Problem 73 - Thomas Calculus

Question

Sphere and plane Find the volume of the smaller region cut from the solid sphere \(\rho \leq 2\) by the plane \(z=1 .\)

Step-by-Step Solution

Answer

The volume of the smaller region is \(\frac{7\pi}{3}\).

Step 1: Identify Region of Interest

The given sphere is defined by the equation \(\rho \leq 2\), which represents a sphere of radius 2 centered at the origin. The plane \(z = 1\) slices through this sphere horizontally. We need to find the volume of the smaller region that is above this plane and inside the sphere.

Step 2: Sphere Equation Transformation

The sphere in spherical coordinates is defined by \(\rho^2 = x^2 + y^2 + z^2 = 4\). In Cartesian coordinates, the region below the sphere is cut off by the plane \(z = 1\). Converting this point in polar coordinates, we use \(z = \rho \cos \phi\). Here, it simplifies the problem to spherical integration to find the volume above the plane.

Step 3: Determine Limits of Integration

For the region above the plane, the limits are: \(z = 1\), thus \(\rho = \sqrt{4 - 1^2} \Rightarrow \rho = \sqrt{3}\). The limits become from \(z = 1\) to \(z = 2\) and in spherical coordinates \(\rho\) from \(1\) to \(2\), \(\phi\) from \(0\) to \(\cos^{-1}(1/2)\), and \(\theta\) from \(0\) to \(2\pi\).

Step 4: Setup the Integral

We set up the integral in spherical coordinates: \[ V = \int_0^{2\pi} \int_0^{\cos^{-1}(1/2)} \int_1^2 \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta. \]

Step 5: Solve The Integral

Begin with the innermost integral over \(\rho\): \[ \int_1^2 \rho^2 \, d\rho = \left[ \frac{\rho^3}{3} \right]_1^2 = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}. \] Then solve the integral over \(\phi\): \[ \int_0^{\cos^{-1}(1/2)} \sin \phi \, d\phi = -\cos \phi \bigg|_0^{\cos^{-1}(1/2)} = - \left( \frac{1}{2} - 1 \right) = \frac{1}{2}. \] Finally, integrate over \(\theta\): \[ \int_0^{2\pi} d\theta = 2\pi. \]

Step 6: Calculate Final Volume

Combine the computed integrals to find the volume: \[ V = \frac{7}{3} \cdot \frac{1}{2} \cdot 2\pi = \frac{7\pi}{3}. \] Thus, the volume of the smaller region cut from the sphere by the plane is \( \frac{7\pi}{3} \).

Key Concepts

Spherical Coordinates

Spherical coordinates provide a way to describe points in a three-dimensional space using three variables:

• \( \rho \) for the radial distance from the origin,
• \( \theta \) for the azimuthal angle in the \( xy \)-plane, and
• \( \phi \) for the polar angle measured from the positive \( z \)-axis.

This system is particularly useful for solving problems involving spheres, as it simplifies calculations and integral setups. The equation \( \rho^2 = x^2 + y^2 + z^2 \) showcases the relationship between spherical and Cartesian coordinates. In this specific problem, converting limits from Cartesian to spherical coordinates made the integration process more straightforward. The integration uses the bounds \( \rho \) from \( 1 \) to \( 2 \), \( \phi \) from \( 0 \) to \( \cos^{-1}(1/2) \), and \( \theta \) from \( 0 \) to \( 2\pi \) as articulated by the region of interest. This transformation leverages the symmetry of the sphere to easily slice out the volume cut by the plane \( z = 1 \).

Integral Calculus

Integral calculus is essential for calculating areas, volumes, and other quantities by summing infinitesimal parts. When faced with the problem of finding the volume of a solid, we employ integrals to accumulate tiny volumes together. In spherical coordinates, the elemental volume is expressed as \( \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \), representing a small chunk of space. To calculate the volume, we integrate this expression across the specified limits. This exercise is a prime example of using triple integrals to find the volume of a defined region, a task best suited for integral calculus. We start our integration from the innermost variable to the outermost—first \( \rho \), then \( \phi \), and finally \( \theta \). These integrations, performed step by step, combine to yield the total volume of the segment sliced by the plane within the sphere.

Volume of Solids

Calculating the volume of solids, especially irregular ones like partial spheres, is a key application of calculus. The problem of finding the volume sliced off by a plane from a sphere is inherently complex due to the shape involved. By employing spherical coordinates and triple integrals, we can precisely calculate this volume.

• First, determine the shape and boundaries of the volume in question—the intersection of the sphere and the plane.
• Next, decide the integration limits, which are crucial for arriving at the correct volume.
• Finally, compute the integral to sum all the infinitesimal parts.

For this particular sphere, the calculated volume is \( \frac{7\pi}{3} \), representing the smaller region cut from the sphere by the plane \( z = 1 \). This approach exemplifies how calculus can extend beyond simple shapes to solve real-world problems involving complex geometries.