Find the volume of the smaller region cut from the solid sphere rho less than or equals 10 by the plane z equals 5.

To find the volume of the smaller region cut from the sphere \( \rho \leq 10 \) by the plane \( z = 5 \), we can set up the problem in spherical coordinates.

### Step 1: Understanding the Setup

The equation \( \rho \leq 10 \) describes a sphere with radius \( 10 \), centered at the origin. The plane \( z = 5 \) cuts this sphere, creating two regions. We want the volume of the smaller region.

### Step 2: Finding the Radius of the Circle of Intersection

1. The plane \( z = 5 \) intersects the sphere at a circle. To find the radius of this circle, substitute \( z = 5 \) into the equation of the sphere.

   \[
   x^2 + y^2 + z^2 = 10^2
   \]
   
   Substituting \( z = 5 \):

   \[
   x^2 + y^2 + 5^2 = 10^2
   \]
   \[
   x^2 + y^2 = 100 - 25 = 75
   \]

   So, the radius of the circle of intersection is \( \sqrt{75} = 5\sqrt{3} \).

### Step 3: Setting Up the Integral in Spherical Coordinates

In spherical coordinates:
- \( \rho \) represents the radial distance, ranging from \( 0 \) to \( 10 \).
- \( \theta \) represents the azimuthal angle, ranging from \( 0 \) to \( 2\pi \).
- \( \phi \) is the polar angle, which will range from \( 0 \) to \( \phi_0 \), where \( \phi_0 \) is the angle at which the plane \( z = 5 \) intersects the sphere.

From spherical coordinates, we know \( z = \rho \cos \phi \). Since \( z = 5 \) on the plane, we get:

\[
5 = \rho \cos \phi
\]

Using \( \rho = 10 \) at the boundary of the sphere:

\[
5 = 10 \cos \phi \Rightarrow \cos \phi = \frac{1}{2} \Rightarrow \phi = \frac{\pi}{3}
\]

So, \( \phi \) ranges from \( 0 \) to \( \frac{\pi}{3} \) in the smaller region.

### Step 4: Volume Integral

The volume element in spherical coordinates is \( dV = \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \).

The volume of the smaller region is given by:

\[
V = \int_0^{2\pi} \int_0^{\pi/3} \int_0^{10} \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta
\]

### Step 5: Evaluating the Integral

1. **Integrate with respect to \( \rho \):**

   \[
   \int_0^{10} \rho^2 \, d\rho = \left[ \frac{\rho^3}{3} \right]_0^{10} = \frac{10^3}{3} = \frac{1000}{3}
   \]

2. **Integrate with respect to \( \phi \):**

   \[
   \int_0^{\pi/3} \sin \phi \, d\phi = \left[ -\cos \phi \right]_0^{\pi/3} = -\cos\left(\frac{\pi}{3}\right) + \cos(0) = -\frac{1}{2} + 1 = \frac{1}{2}
   \]

3. **Integrate with respect to \( \theta \):**

   \[
   \int_0^{2\pi} d\theta = 2\pi
   \]

### Step 6: Combining Results

The volume \( V \) is:

\[
V = \frac{1000}{3} \cdot \frac{1}{2} \cdot 2\pi = \frac{1000 \pi}{3}
\]

### Final Answer

The volume of the smaller region cut from the solid sphere by the plane \( z = 5 \) is:

\[
V = \frac{1000 \pi}{3}
\]

Would you like more details, or do you have any questions?

---

Here are some questions that build on this solution:

1. How would the solution change if the sphere had a different radius?
2. What if the plane were positioned at a different \( z \)-value?
3. How would this problem differ if we wanted the volume of the larger region?
4. Can we solve this problem using Cartesian coordinates instead?
5. How would the volume change if the sphere were intersected by two planes?

**Tip:** When setting up problems in spherical coordinates, always clearly define the ranges for \( \rho \), \( \phi \), and \( \theta \) based on the geometry of the region in question.

Calculate the volume of the smaller region formed when a plane intersects a sphere with radius 10 at z = 5. This detailed solution covers setting up the volume integral in spherical coordinates, determining limits for integration, and computing the final volume for advanced calculus 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