Problem 12 - Thomas Calculus

Question

In Exercises \(11-16,\) integrate \(f\) over the given region. \(\begin{array}{ll}{\text { Square } f(x, y)=1 /(x y)} & {\text { over } \quad \text { the square } \quad 1 \leq x \leq 2} \\ {1 \leq y \leq 2}\end{array}\)

Step-by-Step Solution

Answer

The integral over the region is \((\ln 2)^2\).

Step 1: Set Up the Double Integral

To find the integral of a function over a given region, we first express it as a double integral. Here, the function is given as \( f(x, y) = \frac{1}{xy} \). We integrate over the square region defined by \( 1 \leq x \leq 2 \) and \( 1 \leq y \leq 2 \). The double integral will be: \[\int_{1}^{2} \int_{1}^{2} \frac{1}{xy} \, dy \, dx.\]

Step 2: Integrate Inner Integrand with Respect to y

First, we integrate the function \( \frac{1}{xy} \) with respect to \( y \) from 1 to 2. Since \( \frac{1}{xy} = \frac{1}{x} \cdot \frac{1}{y} \), we can take \( \frac{1}{x} \) outside the inner integral. The inner integral becomes:\[\int_{1}^{2} \frac{1}{y} \, dy = \left[ \ln |y| \right]_{1}^{2} = \ln |2| - \ln |1| = \ln 2.\]

Step 3: Integrate Outer Integrand with Respect to x

Now, substitute \( \ln 2 \) back into the outer integral and integrate with respect to \( x \):\[\int_{1}^{2} \frac{\ln 2}{x} \, dx.\]\( \ln 2 \) is a constant, so it can be factored out:\[\ln 2 \int_{1}^{2} \frac{1}{x} \, dx = \ln 2 \left[ \ln |x| \right]_{1}^{2} = \ln 2 (\ln 2 - \ln 1) = (\ln 2)^{2}.\]

Step 4: Final Result

After computing both integrals, we conclude that the value of the double integral is \((\ln 2)^2\). This represents the integral of \( \frac{1}{xy} \) over the square region from \( (1, 1) \) to \( (2, 2) \).

Key Concepts

Region of Integration

The concept of a region of integration refers to the specific area over which we are calculating the double integral. In the given problem, our region of integration is a square. This square is bounded by the lines \(x = 1\) to \(x = 2\) and \(y = 1\) to \(y = 2\).
Imagine a simple square on a graph where the bottom left corner is at \((1,1)\) and the top right corner is at \((2,2)\). This is the area we are concerned with.

To find a double integral over such a region, you will first look at these bounds and integrate the function within them. Integration can be thought of as "adding up" small rectangles within this square to find the total area under the curve defined by the function. In this case, the function is \(f(x, y) = \frac{1}{xy}\).
The region of integration is fundamental because it determines the limits of both the inner and outer integrals.

Iterated Integrals

Iterated integrals refer to the process of computing a double integral by performing one integral at a time, in sequence. In essence, we are breaking a two-dimensional integral into two, one-dimensional integrals.

In our example, we evaluate the integral by first integrating with respect to \( y \) while treating \(x\) as a constant, and then with respect to \(x\).

• First, you compute the inner integral: \( \int_{1}^{2} \frac{1}{y} \, dy \).
• This gives us \( \ln 2 \) as the result of the inner integration.

Next, you substitute back into the outer integral, keeping \( \ln 2 \) as a constant factor:

• Then, compute the outer integral: \( \int_{1}^{2} \frac{\ln 2}{x} \, dx \).
• This results in \((\ln 2)^{2}\).

By iterating over two levels of integration, we can find the total value of the integral over the region. Iterated integrals simplify the process by handling one variable at a time.

Logarithmic Functions

Logarithmic functions, like the natural logarithm \( \ln x \), play an essential role in integration. In this problem, they arise naturally when integrating functions of the form \( \frac{1}{x} \).

Understanding properties of logarithms is crucial. When you integrate \( \frac{1}{x} \) from 1 to 2, you use the fundamental theorem of calculus to express the result as \( [ \ln |x| ]_{1}^{2} = \ln 2 - \ln 1 \).
This subtraction simplifies to \( \ln 2 \), since \( \ln 1 = 0 \).

Logarithms help translate multiplicative processes into additive ones. In our problem, it helps simplify the integration process by turning the inner integral into a familiar form. Also, constants like \( \ln 2 \) can be easily managed when carrying over into the next level of integration in iterated integrals. Logarithmic functions ensure that complex expressions can be manipulated more easily, leading to simpler results like \( (\ln 2)^2 \).