sqrt{2}}, u = 9 - \frac{9}{2} = \frac{9}{2} \):
     \[
     \int_9^{\frac{9}{2}} -\frac{1}{2} \sqrt{u} \, du = \frac{1}{2} \left[ \frac{2}{3} u^{3/2} \right]_9^{\frac{9}{2}} = \frac{1}{3} \left( \left( \frac{9}{2} \right)^{3/2} - 9^{3/2} \right).
     \]

   - For the second integral:
     \[
     \int_0^{\frac{3}{\sqrt{2}}} r^2 \, dr = \left[ \frac{r^3}{3} \right]_0^{\frac{3}{\sqrt{2}}} = \frac{1}{3} \left( \frac{27}{2\sqrt{2}} \right).
     \]

5. **Combine the results:**
   - The total volume becomes:
     \[
     V = \int_0^{2\pi} \left( \text{result of the inner integral} \right) d\theta.
     \]
   - Since the angular integral \( \int_0^{2\pi} d\theta = 2\pi \), we can multiply the result of the inner integral by \( 2\pi \).

After performing these calculations, you will get the volume of the solid. If you'd like me to compute the numerical result for the inner integral or need further details on any step, let me know! 

### Follow-up Questions:
1. Would you like to see the numerical evaluation of the integrals?
2. Do you want a more detailed breakdown of the substitution method used?
3. Are you interested in exploring different coordinate systems for this problem?
4. Would you like to learn about the geometric interpretation of the volume?
5. Do you have other similar volume problems to work on? 

### Tip:
When working with polar coordinates, always sketch the region of integration to clearly visualize the limits.

Use polar coordinates to find the volume of the solid above the cone z = sqrt(x^2 + y^2) and below the sphere x^2 + y^2 + z^2 = 9.

Explore the method to find the volume of a solid located above the cone z = sqrt(x^2 + y^2) and below the sphere x^2 + y^2 + z^2 = 9 using polar coordinates. This comprehensive guide includes step-by-step calculations, key formulas, and integral evaluations suitable for advanced high school 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