Problem 6 - Calculus Early Transcendentals: Pearson New International Edition

Question

The part of the paraboloid \(z=x^{2}+y^{2}\) that is cut off by the plane \(z=4\)

Step-by-Step Solution

Answer

The cut-off portion is a paraboloid section inside the disk \(x^2 + y^2 \leq 4\).

Step 1: Understand the Problem

We need to identify the portion of the paraboloid given by \(z = x^2 + y^2\) that lies below the plane \(z = 4\). This means we have to find points \((x, y, z)\) satisfying the paraboloid equation with \(z\) less than or equal to 4.

Step 2: Set the Intersection Criteria

The intersection of the paraboloid \(z = x^2 + y^2\) with the plane \(z = 4\) happens when \(x^2 + y^2 = 4\). Thus, the curve of intersection is a circle centered at the origin in the \(xy\)-plane with radius 2.

Step 3: Understand the Domain

The domain consists of all points \((x, y)\) such that \(x^2 + y^2 \leq 4\). These points form a disk of radius 2 centered at the origin in the \(xy\)-plane.

Step 4: Analyze the Portion of Paraboloid

Above this disk, the part of the paraboloid is given by \(z = x^2 + y^2\) where \(0 \leq z \leq 4\). Therefore, the portion of the paraboloid below the plane is everything below this circular boundary.

Key Concepts

Intersection with Plane

When analyzing intersections, consider a paraboloid, which is a three-dimensional surface, and a plane, which is a flat, two-dimensional surface. Specifically, we are looking at how the paraboloid described by the equation \(z = x^2 + y^2\) intersects with the plane \(z = 4\). The intersection takes place where the values of \(z\) for both equations are equal, forming a curve in three-dimensional space.

For this example, setting \(z = x^2 + y^2\) equal to \(z = 4\) simplifies to \(x^2 + y^2 = 4\), which describes a circle. This circle represents the boundary where the paraboloid is exactly at height 4 along the \(z\)-axis. It's important to visualize that this circle is where the surface of the paraboloid "touches" or "cuts through" the plane.

Paraboloid Equation

A paraboloid is a surface that can be described by a quadratic equation, similar to how a parabola is a curve. In our case, the equation \(z = x^2 + y^2\) forms a paraboloid. This particular type of paraboloid is called an "elliptic paraboloid" because its cross-sections parallel to the \(xy\)-plane are ellipses or circles.

The paraboloid opens upwards, similar to a bowl, with its vertex, or the bottom point, located at the origin \((0, 0, 0)\). As \(x\) and \(y\) increase, \(z\) also increases, forming a surface that rises infinitely unless intersected by a shape like a plane.

Domain of Function

The domain of a function refers to all the possible input values (or \(x, y\) pairs in this context) for which the function is defined. For the function \(z = x^2 + y^2\), the domain consists of all points in the \(xy\)-plane. The problem specifically asks us to consider where \(z \leq 4\). This means we're interested in the region where \(x^2 + y^2 \leq 4\).

This condition forms a disk centered at the origin with a radius of 2 in the \(xy\)-plane. All points \((x, y)\) within and on this boundary satisfy the equation \(x^2 + y^2 \leq 4\). When visualizing, imagine a flat circle at the base of our paraboloid, reaching out to all points 2 units away from the origin.

Circle in XY-Plane

The circle formed in the \(xy\)-plane as a result of the intersection is crucial. It's defined by the equation \(x^2 + y^2 = 4\), which specifies a circle with a radius of 2 centered at the origin \((0, 0)\).

This circle signifies where the plane \(z = 4\) precisely meets the paraboloid \(z = x^2 + y^2\). Any point inside this circle lies below the plane on the paraboloid. This is because the paraboloid rises from the origin outward, reaching \(z = 4\) precisely at the circle's edge.

As a visualization tip, picture this circle as slicing horizontally through a bowl, marking the edge where the bowl's surface is 4 units high.