Problem 18 - Calculus Early Transcendentals: Pearson New International Edition

Question

Sketch the solid S. Then write an iterated integral for $$ \iiint_{S} f(x, y, z) d V $$ \(S\) is the region in the first octant bounded by the surface \(z=9-x^{2}-y^{2}\) and the coordinate planes.

Step-by-Step Solution

Answer

The iterated integral is \( \int_{0}^{3} \int_{0}^{\sqrt{9-x^2}} \int_{0}^{9-x^2-y^2} f(x, y, z) \, dz \, dy \, dx \).

Step 1: Understand the Region

The solid \( S \) is bounded by the surface \( z = 9 - x^2 - y^2 \) and the coordinate planes. This is a paraboloid opening downward, with its vertex at \((0, 0, 9)\), and it extends to where this surface meets the \( xy \)-plane \((z=0)\). The region is confined to the first octant, meaning \( x \geq 0, y \geq 0, z \geq 0 \).

Step 2: Determine Limits of Integration for x, y, and z

Since the region is bounded by the coordinate planes and \( z = 9 - x^2 - y^2 \), we determine the bounds by observing:- \( z \) runs from 0 to \( 9 - x^2 - y^2 \).- In the \( xy \)-plane, solve \( 9 = x^2 + y^2 \) to find the circle of radius 3. Since we are in the first octant, \( x \) ranges from 0 to 3 and for each \( x \), \( y \) ranges from 0 to \( \sqrt{9 - x^2} \).

Step 3: Write the Iterated Integral

Based on the region's bounds, the iterated integral is defined as:\[\int_{0}^{3} \int_{0}^{\sqrt{9-x^2}} \int_{0}^{9-x^2-y^2} f(x, y, z) \, dz \ dy \, dx.\]This integral represents the volume within the solid \( S \) where the bounding surface is \( z = 9 - x^2 - y^2 \) and it is integrated in that order: first \( z \), then \( y \), and finally \( x \).

Step 4: Visualize the Solid

Imagine the paraboloid interfacing with the first octant: the top surface curves downward from the point \((0, 0, 9)\), touching the \( xy \)-plane in a quarter circle with radius 3. This forms a dome-like shape within the first octant. Visualizing helps ensure the correct bounded region for integration.

Key Concepts

Iterated Integrals

Iterated integrals are a methodical way to evaluate multiple integrals. They are especially useful when finding the volume under a surface described by a function in a given region. In three dimensions, an iterated integral involves performing the integration step-by-step over each variable.

• First, choose the order of integration, usually as \(z\), \(y\), and \(x\). However, the order can vary depending on the function and the region's constraints.
• Next, determine the limits for the innermost integral based on the region's boundaries.
• Subsequently, repeat the integration for each variable, working outward to resolve the full triple integral.

In the case of the provided problem, you integrate a function \(f(x, y, z)\) through the elaborately defined region \(S\) using iterated integrals. This step-by-step dissection allows efficient computation and a clearer understanding of the volume under complex surfaces, such as paraboloids.

Paraboloid

A paraboloid is a 3D surface created when a parabola is revolved around its axis. In this exercise, the surface \(z = 9 - x^2 - y^2\) defines a paraboloid open downward.

• Its main characteristic is its symmetrical nature, with a vertex serving as the highest or lowest point.
• Here, the vertex is at the point \(0, 0, 9\).
• The paraboloid intersects the \(xy\)-plane at the circle \(x^2 + y^2 = 9\), revealing a radius of 3.

This geometry tells us that while z-value decreases, the paraboloid smoothly spreads horizontally, shaping a dome within the first octant. Understanding its geometry is crucial as it helps to accurately establish the region's bounds for integration within iterated integrals.

Coordinate Systems

Different coordinate systems enable us to describe and solve problems involving multiple integrals more efficiently.

• The Cartesian coordinate system, using axes for \(x\), \(y\), and \(z\), suits rectangular and polyline regions of integration.
• In this exercise, the iterative triple integral on the paraboloid is initially simplified using Cartesian coordinates.

For complex regions, such as when equations involve radii and angles, polar or cylindrical coordinates simplify the process.

• They replace Cartesian coordinates with radial measures, facilitating the integration of circular regions like \(x^2 + y^2 = 9\).
• Converted limits often become trivial to compute, especially if dealing with sectors or curves.

Switching coordinate systems can thus transform intricate integrals into more manageable calculations, underscoring the adaptability and power of choosing the right system when solving multiple integrals.

First Octant Region

The first octant refers to the section of 3D space where all coordinates are non-negative: \(x \geq 0, y \geq 0, z \geq 0\).

• This limits the area of integration and consideration within geometric problems.
• In the given problem, the region defined by the paraboloid lies entirely within the first octant, simplifying the integration by excluding any negative coordinate components.

By focusing on the first octant, we streamline the process of establishing boundary limits for our integrals. Only positive values for \(x\), \(y\), and \(z\) need consideration, as the region adequately prevails where \(z \geq 0\) over the intersection circle in the \(xy\)-plane. Recognizing this region's scope is essential for addressing integrals correctly as it refines the area of computation, dictating the limits applied within the iterated integrals.