Problem 24 - Thomas Calculus

Question

Sketch the surfaces in Exercises \(13-44.\) PARABOLOIDS AND CONES $$y=1-x^{2}-z^{2}$$

Step-by-Step Solution

Answer

The surface is an elliptic paraboloid opening downwards centered at \((0, 1, 0)\).

Step 1: Identify the Surface Type

The given equation is \( y = 1 - x^2 - z^2 \). This equation resembles the standard form of an elliptic paraboloid, which is given as \( y = a - x^2 - z^2 \). Here, the vertex of the paraboloid is at \((0, 1, 0)\) and it opens downwards along the y-axis.

Step 2: Set Up the Cross Sections

To better understand the surface, consider cross-sections by setting one variable constant at a time. 1. If \( y = c \) (a constant), then \( 1 - c = x^2 + z^2 \), giving you circles in the \(xz\)-plane centered at the origin with radius \(\sqrt{1-c}\), valid when \(c \leq 1\).2. If \( x = c \), then \( y = 1 - c^2 - z^2 \) describes a parabola in the \( yz\)-plane.3. Similarly, if \( z = c \), \( y = 1 - x^2 - c^2 \) describes a parabola in the \( xy\)-plane.

Step 3: Determine the Domain

The domain of the surface is for those \(x, y, z\) values that satisfy \(y = 1 - x^2 - z^2\). This means that \(x^2 + z^2 \leq 1\), putting bounds on \(x\) and \(z\) within a circle of radius 1 centered at the origin.

Step 4: Plot the Surface

Start by plotting the vertex at \((0, 1, 0)\). As \(y\) decreases from 1 to 0, plot circles with decreasing y value and increasing radii in the \(xz\)-plane. At \(y = 0\), the equation becomes \(0 = 1 - x^2 - z^2\), where \(x^2 + z^2 = 1\), thus a circle of radius 1 is obtained. The surface will look like a cap opening downwards.

Key Concepts

Three-Dimensional Surfaces

Three-dimensional surfaces are fascinating visualizations in mathematics, which help us understand complex equations in a spatial form. An elliptic paraboloid is a type of three-dimensional surface described by a specific algebraic equation. It resembles a parabola, but in multiple directions.
In the given exercise, the equation provided is \( y = 1 - x^2 - z^2 \). This represents an elliptic paraboloid where:

• The vertex is at \(0, 1, 0\)
• It opens downwards along the y-axis
• The surface extends infinitely in the x and z directions

Visualizing such surfaces requires understanding how the vertices and axes contribute to the shape, much like carving a three-dimensional shape from a square piece of paper by folding it at key points.

Cross-Sections

Cross-sections are essentially slices of a three-dimensional surface that can be examined to gain insights into the shape's structure. Imagine cutting through the elliptic paraboloid horizontally or vertically to see its profile.
Let's consider the cross-sections for our surface:

• When \(y = c \), we get circles on the xz-plane, centered at the origin, with a radius of \(\sqrt{1-c}\). This happens for \(c \leq 1\).
• When \(x = c \), the surface shows as a parabola in the yz-plane, opening downward as the y-axis is the axis of symmetry.
• Similarly, when \(z = c \), we also see a parabolic shape in the xy-plane.

These cross-sections demonstrate the symmetry and structure of the paraboloid, with each slice giving different perspectives depending on which axis we hold constant.

Domain of Functions

Understanding the domain of functions is crucial for knowing where a surface can exist within three-dimensional space. It tells us which values of \(x\), \(y\), and \(z\) make the equation true.
For the paraboloid \( y = 1 - x^2 - z^2 \), the domain is determined by the constraint \( x^2 + z^2 \leq 1 \). This constraint defines a circular boundary within the xz-plane:

• The maximum range for \(x\) and \(z\) is within a circle of radius 1 centered at the origin.
• \(y\) varies depending on the values of \(x\) and \(z\), reaching a maximum of 1 at the origin \((0, 1, 0)\).

The function is valid everywhere inside this circle. Comprehending this domain helps in sketching the boundary and understanding the extent and limitations of the paraboloid surface.