Problem 15 - Calculus Early Transcendentals: Pearson New International Edition

Question

The part of \(z=9-x^{2}-y^{2}\) above the plane \(z=5\).

Step-by-Step Solution

Answer

A circular disk of radius 2 centered at the origin in the x-y plane.

Step 1: Identify the Region

We need to find the region where the surface defined by the equation, \[ z = 9 - x^2 - y^2, \]is above the plane \( z = 5 \). This means we are looking for solutions to the inequality:\[ 9 - x^2 - y^2 > 5. \]

Step 2: Simplify the Inequality

Subtract 5 from both sides of the inequality:\[ 9 - 5 > x^2 + y^2. \]This simplifies to:\[ 4 > x^2 + y^2. \]

Step 3: Interpret the Inequality

The inequality \( 4 > x^2 + y^2 \) describes a circle of radius 2 centered at the origin in the x-y plane. This means that the region is a circular disk with radius 2 around the origin (0, 0).

Step 4: Describe the Solid Region

Thus, the part of the surface \( z = 9 - x^2 - y^2 \) that lies above the plane \( z = 5 \) is a cup-shaped surface over this circular disk of radius 2. The top boundary of this region is the surface where \( z = 9 - x^2 - y^2 \) and \( z > 5 \). The plane \( z = 5 \) serves as the lower boundary.

Key Concepts

Inequality

In the context of calculus, an inequality such as the one we encounter here is a way to define a range of points or a region where a certain condition holds true. Consider the inequality \( 9 - x^2 - y^2 > 5 \). This means we are looking for all points \((x, y)\) where the value of \(9 - x^2 - y^2\) exceeds 5.
This can be simplified by subtracting 5 from both sides, resulting in \(4 > x^2 + y^2\). What this simplification tells us is that the sum of the squares of \(x\) and \(y\) must be less than 4, defining a region of space where this condition is satisfied.
Inequalities like this are often used to describe the boundaries and extents of geometric shapes in mathematics.

Circular Region

A circular region in the xy-plane is defined by an equation like \(x^2 + y^2 < r^2\), where \(r\) is the radius of the circle. In our case, the inequality \(4 > x^2 + y^2\) represents such a circle with a radius of 2, centered at the origin \((0, 0)\). This can be visualized as all the points that lie within a circle with a 2-unit radius in the xy-plane.
The structure of the inequality indicates that \(x\) and \(y\) are not allowed to lie on or outside the boundary where \(x^2 + y^2 = 4\), because they need to remain smaller than 4 to satisfy \(4 > x^2 + y^2\). This means every point \((x, y)\) within a 2-unit distance from the origin satisfies this inequality.
This circular region is crucial in 3D visualization as it acts as the base for the cup-shaped surface extending above in the subsequent concept.

Solid Region

In this example, the solid region is defined by the vertical distance between the 3D surface \(z = 9 - x^2 - y^2\) and the horizontal plane \(z = 5\). We have determined that above this plane lies a part of the paraboloid surface whose edges are formed within the circular disk of radius 2.
So, in essence, the solid region is the space contained above the plane \(z = 5\) and below the surface \(z = 9 - x^2 - y^2\), only above this disk. The radius 2 limit from the circular region ensures that this solid region is properly bounded horizontally.
Thus, any point within this solid must satisfy both the surface's equation: \(z = 9 - x^2 - y^2 > 5\), with \(4 > x^2 + y^2\), verifying it remains within the prescribed bounds.

3D Surfaces

3D surfaces are often described using equations that relate x, y, and z coordinates. Here, \(z = 9 - x^2 - y^2\) sketches an inverted paraboloid in space, which would typically produce a shape resembling an umbrella.
To understand the area above the plane \(z = 5\), imagine slicing this paraboloid horizontally at \(z = 5\). The slice reveals a depth or thickness to the curved surface above this plane, creating a hollow area beneath the original paraboloid and above the plane, limited within a circle of radius 2 in the x-y plane.
This visualization is quite similar to understanding how partial objects can be viewed in 3D modeling, constructing a surface region or solid that adheres to given constraints and boundaries.