Problem 32 - Calculus Early Transcendentals

Question

The surface of an island is defined by the following functions over the region on which the function is non-negative. Find the volume of the island. $$z=\frac{20}{1+x^{2}+y^{2}}-2$$

Step-by-Step Solution

Answer

The volume of the island can be found by calculating the double integral of the function over the region where it is non-negative. The region in the xy-plane where the function is non-negative is determined by the inequality $$x^2 + y^2 \leq 10$$. By converting to polar coordinates and setting up the double integral, the volume is found to be: $$V = 20\pi \left[\arctan(\sqrt{10}) - \frac{1}{3}\right]$$

Step 1: Determine the region of integration

First, we want to find the region in the \(xy\)-plane where the function is non-negative, i.e., \(z \geq 0\). So, we need to solve the inequality: $$\frac{20}{1 + x^2 + y^2} - 2 \geq 0$$ Now, we isolate the radical term: $$\frac{20}{1 + x^2 + y^2} \geq 2$$ And rearrange: $$1 + x^2 + y^2 \leq 10$$ This inequality represents the interior of a circle with radius \(\sqrt{10}\) centered at the origin in the \(xy\)-plane: $$x^2 + y^2 \leq 10$$

Step 2: Set up the integral in polar coordinates

Since the region of integration is a circle, it is convenient to use polar coordinates to evaluate the integral for the volume. We can convert \(x^2 + y^2 \leq 10\) to polar coordinates, and we get: $$r^2 \leq 10$$ So, \(0 \leq r \leq \sqrt{10}\) and \(0 \leq \theta \leq 2\pi\). Now we can set up the double integral for the volume: $$V = \int_{0}^{2\pi} \int_{0}^{\sqrt{10}} \left(\frac{20}{1 + r^2}-2\right) r \, dr \, d\theta$$

Step 3: Evaluate the integral

To evaluate the integral, we first integrate with respect to \(r\): $$V(\theta) = \int_{0}^{2\pi} \left[10\arctan(r) - r^2\right]_{0}^{\sqrt{10}} d\theta$$ Now integrate with respect to \(\theta\): $$V = \left[10\theta \arctan(\sqrt{10}) - \frac{10}{3}\theta\right]_{0}^{2\pi}$$ Evaluate the expression at the limits of integration: $$V = \left[20\pi \arctan(\sqrt{10}) - \frac{20\pi}{3}\right] - [0]$$ So, the volume of the island is: $$V = 20\pi \left[\arctan(\sqrt{10}) - \frac{1}{3}\right]$$

Key Concepts

Double Integral

The concept of a double integral is central to finding the volume under a surface in mathematics. When we are given a function of two variables, say, a surface defined by a function of the form z = f(x, y), the double integral helps us compute the aggregation of infinitesimal elements of volume over a certain region. Double integrals are denoted as \( \int\int_R f(x,y) \, dx \, dy \), where R specifies the region of integration in the xy-plane.

Understanding double integrals involves recognizing that we are adding up all the 'columns' or 'pillars' of volume, where each represents the volume element \(f(x, y) \, dx \, dy\). In the context of the given exercise, the double integral is used to calculate the total volume of a three-dimensional island above the described region in the plane.

Polar Coordinate System

The polar coordinate system is an alternative to the Cartesian coordinate system. It represents a point in the plane by how far away it is from a reference point (the origin, represented by the radius r) and the angle (\( \theta \)) formed with the positive x-axis. This system has particular utility in modeling scenarios where symmetry about a point makes the situation easier to analyze.

In polar coordinates, a point is expressed as (r, \( \theta \)). The polar coordinate system simplifies calculations for regions that are circular or sectors of a circle, like the one in our exercise. This system links closely with double integrals, as we often convert Cartesian coordinates to polar coordinates to make the integration process more feasible.

Volume Calculation

Volume calculation using integrals is a staple of multivariable calculus. To find the volume under a surface, we integrate a height function over a particular region. When the surface is simple and the region of interest is rectangular, Cartesian coordinates work well. But when dealing with circular or polar symmetrical regions, it is often easier to convert the volume integral into polar coordinates, resulting in an expression like this: \( V = \int \int_R f(r, \theta)\, r \, dr \, d\theta \), where f(r, \( \theta \)) is the function in polar coordinates and r is the radial distance from the origin.

In the exercise solution, the double integral in polar coordinates effectively simplifies the volume calculation of the island by considering the symmetry of the region and the form of the function.

Non-negative Function Region

In many real-world applications, including our textbook problem, we are interested in the volume above a region where the function describing the surface is non-negative. This implies that for each point (x, y) within the region R, the function's value is zero or positive, which corresponds to z being on or above the xy-plane.

The region where the function is non-negative must be specified or determined as part of solving the problem. In the provided solution, this region was shown to be a circular disk in the xy-plane with a radius determined by setting the function equal to zero and solving for x and y. It is the base over which we calculate the volume when doing the double integral, and is crucial for accurately determining the volume of the shape described by the function.