Problem 22 - Thomas Calculus

Question

In Exercises \(17-24\) , evaluate the double integral over the given region \(R .\) $$\iint_{R} x y e^{\mathrm{y}^{2}} d A, \quad R : \quad 0 \leq x \leq 2, \quad 0 \leq y \leq 1$$

Step-by-Step Solution

Answer

The value of the double integral is \( e - 1 \).

Step 1: Define the Region of Integration

The region of integration \( R \) is defined by the inequalities \( 0 \leq x \leq 2 \) and \( 0 \leq y \leq 1 \). This region is rectangular in the \( xy \)-plane.

Step 2: Set Up the Double Integral

The given double integral can be set up with the integrand \( xy e^{y^2} \) and the defined region \( R \). Hence, the double integral is \[ \int_{0}^{1} \int_{0}^{2} xy e^{y^2} \, dx \, dy. \] The outer integral represents the integral with respect to \( y \), and the inner one is with respect to \( x \).

Step 3: Evaluate the Inner Integral

Evaluate \( \int_{0}^{2} xy e^{y^2} \, dx \). Notice that \( xy e^{y^2} \) treats \( y \) as a constant while integrating with respect to \( x \). So, the integral becomes \[ xy e^{y^2} \cdot \frac{x^2}{2} \bigg|_0^2 = y e^{y^2} \cdot \frac{2^2}{2} = 2y e^{y^2}. \]

Step 4: Evaluate the Outer Integral

The outer integral to evaluate is \( \int_{0}^{1} 2y e^{y^2} \, dy \). Notice that the integrand suggests a substitution where \( u = y^2 \), which implies \( du = 2y \, dy \). Thus, the integral in terms of \( u \) becomes \[ \int e^u \, du = e^u + C. \] Transform back in terms of \( y \), we have \( e^{y^2} \bigg|_0^1 \).

Step 5: Compute the Final Result

Evaluate the expression \( e^{y^2} \) from \( 0 \) to \( 1 \), giving \( e^1 - e^0 = e - 1 \).

Step 6: Conclude the Calculation

The result of the original double integral is \( e - 1 \).

Key Concepts

Region of Integration

In calculus, when dealing with double integrals, it is crucial to first define the region of integration. The region of integration refers to the specific part of the plane over which you will integrate your function. In this exercise, the region of integration \( R \) is defined by the inequalities \( 0 \leq x \leq 2 \) and \( 0 \leq y \leq 1 \). This essentially describes a rectangle in the \( xy \)-plane.

Understanding the limits of integration:

• The range of \( x \) values goes from 0 to 2, creating the width of the rectangle.
• The range of \( y \) values goes from 0 to 1, forming the height of the rectangle.

Knowing this helps you visualize the area over which you want to integrate the function. It's like setting up the boundaries within which you will calculate the integral. It's one of the steps that lay the groundwork for solving the integral correctly.

Substitution Method

The substitution method is a helpful technique used to simplify the process of integration, especially with complex integrals. This method is akin to reverse differentiation or applying the chain rule in reverse.

In this step of the exercise, after evaluating the inner integral, we encounter the outer integral \( \int_{0}^{1} 2y e^{y^2} \, dy \). The integrand \( 2y e^{y^2} \) can appear complicated at first glance, but substitution can simplify it.

Key steps in substitution:

• Identify a part of the integrand whose derivative appears elsewhere in the integral. Here, let \( u = y^2 \) since the derivative \( du = 2y \, dy \) matches part of the integrand.
• Replace \( 2y \, dy \) with \( du \) and change the variable from \( y \) to \( u \). The integral in terms of \( u \) then becomes \( \int e^u \, du \).
• Integration now simplifies to expressing as \( e^u + C \). Finally, substitute back \( u = y^2 \) to get \( e^{y^2} \).

Using substitution helps to streamline the evaluation process, making integration easier and more intuitive.

Rectangular Coordinates

Rectangular coordinates, sometimes known as Cartesian coordinates, use two numerical values to specify each point's horizontal and vertical position within a plane. This system is based on two intersecting lines (axes), usually the x-axis and y-axis, that divide the plane into quadrants.

In double integrals like the one in this exercise, rectangular coordinates are indispensable for defining and setting up the region of integration.

• The x-coordinate (horizontal axis) helps define the left and right boundaries of the rectangular region you’re integrating over, e.g., from \( 0 \) to \( 2 \).
• The y-coordinate (vertical axis) outlines the top and bottom limits, for instance, from \( 0 \) to \( 1 \).

By using rectangular coordinates, mathematicians can easily set boundaries and visualize the region of integration when dealing with multivariable functions. This approach becomes the foundation for more advanced calculus topics, making it a critical tool for students to master.