Problem 21 - Thomas Calculus

Question

Suppose that \(f(x, y)\) can be written as a product \(f(x, y)=F(x) G(y)\) of a function of \(x\) and a function of \(y\) . Then the integral of \(f\) over the rectangle \(R : a \leq x \leq b, c \leq y \leq d\) can be evaluated as a product as well, by the formula $$ \iint_{R} f(x, y) d A=\left(\int_{a}^{b} F(x) d x\right)\left(\int_{c}^{d} G(y) d y\right) $$ The argument is that $$ \begin{aligned} \iint_{R} f(x, y) d A &=\int_{c}^{d}\left(\int_{a}^{b} F(x) G(y) d x\right) d y \\ &=\int_{c}^{d}\left(G(y) \int_{a}^{b} F(x) d x\right) d y \\ &=\int_{c}^{d}\left(\int_{a}^{b} F(x) d x\right) G(y) d y \\\ &=\left(\int_{a}^{b} F(x) d x\right) \int_{c}^{d} G(y) d y \end{aligned} $$ a. Give reasons for steps (i) through (v). When it applies, Equation \((1)\) can be a time saver. Use it to evaluate the following integrals. $$ \text { b. }\int_{0}^{\ln 2} \int_{0}^{\pi / 2} e^{x} \cos y d y d x \quad \text { c. } \int_{1}^{2} \int_{-1}^{1} \frac{x}{y^{2}} d x d y $$

Step-by-Step Solution

Answer

The evaluated integral for (b) is 1, but (c) needs reevaluation due to improper conditions.

Step 1: Understand the Integral Problem

The problem is about evaluating an integral of the function \(f(x, y)=F(x)G(y)\) over a rectangle \(R: a \leq x \leq b, c \leq y \leq d\). Given that \(f(x, y)\) is separable into functions of \(x\) and \(y\), this integral can be calculated as a product of two separate integrals: \[\iint_{R} f(x, y) dA = \left(\int_{a}^{b} F(x) dx\right)\left(\int_{c}^{d} G(y) dy\right)\].

Step 2: Explanation of the Derivation

The derivation explains how the double integral is separated into single integrals due to the function's separability. 1. The integral \(\iint_{R} f(x, y) dA\) is expressed as \(\int_{c}^{d}\left(\int_{a}^{b} F(x) G(y) dx\right) dy\), which means you're integrating over \(x\) first while \(y\) is constant.2. Inside, \(F(x)\) is integrated over \([a, b]\) giving \(G(y)\left(\int_{a}^{b} F(x) dx\right)\).3. The structure allows us to factor out the integral of \(F(x)\), reducing it to the integral of \(G(y)\) over \([c, d]\), i.e., \(\left(\int_{a}^{b} F(x) dx\right)\int_{c}^{d} G(y) dy\).

Step 3: Solving Problem (b)

For \(\int_{0}^{\ln 2} \int_{0}^{\pi / 2} e^{x} \cos y \, dy \, dx\):Recognize that \(f(x, y) = e^x \cos y\) which is separable, where \(F(x) = e^x\) and \(G(y) = \cos y\). The integral can be split into a product:\[\left(\int_{0}^{\ln 2} e^x \, dx\right) \left(\int_{0}^{\pi / 2} \cos y \, dy\right)\]. Calculate each separately:1. \(\int_{0}^{\ln 2} e^x \, dx = [e^x]_0^{\ln 2} = e^{\ln 2} - e^0 = 2 - 1 = 1\).2. \(\int_{0}^{\pi/2} \cos y \, dy = [\sin y]_0^{\pi/2} = \sin(\pi/2) - \sin(0) = 1 - 0 = 1\).Thus, the value is 1.

Step 4: Solving Problem (c)

For \(\int_{1}^{2} \int_{-1}^{1} \frac{x}{y^{2}} \, dx \, dy\):Here, \(f(x, y) = \frac{x}{y^2}\), separable into \(F(x) = x\) and \(G(y) = \frac{1}{y^2}\). The integral can be rearranged as:\[\left(\int_{1}^{2} x \, dx\right) \left(\int_{-1}^{1} \frac{1}{y^2} \, dy\right)\].Compute each:1. \(\int_{1}^{2} x \, dx = \left[\frac{x^2}{2}\right]_1^2 = \frac{4}{2} - \frac{1}{2} = 2 - \frac{1}{2} = \frac{3}{2}\).2. \(\int_{-1}^{1} \frac{1}{y^2} \, dy\) is improper because of division by zero at \(y = 0\), traditionally split and considered as limits approaching \(y = 0\) from left and right.Therefore, further reevaluation or treatment as a limit is needed for the validity of this integral.

Key Concepts

Separability of Functions

In calculus, many functions of two or more variables can be expressed as the product of two individual functions. This concept is known as the separability of functions. When a function of two variables, say \( f(x, y) \), can be expressed as \( f(x, y) = F(x) \cdot G(y) \), it becomes easier to evaluate the integral over a given domain.

This characteristic enables us to break down complex double integrals into simpler, manageable parts. Specifically, it turns the double integral into the product of two single integrals. This simplification is incredibly useful in solving problems related to physics and engineering, where such functions frequently appear. When dealing with a separable function over a rectangle \( R: a \leq x \leq b, \ c \leq y \leq d \), the integral is given by:

• \( \int_{a}^{b} F(x) \, dx \), integrating over \( x \) only, and
• \( \int_{c}^{d} G(y) \, dy \), integrating over \( y \) only.

This property dramatically reduces the effort in calculating multi-variable integrals, allowing complex problems to be handled more efficiently.

Iterated Integrals

Iterated integrals are a method used to compute double integrals by performing integration sequentially. When you encounter a double integral like \( \iint_R f(x, y) \, dA \), it indicates the need to integrate over two dimensions.

First, one performs the integration with respect to one variable, holding the other constant; this is known as the inner integral. Once the inner integration is complete, you proceed to the outer integral, which involves the other variable. This step-by-step approach is what characterizes iterated integrals. The usual format for such an integral over a rectangle \( R \) is:

• \( \int_{c}^{d} \left( \int_{a}^{b} f(x, y) \, dx \right) \, dy \) or equivalently
• \( \int_{a}^{b} \left( \int_{c}^{d} f(x, y) \, dy \right) \, dx. \)

In both cases, you first handle one dimension, transforming what may appear as an overwhelming task into more approachable segments. When functions are separable, this process becomes even more straightforward, as it allows for factoring out the integrals of \( F(x) \) and \( G(y) \) as individual components.

Improper Integrals

Improper integrals occur when the integrand becomes undefined, such as when division by zero might occur at some points in the interval, or when limits extend to infinity. For example, taking the integral \( \int_{-1}^{1} \frac{1}{y^2} \, dy \) presents a situation where the function becomes undefined at \( y = 0 \).

Such scenarios require special treatment by integrating around the undefined point and considering limits. This typically involves splitting the integral at the point of discontinuity and taking limits from both directions. In formal terms, for an integral over a discontinuous point \( c \) within an interval \( [a, b] \), one would evaluate:

• \( \lim_{t \to c^-} \int_{a}^{t} f(y) \, dy \)
• \( \lim_{t \to c^+} \int_{t}^{b} f(y) \, dy \)

By using limits, we capture the behavior of the function near the discontinuity. Improper integrals play a significant role in analysis and applications, especially in contexts where outcomes are sensitive to behavior around singularities or when quantities extend towards infinity.