Problem 26 - Thomas Calculus

Question

Use a parametrization to express the area of the surface as a double integral. Then evaluate the integral. (There are many correct ways to set up the integrals, so your integrals may not be the same as those in the back of the book. They should have the same values, however.) Spherical band The portion of the sphere \(x^{2}+y^{2}+z^{2}=4\) between the planes \(z=-1\) and \(z=\sqrt{3}\)

Step-by-Step Solution

Answer

The surface area of the spherical band is \( 4\pi + 4\pi\sqrt{3} \).

Step 1: Understand the problem

We need to find the area of a spherical band on the sphere defined by \( x^2 + y^2 + z^2 = 4 \), which is a sphere of radius 2 centered at the origin. The spherical band is bounded between the planes \( z = -1 \) and \( z = \sqrt{3} \).

Step 2: Parametrize the surface

A common parametrization for a sphere is to use spherical coordinates. Using spherical coordinates, \( x = 2 \sin \phi \cos \theta \), \( y = 2 \sin \phi \sin \theta \), and \( z = 2 \cos \phi \), where the radius is 2, \( \phi \) is the polar angle, and \( \theta \) is the azimuthal angle. The range for \( \theta \) is \( 0 \leq \theta \leq 2\pi \). Determine the range of \( \phi \) corresponding to the planes \( z = -1 \) and \( z = \sqrt{3} \).

Step 3: Determine the bounds for \( \phi \)

Since \( z = 2 \cos \phi \), substitute the bounds for \( z \):- For \( z = -1 \), we get \( 2 \cos \phi = -1 \) or \( \cos \phi = -\frac{1}{2} \). Thus, \( \phi = \frac{2\pi}{3} \).- For \( z = \sqrt{3} \), we get \( 2 \cos \phi = \sqrt{3} \) or \( \cos \phi = \frac{\sqrt{3}}{2} \). Thus, \( \phi = \frac{\pi}{6} \).So, the range for \( \phi \) is \( \frac{\pi}{6} \leq \phi \leq \frac{2\pi}{3} \).

Step 4: Set up the double integral for surface area

The element of surface area in spherical coordinates is given by \( dS = R^2 \sin \phi \, d\phi \, d\theta \), where \( R = 2 \) is the radius. Therefore, the integral to find the surface area is:\[ \text{Area} = \int_0^{2\pi} \int_{\frac{\pi}{6}}^{\frac{2\pi}{3}} 4 \sin \phi \, d\phi \, d\theta \]

Step 5: Evaluate the inner integral with respect to \( \phi \)

Firstly, calculate the inner integral:\[ \int_{\frac{\pi}{6}}^{\frac{2\pi}{3}} 4 \sin \phi \, d\phi = \left[ -4 \cos \phi \right]_{\frac{\pi}{6}}^{\frac{2\pi}{3}} = -4 (\cos \frac{2\pi}{3} - \cos \frac{\pi}{6}) \]Calculating the values:\( \cos \frac{2\pi}{3} = -\frac{1}{2} \) and \( \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} \)Thus, it becomes:\[ -4 \left(-\frac{1}{2} - \frac{\sqrt{3}}{2} \right) = 2 + 2\sqrt{3} \]

Step 6: Evaluate the outer integral with respect to \( \theta \)

Now integrate the result from the concrete result of the inner integral:\[ \int_0^{2\pi} (2 + 2\sqrt{3}) \, d\theta = (2 + 2\sqrt{3}) \left[\theta \right]_0^{2\pi} = (2 + 2\sqrt{3})(2\pi) \]

Step 7: Compute the final result

Finally, compute the overall surface area.\[ \text{Area} = 2\pi (2 + 2\sqrt{3}) = 4\pi + 4\pi\sqrt{3} \]

Key Concepts

Spherical Coordinates

Spherical coordinates are a way of representing points in three-dimensional space using three parameters: radius, polar angle, and azimuthal angle. Unlike the more common Cartesian coordinates (x, y, z), spherical coordinates are designed to work efficiently for problems involving spheres and circular symmetry.

- **Radius (r):** In spherical coordinates, the radius is the distance from the origin to the point in space. In our case, the sphere's radius is constant at 2.
- **Polar Angle (φ):** Also known as the inclination angle, it is measured from the positive z-axis. For our exercise, φ varies between \( \frac{\pi}{6} \) and \( \frac{2\pi}{3} \) to form the spherical band.
- **Azimuthal Angle (θ):** It represents rotation around the z-axis and is highlighted as varying from \( 0 \) to \( 2\pi \), making a complete circle around the z-axis.

These parameters excel when describing spheres, minimizing the complexity of equations like the surface area of a spherical band.

Parametrization

Parametrization is a technique used to express geometrical shapes with sets of functions dependent on one or two parameters. In the context of spheres, it involves converting spherical coordinates into Cartesian coordinates.
For the sphere of radius 2, we parametrize by using the equations:
- \( x = 2 \sin \phi \cos \theta \)
- \( y = 2 \sin \phi \sin \theta \)
- \( z = 2 \cos \phi \)

These equations reflect how a point is placed on the sphere:- The height (z) depends on \( \cos \phi \), spanning from the top of the sphere to its equator.
- The radial length in the xy-plane is determined by \( \sin \phi \), controlling how far out from the z-axis a point goes.
- Finally, \( \theta \) rotates the point around the z-axis.
Parametrization helps solve problems by reducing complex 3D structures to simpler integrals, utilizing the symmetry of the sphere.

Double Integral

The double integral extends the concept of integration to two-dimensional areas. It calculates things like volume under a surface or the area of complex shapes.
For our exercise, we're interested in finding the surface area of part of a sphere, achieved through a double integral. We set up the double integral like this:
- The inner integral \( \int \) with respect to \( \phi \)calculates the sphere's cross-sectional segment for each \( \theta \), expressing the vertical band on the sphere.
- The outer integral \( \int \) with respect to \( \theta \) collects all these band segments to cover the full area in the desired range of azimuthal rotation.
The element of surface area, \( dS \), based on our parametrization becomes:\[ dS = R^2 \sin \phi \, d\phi \, d\theta. \]
In essence, this surface element is integrated across both angles to compute the total surface area. This makes dealing with the sphere's curvature quite straightforward.

Sphere

A sphere is a perfectly symmetrical three-dimensional shape, where all points on the surface are equidistant from the center. In mathematical terms, its definition is simplified.- **Equation:** The equation \( x^2 + y^2 + z^2 = R^2 \) defines a sphere, where R represents the radius. Our specific sphere in the problem has a radius of 2, giving the equation \( x^2 + y^2 + z^2 = 4 \).
### Spherical BandsA spherical band is a segment of a sphere bounded by planes. Imagine slicing a sphere vertically at different heights; the slice between two such cuts makes up a spherical band. For our problem:
- The sphere is cut between \( z = -1 \) and \( z = \sqrt{3} \).
- These planes create the boundaries for the band, essentially a bracelet-like strip around the sphere.
The sphere, with its continuous curves, makes it imperative to use spherical coordinates, simplifying not just the mathematical description but also calculations like tracing bands or arcs.