Problem 41 - University Calculus: Early Transcendentals

Question

Let \(D\) be the smaller cap cut from a solid ball of radius 2 units by a plane 1 unit from the center of the sphere. Express the volume of \(D\) as an iterated triple integral in (a) spherical, (b) cylindrical, and (c) rectangular coordinates. Then (d) find the volume by evaluating one of the three triple integrals.

Step-by-Step Solution

Answer

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

Step 1: Define the sphere and plane

The solid ball is a sphere with a radius of 2 units, centered at the origin. The equation of the sphere is \(x^2 + y^2 + z^2 = 4\). The plane that cuts the sphere is given by \(z = 1\). The smaller cap is above this plane.

Step 2: Set up the integral in spherical coordinates

In spherical coordinates, the variables are \((\rho, \theta, \phi)\), where \(\rho^2 = x^2 + y^2 + z^2\), \(\theta\) is the azimuthal angle, and \(\phi\) is the polar angle. The volume element is \(dV = \rho^2 \sin(\phi) d\rho d\theta d\phi\). The limits are \(\rho: 1 \leq \rho \leq 2\), \(\theta: 0 \leq \theta \leq 2\pi\), and \(\phi: 0 \leq \phi \leq \arccos(1/2)\).

Step 3: Set up the integral in cylindrical coordinates

In cylindrical coordinates, the variables are \((r, \theta, z)\). The volume element is \(dV = r \, dr \, d\theta \, dz\). The limits are \(r: 0 \leq r \leq \sqrt{3}\), \(\theta: 0 \leq \theta \leq 2\pi\), and \(z: 1 \leq z \leq \sqrt{4 - r^2}\).

Step 4: Set up the integral in rectangular coordinates

In rectangular coordinates \((x, y, z)\), the equation of the sphere is \(x^2 + y^2 + z^2 = 4\) and the plane is \(z = 1\). The limits for \(z\) are \(1 \leq z \leq \sqrt{4 - x^2 - y^2}\). For \(x\) and \(y\), considering a circle projected onto the \((x, y)\)-plane: \(x^2 + y^2 \leq 3\). So, the limits are \(y: -\sqrt{3 - x^2} \leq y \leq \sqrt{3 - x^2}\) and \(x: -\sqrt{3} \leq x \leq \sqrt{3}\).

Step 5: Evaluate the spherical coordinate integral

Choose to evaluate the spherical coordinate integral. The integral is:\[\int_{0}^{2\pi} \int_{0}^{\arccos(1/2)} \int_{1}^{2} \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta\]Calculate the integral step by step:1. Integrate with respect to \(\rho\) from 1 to 2: \[\int_{1}^{2} \rho^2 \, d\rho = \left[ \frac{\rho^3}{3} \right]_{1}^{2} = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}\]2. Integrate with respect to \(\phi\):\[\int_{0}^{\arccos(1/2)} \frac{7}{3} \sin\phi \, d\phi = \frac{7}{3} \left[-\cos\phi\right]_{0}^{\arccos(1/2)} = \frac{7}{3} \left(-\frac{1}{2} + 1\right) = \frac{7}{6}\]3. Integrate with respect to \(\theta\):\[\int_{0}^{2\pi} \frac{7}{6} \, d\theta = \frac{7}{6} [\theta]_{0}^{2\pi} = \frac{7}{6} \times 2\pi = \frac{7\pi}{3}\]

Step 6: Final Answer: Volume of the cap

The volume of the smaller cap cut from the sphere is \(\frac{7\pi}{3}\) cubic units.

Key Concepts

Spherical Coordinates

Spherical coordinates provide a handy way to address problems dealing with spheres or spherical bodies. In this system, a point in space is identified by three values:

• \( \rho \): the radial distance from the origin.
• \( \theta \): the azimuthal angle, which measures the angle from the positive x-axis in the \((x, y)\) plane.
• \( \phi \): the polar angle, defined as the angle from the positive z-axis toward the \((x, y)\)-plane.

The conversion formulas for transforming Cartesian coordinates to spherical coordinates are:

• \( x = \rho \sin\phi \cos\theta \)
• \( y = \rho \sin\phi \sin\theta \)
• \( z = \rho \cos\phi \)

The volume element, \( dV \), in spherical coordinates is \( \rho^2 \sin\phi \, d\rho \, d\theta \, d\phi \), helpful for integrating over spheres. In our exercise, the boundaries for \( \rho \), \( \theta \), and \( \phi \) define the specific cap of the sphere that is desired.

Cylindrical Coordinates

Cylindrical coordinates are perfect for problems involving circles and symmetry around an axis, like a cylinder. Here, points in space are represented with:

• \( r \): the radial distance from the z-axis.
• \( \theta \): the azimuthal angle, as in spherical coordinates, but restricted to the \((x, y)\)-plane.
• \( z \): the height above the \((x, y)\)-plane.

The cylindrical to Cartesian conversion equations are:

• \( x = r \cos\theta \)
• \( y = r \sin\theta \)
• \( z = z \)

The element of volume, \( dV \), is denoted as \( r \, dr \, d\theta \, dz \) in these coordinates. This system simplifies problems involving rotational symmetry. For the cap from the sphere, the chosen bounds allow us to evaluate from the flat plane \( z = 1 \) to the spherical surface, converting the spherical boundary to an analogous form in cylindrical terms.

Rectangular Coordinates

Rectangular coordinates, or Cartesian coordinates, are the most straightforward system familiar to all. They identify points using:

• \( x \): horizontal distance (left/right).
• \( y \): horizontal distance (forward/backward).
• \( z \): vertical distance (up/down).

While they aren't always the most efficient for spherical problems, they provide a basis for building intuitions about spatial integration. The volume element remains \( dV = dx \, dy \, dz \). During integration for the cap of the sphere, the bounds for each variable are derived from the equations defining the sphere and the intersecting plane, creating a manageable form for computation via triple integration within those fixed limits.

Volume of a Sphere

Finding the volume of a sphere, specifically spherical caps or segments, often leads to employing integrals in varying coordinate systems. A sphere of radius \( R \) has the volume formula: \[ V = \frac{4}{3}\pi R^3 \]For a cap, where a plane intersects a sphere, integration becomes imperative. In our context, iterating through the limits of applicable coordinates assists in deriving the accurate piece-volume. Each coordinate system provides unique simplifications and insights:

• Spherical coordinates directly reflect the spherical shape for easier integration over angles and radius.
• Cylindrical coordinates treat the cross-sections parallel to the \((x, y)\)-plane as manageable circles.
• Rectangular coordinates treat the problem using familiar x-y-z geometry, adaptable for computational approaches.

Thus, being able to transition between these forms is powerful for approaching different geometrical conditions.

Iterated Integrals

Iterated integrals offer a methodical way to calculate volumes, particularly in multivariable calculus settings. By systematically nesting integrations, one can solve complex volumetric challenges involving multiple integral layers.Here's how iterated integrals work in our sphere cap context:

• First, align limits and functions for each coordinate system.
• Begin with the innermost integral, dealing with the variable having the tightest bounds (often \( \rho \) or \( r \)).
• Progress outward, solving each successive integral dependent on those inner results.

Using iterated integrals, we evaluate sequences of nested integrals corresponding to the bounds of the volume we seek. This process integrates complexity piece by piece, ensuring manageable computational steps for even intricate 3D structures like our spherical cap. The triple integral, especially in spherical coordinates for volume calculation, iterates first over \( \rho \), then \( \phi \), and finally \( \theta \), as it naturally aligns with this problem's symmetry.