Question

A domain \(D\) in space is given. Parametrize each of the bounding surfaces of \(D\). \(D\) is the domain bounded by \(z=2 y, y=4-x^{2}\) and \(z=0\).

Short answer

Parametrize the bounding surfaces with \(\mathbf{r}(u, v) = (u, 4 - u^2, v)\) for \(-2 \leq u \leq 2\) and \(0 \leq v \leq 8 - 2u^2\).

Step-by-step solution

Understand the Domain

The domain, \(D\), is bounded by three surfaces: \(z = 2y\), \(y = 4 - x^2\), and \(z = 0\). This suggests a volume under the paraboloid \(y = 4 - x^2\) and above the xy-plane \(z = 0\), with the top surface bounded by \(z = 2y\).

Identify the Intersection Curve

To find the boundary of the domain, determine the intersection of the surfaces. Setting \(z=2y\) and \(z=0\), we find that \(2y = 0\), thus \(y = 0\). The curve where \(y = 4 - x^2\) intersects \(z=2y\) is found by substituting \(z = 2y\) and \(y = 4 - x^2\) into \(z = 2(4 - x^2) = 8 - 2x^2\).

Parametrize the Surface in xy-plane

To parametrize the surface \(y = 4 - x^2\) in the xy-plane, solve for \(x\) and \(y\). The domain is within \(y = 0\) and \(y = 4 - x^2 = 0\) which implies \(-2 \leq x \leq 2\). Set \(x = u\), where \(-2 \leq u \leq 2\), and \(y = 4 - u^2\).

Parametrize the Surface for z-values

The z-values are bounded between \(z = 0\) and \(z = 2y\). Given \(y = 4 - u^2\), it becomes \(z = 0\) to \(z = 2(4-u^2)\). Use \(v\) to sweep \(z\) from top to bottom within these limits: \(z = v\), \(0 \leq v \leq 2(4 - u^2)\).

Formulate the Parametrization

The parametrization becomes \(\mathbf{r}(u, v) = (u, 4 - u^2, v)\) for the entire surface with \(-2 \leq u \leq 2\) and \(0 \leq v \leq 8 - 2u^2\).

Verify Parametrization

Check if the parametrization adheres to the original surfaces. Substitute back into the equations of the surfaces \(z = 2y\) and \(y = 4 - x^2\) and ensure \(z = 0\) is satisfied within defined limits. The form \(\mathbf{r}(u, v)\) satisfies all boundary conditions.

Key concepts

Intersection of Surfaces

Understanding how surfaces intersect in space is a pivotal aspect of geometry. In our given problem, we study the intersection of surfaces to understand the domain. The domain is determined by the boundary surfaces: \(z = 2y\), \(y = 4 - x^2\), and \(z = 0\). These surfaces enclose a volume, essentially defining where they meet and form boundaries.
To explore the intersection points, we equate the surfaces. For instance, by setting \(z = 2y\) and \(z = 0\), we find \(2y = 0\), leading to \(y = 0\). Similarly, by intersecting \(y = 4 - x^2\) with \(z = 2y\), we substitute \(y\) and solve for \(z\), confirming \(z = 8 - 2x^2\). This gives us a curve that further shapes the domain in space.

Parametrization in Space

Parametrization is like creating a set of instructions to navigate a surface in space. It helps describe curves and surfaces using variables. In simple terms, we use parameters to trace out a path or surface.
In our scenario, to parametrize the paraboloid \(y = 4 - x^2\) in the \(xy\)-plane, we express both \(x\) and \(y\) in terms of a parameter \(u\). Here, by setting \(x = u\), we span values from \(-2\) to \(2\). For each \(u\), \(y\) is given by the equation \(y = 4 - u^2\). This neatly describes our curve in the \(xy\)-plane.

Bounding Surfaces

Bounding surfaces are the walls of a domain in space, enclosing a volume. They ensure that every point within a domain is contained between these surfaces.
In this exercise, the bounding surfaces are \(z=0\), which is the lower boundary, and \(z=2y\), which is the upper boundary. Additionally, the shape is constrained by the paraboloid \(y = 4 - x^2\), effectively setting side walls.
Think of it as a room with a floor, a ceiling, and curved walls. The parametrization defines every internal point that stays within the constraints of these surfaces. It ensures there's no escape outside the area confinements as dictated by the equations.

Coordinate Systems

Choosing the appropriate coordinate system is crucial in parametrization. It offers a simplified way to handle complex surfaces by rearranging parameters.
In this example, we primarily use the Cartesian system, expressed in terms of \(x\), \(y\), and \(z\). By treating \(x\) as a controllable parameter, \(y\) and \(z\) are modeled according to \(x\)'s choice. We use \(u\) and \(v\) as parameters, relating directly to \(x\) as \(x = u\) and \(z\) as \(z = v\). By doing so, we can represent complex shapes in space clearly and mathematically efficiently.
Mastering the use of coordinate systems streamlines the understanding and application of parametrization techniques to capture the essence of spatial domains.