Problem 6 - Thomas Calculus

Question

Find a parametrization of the surface. (There are many correct ways to do these, so your answers may not be the same as those in the back of the book.) Spherical cap The portion of the sphere \(x^{2}+y^{2}+z^{2}=4\) in the first octant between the \(x y\) -plane and the cone \(z=\sqrt{x^{2}+y^{2}}\)

Step-by-Step Solution

Answer

Use spherical coordinates with \(\rho=2\), \(0 \leq \phi \leq \frac{\pi}{4}\), \(0 \leq \theta \leq \frac{\pi}{2}\).

Step 1: Analyze the Problem

We need to parametrize the part of a sphere that's in the first octant and lies above the cone. The sphere's equation is \(x^2 + y^2 + z^2 = 4\) and the cone's equation is \(z = \sqrt{x^2 + y^2}\). The first octant requires \(x, y, z \geq 0\).

Step 2: Solve for Spherical Coordinates

We can use spherical coordinates where \(x = \rho \sin \phi \cos \theta\), \(y = \rho \sin \phi \sin \theta\), and \(z = \rho \cos \phi\). Since the sphere's radius is 2, \(\rho = 2\).

Step 3: Determine Bounds for Parameters

The surface exists in the first octant with \(z = \sqrt{x^2 + y^2}\) providing the lower boundary, so \(z = \rho \cos \phi = \rho \sin \phi\). From this, \(\sin \phi = \cos \phi\), yielding \(\phi = \frac{\pi}{4}\). For the first octant \(\phi\) should go from 0 to \(\frac{\pi}{4}\). \(\theta\) should range from 0 to \(\frac{\pi}{2}\).

Step 4: Write the Parametrization

Substitute \(\rho = 2\), \(\phi\), and \(\theta\) into the spherical coordinates equations. This yields: \(x = 2 \sin \phi \cos \theta\), \(y = 2 \sin \phi \sin \theta\), \(z = 2 \cos \phi\). These give the parametrization of the surface.

Key Concepts

Spherical Coordinates

Understanding spherical coordinates is crucial when dealing with three-dimensional problems in calculus and geometry. Spherical coordinates use three parameters: \( \rho \), \( \phi \), and \( \theta \). These represent the radius, the inclination, and the azimuth respectively. By converting Cartesian coordinates to spherical coordinates, we can simplify the parametrization of surfaces like spheres and cones. Here's how each parameter is used:

• \( \rho \) denotes the distance from the origin to the point in space. For a sphere of radius 2, \( \rho \) is 2.
• \( \phi \) is the angle measured from the positive z-axis to the line connecting the origin to the point. It helps to measure the vertical inclination.
• \( \theta \) is the azimuthal angle in the xy-plane from the positive x-axis.

Spherical coordinates offer a more natural way to describe surfaces beyond rectangular systems, especially for objects like spheres.

Sphere Equation

A sphere is defined in 3D space with an equation \( x^2 + y^2 + z^2 = r^2 \), where \( r \) is the radius. In our problem, the sphere has a radius 2, leading to the equation \( x^2 + y^2 + z^2 = 4 \). This equation includes all points that are exactly 2 units away from the origin.

To visualize this, imagine a globe centered at the origin. Each point equates to a spot on the sphere's surface. Importantly, when we work with spherical coordinates for this sphere, \( \rho = r = 2 \). This is vital because it simplifies calculations and ensures all points remain on the sphere's surface as parametrization occurs.

Surface in First Octant

The first octant in 3D geometry refers to the space where all three coordinates—\( x \), \( y \), and \( z \)—are positive. This gives a framework for visualizing the portion of the sphere we're interested in. Often in mathematics and engineering, problems are simplified by focusing only on the first octant, reducing the complexity by a factor of eight.

In this exercise, only the part of the sphere existing in the first octant is needed. We apply the condition \( x, y, z \geq 0 \) to our spherical parametrization to constrain the surface to just this octant. This is important for accurately defining the region of interest and ensuring the relevant bounds for the parameters \( \phi \) and \( \theta \). For our sphere, this results in parameters \( \phi \) ranging from 0 to \( \frac{\pi}{4} \), and \( \theta \) going from 0 to \( \frac{\pi}{2} \).

Cone Equation

The cone's equation in this exercise is \( z = \sqrt{x^2 + y^2} \). This is characteristic of a cone where the vertical height \( z \) is equivalent to the radial distance from the z-axis in the xy-plane.

In simplified terms:

• All points where \( z \) equals the distance to the z-axis lie on the cone.
• The intersection of our sphere and cone requires that the sphere lies above this cone.
• In spherical coordinates, this condition can be expressed as \( \cos \phi = \sin \phi \).

Transforming this into normal angles, the result is \( \phi = \frac{\pi}{4} \), representing the angle of intersection between the cone and the sphere. The surface part above the cone in the first octant is what we're interested in parametrizing.