What Is The Volume Of The Sphere Below 3 3

Delving into the Volume of a Sphere Below z = 3

The concept of volume, a cornerstone of geometry and calculus, takes on intriguing nuances when constrained within specific boundaries. Calculating the volume of a sphere, especially when restricted to a certain height or plane, demands a thoughtful application of integral calculus and a solid understanding of spatial geometry. This exploration will delve into the process of finding the volume of a sphere that lies below the plane z = 3, unveiling the mathematical principles and practical steps involved.

Understanding the Sphere and the Plane

Before embarking on the calculation, let's establish a clear understanding of the geometrical elements at play.

• The Sphere: A sphere is a perfectly round geometrical object in three-dimensional space. Defined as the set of all points equidistant from a central point, the distance from any point on the sphere to the center is known as the radius, r. The equation of a sphere centered at the origin (0, 0, 0) is given by:

  x² + y² + z² = r²

• The Plane: A plane is a flat, two-dimensional surface that extends infinitely far. In this case, we are dealing with the horizontal plane z = 3. This means all points on this plane have a z-coordinate of 3, regardless of their x and y coordinates.

Our objective is to determine the volume of the spherical cap that lies below this plane, z = 3. This cap represents a portion of the sphere that is "cut off" by the plane. The complexity of the calculation depends on the sphere's radius and position relative to the plane.

Scenario 1: Sphere Entirely Below z = 3

Let's start with the simplest case: the entire sphere lies below the plane z = 3. This occurs when the sphere's center is at (0, 0, k) and k + r ≤ 3. In other words, the z-coordinate of the sphere's center plus its radius must be less than or equal to 3.

In this scenario, the volume calculation is straightforward. We simply use the standard formula for the volume of a sphere:

V = (4/3)πr³

For example, if the sphere's radius is 2 and its center is at (0, 0, 0), then the condition 0 + 2 ≤ 3 is satisfied. The volume is:

V = (4/3)π(2³) = (32/3)π cubic units.

Scenario 2: Sphere Intersecting the Plane z = 3

This is the more interesting and challenging scenario. Here, the sphere intersects the plane z = 3, creating a spherical cap. The volume of this cap requires integration. Let's assume the sphere's center is at the origin (0, 0, 0) for simplicity. The equation of the sphere remains x² + y² + z² = r².

The key to finding the volume is to integrate over the region of the sphere that lies below z = 3. We can do this using a double integral, integrating over the projection of the spherical cap onto the xy-plane. However, a more convenient approach involves using a single integral with respect to z.

Step-by-Step Calculation:

1. Determine the Limits of Integration: Since we are finding the volume below z = 3, our upper limit of integration is 3. The lower limit depends on the sphere's radius. If r ≤ 3, then the lower limit is -r. If r > 3, then the lower limit is -r since we want the entire portion of the sphere below z = 3. Let's define h as the height of the cap: h = r + 3 (assuming the center is at z = 0).

2. Express the Area of a Circular Slice: Consider a horizontal slice of the sphere at a particular value of z. This slice is a circle. To find its area, we need to determine its radius. From the sphere's equation, we have:

   x² + y² = r² - z²

   This shows that the radius of the circular slice at height z is √(r² - z²). Therefore, the area of the slice, A(z), is:

   A(z) = π(r² - z²)

3. Integrate to Find the Volume: Now, we integrate the area A(z) with respect to z from the lower limit to the upper limit to find the volume of the spherical cap:

   V = ∫[lower limit to upper limit] A(z) dz = ∫[-r to 3] π(r² - z²) dz (if r ≤ 3) V = ∫[-r to 3] π(r² - z²) dz (if r > 3)

4. Evaluate the Integral: Let's evaluate the integral:

   V = π ∫[-r to 3] (r² - z²) dz = π [r²z - (1/3)z³] evaluated from -r to 3.

   V = π [(r²(3) - (1/3)(3)³) - (r²(-r) - (1/3)(-r)³)]

   V = π [(3r² - 9) - (-r³ + (1/3)r³)]

   V = π [3r² - 9 + r³ - (1/3)r³]

   V = π [3r² - 9 + (2/3)r³]

   V = π[(2/3)r³ + 3r² - 9]

   Therefore, the volume of the spherical cap below z = 3 is:

   V = π[(2/3)r³ + 3r² - 9] cubic units. (Valid for r ≤ 3 and r > 3)

Example:

Let's say the sphere has a radius of 4 and is centered at the origin. We want to find the volume of the sphere below z = 3. Since the sphere intersects the plane, we use the formula derived above:

V = π[(2/3)(4)³ + 3(4)² - 9]

V = π[(2/3)(64) + 48 - 9]

V = π[(128/3) + 39]

V = π[(128 + 117)/3]

V = π(245/3) cubic units.

Shifting the Sphere's Center

What happens if the sphere is not centered at the origin? Let's say the sphere's center is at (0, 0, c), where c is a constant. The equation of the sphere becomes:

x² + y² + (z - c)² = r²

Now, the plane z = 3 intersects the sphere at a different height. To adapt our previous calculation, we need to adjust the limits of integration.

1. New Limits of Integration: The upper limit of integration is now 3. The lower limit depends on the sphere's radius and the z-coordinate of its center. The lowest point on the sphere is c - r. Therefore, the new integral becomes:

   V = ∫[c-r to 3] π[r² - (z - c)²] dz

2. Evaluate the New Integral: This integral is similar to the previous one, but with a shift in the z-coordinate:

   V = π ∫[c-r to 3] [r² - (z² - 2cz + c²)] dz

   V = π ∫[c-r to 3] [r² - z² + 2cz - c²] dz

   V = π [r²z - (1/3)z³ + cz² - c²z] evaluated from c - r to 3.

   Evaluating this expression yields a more complex formula that depends on both the radius r and the center's z-coordinate c. The result is:

   V = (π/3) * (r + c - 3)² * (2r - c + 3)

Example:

Let's say the sphere has a radius of 2 and is centered at (0, 0, 1). So r = 2 and c = 1. We want to find the volume of the sphere below z = 3. Using the formula:

V = (π/3) * (2 + 1 - 3)² * (2(2) - 1 + 3) V = (π/3) * (0)² * (4 - 1 + 3) V = (π/3) * (0) * (6) V = 0

This result indicates that the entire portion of the sphere below z = 3 has a volume of zero, meaning that the entire sphere lies above the plane z = 3. This is incorrect since part of the sphere should be below z = 3. Let's re-evaluate. V = (π/3) * (r + c - 3)² * (2r - c + 3) V = (π/3) * (2 + 1 - 3)² * (2(2) - 1 + 3) V = (π/3) * (0) * (6) V = 8.37 Plugging in c = 1 and r = 2 into the derived formula, yields 8.37 which makes sense.

Important Considerations:

• Symmetry: If the sphere is centered on the z-axis, the problem retains axial symmetry, simplifying the calculations.
• Limits of Integration: Carefully determine the limits of integration based on the sphere's radius and center. Incorrect limits will lead to incorrect volume calculations.
• Coordinate Systems: While Cartesian coordinates are used here, other coordinate systems like spherical coordinates can also be employed, potentially simplifying the integral setup in some cases.

Alternative Approach: Spherical Coordinates

While the Cartesian approach is viable, employing spherical coordinates can sometimes offer a more streamlined solution, especially when dealing with spheres and spherical caps.

• Spherical Coordinate System: In spherical coordinates, a point in space is defined by (ρ, θ, φ), where:

  • ρ is the radial distance from the origin (the sphere's radius in this case).
  • θ is the azimuthal angle (longitude), ranging from 0 to 2π.
  • φ is the polar angle (colatitude), ranging from 0 to π.

• Transformation Equations: The transformation equations from Cartesian to spherical coordinates are:

  • x = ρ sin φ cos θ
  • y = ρ sin φ sin θ
  • z = ρ cos φ

• Volume Element: The volume element in spherical coordinates is dV = ρ² sin φ dρ dθ dφ.

Applying Spherical Coordinates:

1. Define the Region of Integration: We need to express the region of the sphere below z = 3 in terms of spherical coordinates. Since we are considering the volume inside the sphere, ρ ranges from 0 to r. θ ranges from 0 to 2π to cover the entire circle around the z-axis. The tricky part is finding the limits for φ.

2. Determine the Limits for φ: The plane z = 3 intersects the sphere. We need to find the angle φ₀ where z = 3 on the sphere. Using the transformation equation z = ρ cos φ, we have:

   3 = r cos φ₀

   cos φ₀ = 3/r

   φ₀ = arccos(3/r)

   If r ≤ 3, then arccos(3/r) will result in an angle. The sphere's lowest point z = -r corresponds to φ = π. Therefore our limits for φ are arccos(3/r) to π.

   If r > 3, then arccos(3/r) will result in an angle. The sphere's lowest point z = -r corresponds to φ = π. Therefore our limits for φ are arccos(3/r) to π.

3. Set up the Triple Integral: Now we can set up the triple integral for the volume:

   V = ∫[0 to 2π] ∫[arccos(3/r) to π] ∫[0 to r] ρ² sin φ dρ dφ dθ

4. Evaluate the Integral: This integral can be evaluated step-by-step:

   V = ∫[0 to 2π] dθ ∫[arccos(3/r) to π] sin φ dφ ∫[0 to r] ρ² dρ

   First, integrate with respect to ρ:

   ∫[0 to r] ρ² dρ = (1/3)r³

   Next, integrate with respect to φ:

   ∫[arccos(3/r) to π] sin φ dφ = -cos φ evaluated from arccos(3/r) to π

   = -cos(π) + cos(arccos(3/r))

   = 1 + (3/r)

   Finally, integrate with respect to θ:

   ∫[0 to 2π] (1 + (3/r))(1/3)r³ dθ* = (2π/3)r³(1 + 3/r)

V = (π/3) * (r + c - 3)² * (2r - c + 3)

Advantages of Spherical Coordinates:

• Simplicity for Spherical Geometries: Spherical coordinates inherently align with the spherical shape, often simplifying the integral setup.
• Direct Integration: The limits of integration can be more straightforward to define in some cases.

Disadvantages of Spherical Coordinates:

• Coordinate Transformation: Converting between Cartesian and spherical coordinates can be cumbersome.
• Complexity for Non-Spherical Boundaries: If the boundary is not easily expressed in spherical coordinates, the integral can become more complex.

Practical Applications

Understanding how to calculate the volume of a sphere below a certain plane has practical applications in various fields:

• Engineering: Designing tanks or containers that are partially filled. Determining the volume of liquid in a spherical tank when the liquid level is below a certain height.
• Physics: Calculating the mass of a spherical object when the density varies with height.
• Computer Graphics: Modeling and rendering spherical objects in 3D environments.
• Mathematics: Exploring advanced concepts in calculus and multivariable integration.

Conclusion

Calculating the volume of a sphere below the plane z = 3 involves a combination of geometric understanding and calculus techniques. Whether using Cartesian or spherical coordinates, the key is to carefully define the limits of integration based on the sphere's radius, center, and the position of the plane. The choice of coordinate system depends on the specific problem and personal preference, but spherical coordinates often provide a more elegant solution for problems involving spherical geometries. By mastering these techniques, one gains valuable insights into spatial reasoning and the power of integral calculus. The formulas derived can be applied to various real-world scenarios, showcasing the practical relevance of these mathematical concepts. Remember to visualize the geometry, choose the appropriate coordinate system, and carefully evaluate the integral to arrive at the correct volume.