Problem 56 - Thomas Calculus

Question

Sketch the region of integration, reverse the order of integration, and evaluate the integral. \begin{equation} \begin{array}{l}{\text { Triangular region } \iint_{R} x y d A \text { where } R \text { is the region bounded }} \\ {\text { by the lines } y=x, y=2 x, \text { and } x+y=2}\end{array}\end{equation}

Step-by-Step Solution

Answer

The reversed integral evaluates to \( \frac{11}{48} \).

Step 1: Identify the Region of Integration

The given region R is bounded by the lines \( y = x \), \( y = 2x \), and \( x+y = 2 \). Start by understanding these line equations: \( y = x \) is a line with a slope of 1 passing through the origin, \( y = 2x \) has a steeper slope of 2 also passing through the origin, and \( x + y = 2 \) is a line with a slope of -1 intersecting the axes at \( (2,0) \) and \( (0,2) \).

Step 2: Sketch the Region

Plot the lines on the xy-plane. The intersection points provide the vertices of the triangular region: \( y=x \) intersects \( x+y=2 \) at (1,1), \( y=2x \) intersects \( x+y=2 \) at (2,0), and \( y=x \) intersects \( y=2x \) at (0,0). The triangular region R is therefore bounded by these points: (0,0), (1,1), and (2,0).

Step 3: Find New Limits of Integration

We are reversing the order of integration from \( dy \, dx \) to \( dx \, dy \). For a fixed \( y \), \( x \) varies from the line \( y=2x \) to \( y=x \). Thus, for fixed \( y \), \( x \) goes from \( \frac{y}{2} \) to \( y \) in the part of the region where \( 0 \leq y \leq 1 \), and from \( 0 \) to \( 2-y \) for \( 1 \leq y \leq 2 \).

Step 4: Set up the Reversed Integral

With the new limits, the double integral is split into two parts: \[ \int_{0}^{1} \int_{\frac{y}{2}}^{y} xy \, dx \, dy + \int_{1}^{2} \int_{0}^{2-y} xy \, dx \, dy \]

Step 5: Evaluate the Inner Integrals

Evaluate the inner integrals for each region:1. \( \int_{\frac{y}{2}}^{y} xy \, dx = \left[ \frac{x^2y}{2} \right]_{\frac{y}{2}}^{y} = \frac{y^3}{2} - \frac{y^3}{8} = \frac{3y^3}{8} \)2. \( \int_{0}^{2-y} xy \, dx = \left[ \frac{x^2y}{2} \right]_{0}^{2-y} = \frac{y(2-y)^2}{2} \)

Step 6: Evaluate the Outer Integrals

Now, compute the outer integrals:1. \( \int_{0}^{1} \frac{3y^3}{8} \, dy = \left[ \frac{3y^4}{32} \right]_{0}^{1} = \frac{3}{32} \)2. \( \int_{1}^{2} \frac{y(2-y)^2}{2} \, dy \): Expand \( (2-y)^2 \) to \( 4 - 4y + y^2 \). Integrate: \( \int_{1}^{2} \frac{y}{2}(4 - 4y + y^2) \, dy = \int_{1}^{2} (2y - 2y^2 + \frac{y^3}{2}) \, dy \) Solve this to get \( \frac{1}{6} \).

Step 7: Sum the Integrals

The total value of the original integral is the sum of the results obtained from both parts: \( \frac{3}{32} + \frac{1}{6} = \frac{11}{48} \).

Key Concepts

Region of Integration

In integration, defining the region of integration is crucial as it specifies the area over which we calculate the integral. In this problem, the region is denoted as \( R \) and is bounded by the lines \( y=x \), \( y=2x \), and \( x+y=2 \). These lines form a triangular region on the xy-plane.

Understanding the equations of these lines is the first step. The line \( y = x \) has a slope of 1, passing through the origin. The line \( y = 2x \) has a steeper slope of 2, also passing through the origin. The last equation \( x + y = 2 \) is a line with a negative slope of -1, crossing the axes at points \( (2,0) \) and \( (0,2) \).

These equations intersect at specific points to form the vertices of the triangle:

• The intersection of \( y=x \) and \( x+y=2 \) is at \( (1,1) \).
• The intersection of \( y=2x \) and \( x+y=2 \) is at \( (2,0) \).
• The intersection of \( y=x \) and \( y=2x \) is at \( (0,0) \).

This triangular region is what we integrate over.

Order of Integration

The order of integration refers to the sequence in which we integrate the variables in a multiple integral. In the problem provided, the integral's initial order is \( dy \, dx \). This means we first integrate with respect to \( y \) and then with respect to \( x \).

However, to solve the problem efficiently, we reverse this order to \( dx \, dy \). By reversing the order, new integration limits are set, which can sometimes simplify the process of solving the integral.

Reversing the integration order can also be necessary if the original order leads to complex or impossible integration limits based on the region's boundaries. With the reversed order, you ensure that evaluations of the integral become more manageable.

Integration Limits

Integration limits define the range for each variable during the integration process. Finding the integration limits is essential to accurately evaluate the integral.

In this problem, once the order of integration is reversed to \( dx \, dy \), the limits for \( y \) are clearly between \( 0 \) and \( 2 \) because this captures the entire height of the triangular region.

The limits for \( x \) change based on the value of \( y \). For \( 0 \leq y \leq 1 \), \( x \) varies from \( \frac{y}{2} \) (derived from \( y=2x \)) to \( y \) (as per \( y=x \)). For \( 1 \leq y \leq 2 \), \( x \) ranges from \( 0 \) to \( 2-y \) (from \( x+y=2 \)).

These limits ensure that we integrate precisely over the triangular region, capturing all the area without overlapping or extending beyond the boundaries provided by the line equations.

Sketching Regions

Sketching the region of integration is a practical step that aids in visualizing the areas over which you're integrating. It can prevent mistakes and clarify complex boundaries, especially with multiple intersecting lines or curves.

In this exercise, sketching the three lines \( y=x \), \( y=2x \), and \( x+y=2 \) on a coordinate plane helps you identify their intersections and thus the vertices of the triangular region. By sketching:

• Draw each line according to its slope and intercepts.
• Identify and mark the intersection points \( (0,0) \), \( (1,1) \), and \( (2,0) \).
• Shade or outline the triangular area formed by these points.

Such a visual representation assists in setting the correct integration limits and understanding the bounds of the integral. It's a simple yet powerful tool to ensure clarity and accuracy in setting up integrals.