Problem 56 - Calculus Volume 3

Question

In the following exercises, the functions \(f_{n}\) are given, where \(n \geq 1\) is a natural number. a. Find the volume of the solids \(S_{n}\) under the surfaces \(z=f_{n}(x, y)\) and above the region \(R\) b. Determine the limit of the volumes of the solids \(S_{n}\) as \(n\) increases without bound. $$ f(x, y)=\frac{1}{x^{n}}+\frac{1}{y^{n}},(x, y) \in R=[1,2] \times[1,2] $$

Step-by-Step Solution

Answer

The volume is \( 2 \cdot \frac{2^{1-n} - 1}{1-n} \), and the limit as \( n \to \infty \) is 0.

Step 1: Set Up the Integral for Volume

The volume of the solid under the surface \( z = f_n(x, y) \) and above the region \( R \) is given by the double integral \( V_n = \int_{1}^{2} \int_{1}^{2} f_n(x, y) \, dx \, dy \). For this problem, \( f_n(x, y) = \frac{1}{x^n} + \frac{1}{y^n} \).

Step 2: Split the Integral

The integral can be split into two separate integrals due to the addition in \( f_n(x, y) \): \[ V_n = \int_{1}^{2} \int_{1}^{2} \frac{1}{x^n} \, dx \, dy + \int_{1}^{2} \int_{1}^{2} \frac{1}{y^n} \, dx \, dy \].

Step 3: Evaluate Each Integral Separately

First, evaluate \( \int_{1}^{2} \frac{1}{x^n} \, dx \) and then multiply the result by the outer integral \( \int_{1}^{2} dy \). Similarly, evaluate \( \int_{1}^{2} \frac{1}{y^n} \, dy \) and multiply by \( \int_{1}^{2} dx \). This will simplify each part separately.

Step 4: Evaluate the Inner Integral for \( \frac{1}{x^n} \)

\[ \int_{1}^{2} \frac{1}{x^n} \, dx = \left[ \frac{x^{1-n}}{1-n} \right]_{1}^{2} = \frac{2^{1-n} - 1^{1-n}}{1-n} = \frac{2^{1-n} - 1}{1-n} \].

Step 5: Evaluate the Outer Integral for the First Part

The integral \( \int_{1}^{2} dy = [y]_{1}^{2} = 1 \). Thus, the volume contribution from the first integral is \( V_{n,1} = 1 \cdot \frac{2^{1-n} - 1}{1-n} \).

Step 6: Evaluate the Integral for \( \frac{1}{y^n} \)

This integral is similar to Step 4: \[ \int_{1}^{2} \frac{1}{y^n} \, dy = \frac{2^{1-n} - 1}{1-n} \].

Step 7: Multiply by Outer Integral for Second Part

Since both the x and y integrals evaluate the same, the outer integral \( \int_{1}^{2} dx = 1 \). Thus, the volume contribution from the second part is \( V_{n,2} = 1 \cdot \frac{2^{1-n} - 1}{1-n} \).

Step 8: Combine Results for Total Volume

Add the results of the separate integrals to find the total volume: \( V_n = V_{n,1} + V_{n,2} = 2 \cdot \frac{2^{1-n} - 1}{1-n} \).

Step 9: Determine the Limit of Volumes as \( n \to \infty \)

Taking the limit of \( V_n = 2 \cdot \frac{2^{1-n} - 1}{1-n} \) as \( n \rightarrow \infty \), notice that \( 2^{1-n} \to 0 \), thus \[ \lim_{n \to \infty} V_n = 2 \cdot \frac{0 - 1}{1-n} = 2 \times 0 = 0 \].

Key Concepts

Volume of Solids

Understanding the volume of solids is key in multivariable calculus. It's about finding how much space a 3D shape occupies. Imagine filling a container with water and measuring the quantity; this is similar to calculating volume mathematically.
In our original exercise, the region under consideration is defined by the function \( z = f_n(x, y) \). This function describes a surface or boundary above the region \( R = [1, 2] \times [1, 2] \). To find the volume of the solid lying under this surface and above \( R \), a double integral is employed.
This requires evaluating \( V_n = \int_{1}^{2} \int_{1}^{2} f_n(x, y) \, dx \, dy \), which essentially sums up small pieces of volume across the defined region. As the function changes, so does the represented solid, illustrating this fascinating union of geometry and calculus.

Double Integrals

Double integrals extend the concept of a single integral to calculate values over a two-dimensional area. They are indispensable for finding areas, masses, and, as in our case, volumes of certain objects.
To perform a double integral, you evaluate the integral in steps. First, compute the inner integral—think of it as summing slices of the solid along one axis. Next, the result is used in the outer integral to sum slices along the other axis.
In the provided exercise, the double integral is split into two parts due to the addition in the function \( f_n(x, y) \):

• \( \int_{1}^{2} \int_{1}^{2} \frac{1}{x^n} \, dx \, dy \)
• \( \int_{1}^{2} \int_{1}^{2} \frac{1}{y^n} \, dx \, dy \)

This separation allows a more straightforward calculation and demonstrates the flexibility and power of double integrals in simplifying complex problems.

Limit of a Sequence

The concept of a limit is fundamental to calculus and serves as a bridge to understanding continuity and derivatives, especially when dealing with series or sequences.
A limit refers to the value that a sequence or function approaches as the variable of interest grows indefinitely. In many calculus problems, finding this behavior provides insights into long-term trends or asymptotic behavior.
In the example, the goal is to determine \( \lim_{n \to \infty} V_n \), the limiting behavior of the volume of solids as \( n \) goes to infinity. As seen, as \( n \) increases, the function \( 2^{1-n} \) approaches zero. Thus, the entire expression simplifies, and the volume \( V_n \) tends to zero, indicating that the solid's volume shrinks as \( n \) grows. This elegantly illustrates how limits can predict behavior in varying systems.