Problem 63 - Thomas Calculus

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 cap is \( \frac{2\pi}{3}(4 \sqrt{3} - 3) \).

Step 1: Identify the problem and requirements

We have a sphere with a radius of 2 units that is intersected by a plane 1 unit from the center. We need to express the volume of the resulting spherical cap as a triple integral in three different coordinate systems: spherical, cylindrical, and rectangular. Then, we will evaluate one of these integrals.

Step 2: Understand the geometry of the problem

The sphere has its center at the origin, and a radius of 2 units. The plane is positioned 1 unit above the origin, cutting through the sphere, forming a smaller cap. We need to determine the appropriate region over which to integrate.

Step 3: Set up the spherical coordinates integral

In spherical coordinates, we have \( x = ho \sin \phi \cos \theta \), \( y = ho \sin \phi \sin \theta \), and \( z = \rho \cos \phi \). The region extends vertically from \( z = 1 \) to \( z = 2 \). The limits in spherical coordinates are: \( 1 \leq \rho \cos \phi \leq 2 \), \( 0 \leq \theta \leq 2\pi \), and finding \( \phi \) range involves solving for the corresponding \( \rho \) range such that \( \cos^{-1}(1/\rho) \leq \phi \leq \pi/2 \). The integral becomes \( \int_0^{2\pi} \int_0^{\cos^{-1}(1/2)} \int_{1/\cos(\phi)}^{2} \rho^2\sin(\phi) \, d\rho \, d\phi \, d\theta \).

Step 4: Set up the cylindrical coordinates integral

In cylindrical coordinates, we have \( x = r \cos \theta \), \( y = r \sin \theta \), and \( z = z \). The equation of the sphere is \( r^2 + z^2 = 4 \). The limits are: \( z \) from 1 to \( \sqrt{4-r^2} \), \( r \) from 0 to \( \sqrt{3} \) (as the cap is at the top, \( z = 1 \) is the plane cutting the sphere), and \( \theta \) from 0 to \( 2\pi \). The integral becomes \( \int_0^{2\pi} \int_0^{\sqrt{3}} \int_1^{\sqrt{4-r^2}} r \, dz \, dr \, d\theta \).

Step 5: Set up the rectangular coordinates integral

In rectangular coordinates, the sphere is \( x^2 + y^2 + z^2 = 4 \) and the plane is \( z = 1 \). We solve for limits: \( z \) is from 1 to \( \sqrt{4-x^2-y^2} \). The 2D region in the \( xy \)-plane is \( x^2 + y^2 \leq 3 \): \( x \) from \(-\sqrt{3} \) to \( \sqrt{3} \) and for each \( x \), \( y \) from \(-\sqrt{3-x^2}\) to \( \sqrt{3-x^2} \). The integral becomes \( \int_{-\sqrt{3}}^{\sqrt{3}} \int_{-\sqrt{3-x^2}}^{\sqrt{3-x^2}} \int_1^{\sqrt{4-x^2-y^2}} \, dz \, dy \, dx \).

Step 6: Evaluate the cylindrical coordinates integral (chosen option)

Using cylindrical coordinates, evaluate \( \int_0^{2\pi} \int_0^{\sqrt{3}} \int_1^{\sqrt{4-r^2}} r \, dz \, dr \, d\theta \). Solve the inner integral \( \int_1^{\sqrt{4-r^2}} r \, dz = r(\sqrt{4-r^2} - 1) \). Now the integral is \( \int_0^{2\pi} \int_0^{\sqrt{3}} r(\sqrt{4-r^2} - 1) \, dr \, d\theta \). Perform a substitution \( u = 4 - r^2 \), and solve the middle integral to get \( \frac{\pi}{3}(4 \sqrt{3} - 3) \). Using \(\theta\) integration, multiply by \(2\pi\) to finish with: \( \frac{2\pi}{3}(4 \sqrt{3} - 3) \).

Step 7: Final result interpretation

The calculated volume of the smaller spherical cap is: \( \frac{2\pi}{3}(4 \sqrt{3} - 3) \). This represents the space occupied within the sphere, below the plane and above the cap.

Key Concepts

Spherical Coordinates

Spherical coordinates offer a streamlined method for solving calculations involving spheres or ball-like objects, especially when looking at three-dimensional scenarios. They provide a way to define a point in 3D space using three parameters: the radius, the polar angle, and the azimuthal angle.

• Radius (\( \rho \)): This measures the distance from the origin to the point in 3D space.
• Polar Angle (\( \phi \)): Typically represents the angle between the positive z-axis and the line connecting the origin and the point.
• Azimuthal Angle (\( \theta \)): This is the angle made with the positive x-axis and is commonly used to define rotation around the z-axis.

In this exercise, spherical coordinates are particularly helpful in setting up and evaluating the volume of a spherical cap since the sphere's symmetry naturally aligns with spherical coordinates. The relationship through which each Cartesian coordinate (\( x, y, z \)) is expressed in terms of spherical coordinates involves:
\( x = \rho \sin \phi \cos \theta \)
\( y = \rho \sin \phi \sin \theta \)
\( z = \rho \cos \phi \)These formulas allow us to convert between spherical and Cartesian coordinates seamlessly, simplifying the integration over spherical volumes.

Cylindrical Coordinates

Cylindrical coordinates are practical for systems involving rotational symmetry around an axis, such as cylinders. They efficiently extend polar coordinates into three dimensions by adding a height parameter to the model.

• Radial distance (\( r \)): The distance from the z-axis to the projection of the point in the xy-plane.
• Azimuthal angle (\( \theta \)): The angle between the x-axis and the line from the z-axis to the point's projection on the xy-plane.
• Height (\( z \)): Corresponds to the height of the point above the xy-plane.

In the context of the sphere with a plane cutting through, cylindrical coordinates effectively simplify complex integrations by taking advantage of the circular symmetry. The transformation equations connecting rectangular coordinates (\( x, y, z \)) to cylindrical coordinates are:
\( x = r \cos \theta \)
\( y = r \sin \theta \)
\( z = z \)This approach simplifies the integration for volumes with fixed radii and variable heights, as seen in the exercise when evaluating the integral for the volume of the spherical cap.

Rectangular Coordinates

Rectangular coordinates, or Cartesian coordinates, are the most straightforward way to specify points in a plane with the x, y, and z-axes, forming a right-angled grid. They are especially useful for problems aligned along these axes.
Though not necessarily the most efficient for spherical problems, they provide a familiar structure for beginners.

• X-coordinate (\( x \)): Measures horizontal distance.
• Y-coordinate (\( y \)): Measures vertical distance.
• Z-coordinate (\( z \)): Measures depth.

For the problem of finding the spherical cap's volume, rectangular coordinates establish a clear framework even if the integration becomes complex. The relation between points and the axes allows direct definition of bounds by solving for a change in z as a function of x and y, which is described with the equation of a sphere: \( x^2 + y^2 + z^2 = 4 \).
Ultimately, for spherical geometries, although rectangular coordinates require more intricate limits, they offer a familiar base from which to explore transformations to other systems.

Spherical Cap

A spherical cap is the portion of a sphere that is "cut off" by a plane. It’s crucial for many geometry-related calculations, especially involving volume and surface area.
This cap can be visualized as a dome or a slice held above or below a sphere’s middle section. In this exercise, the spherical cap results from intersecting a sphere centered at the origin with radius 2 and a horizontal plane positioned 1 unit above its center.

• The "cap" is characterized by the plane cutting through the sphere, forming a circular top.
• The height (\( h \)) of our spherical cap is defined in terms of the sphere’s radius and the vertical distance from its center to the plane.
• In general, the volume formula for a spherical cap when radius (\( R \)) and height are known is, \( V = \frac{1}{3}\pi h^2(3R - h) \). Here, spherical coordinates can directly help in integrating over this cap volume.

Understanding the geometry of the cap is essential for setting limits in integral forms and understanding the result obtained.

Volume Calculation

Calculating the volume of a geometric object using integrals is a fundamental task in multivariable calculus. Triple integrals extend this idea into three dimensions, allowing the specification of volume in space.
In this exercise, we've employed triple integrals in spherical, cylindrical, and rectangular coordinates to determine the spherical cap's volume. To grasp volume calculation in 3D:

• Identify the region: Integration bounds must match the dimensions and constraints of the geometry you’re evaluating.
• Select the appropriate coordinate system: Choose between spherical, cylindrical, and rectangular based on symmetry and axis alignment.
• Evaluate the integral: Compute the iterated integral by solving it in steps. Often, simplifications or symmetries can aid in reducing complexity.

Ultimately, once the limits and conversions from other coordinates are sound, plug into the integral and solve. The exercise effectively highlighted how changing coordinate systems simplifies the calculations and leads to the same physical quantity, reinforcing the importance of choosing the correct system for multivariate calculus tasks.