Problem 17 - University Calculus: Early Transcendentals

Question

The integrals and sums of integrals in Exercises \(13-18\) give the areas of regions in the \(x y\) -plane. Sketch each region, label each bounding curve with its equation, and give the coordinates of the points where the curves intersect. Then find the area of the region. $$\int_{-1}^{0} \int_{-2 x}^{1-x} d y d x+\int_{0}^{2} \int_{-x / 2}^{1-x} d y d x$$

Step-by-Step Solution

Answer

The total area is 3.

Step 1: Understand the Integral Setup

The given problem involves two double integrals with defined limits. These integrals define regions in the xy-plane. The first integral is \( \int_{-1}^{0} \int_{-2x}^{1-x} dy \, dx \), which covers the region from \( x = -1 \) to \( x = 0 \). For each \( x \), \( y \) ranges from \( y = -2x \) to \( y = 1 - x \). The second integral is \( \int_{0}^{2} \int_{-x/2}^{1-x} dy \, dx \), which covers the region from \( x = 0 \) to \( x = 2 \). For each \( x \), \( y \) ranges from \( y = -x/2 \) to \( y = 1 - x \).

Step 2: Identify Bounding Curves

For both integrals, identify the bounding curves. For the first integral, the bounding curves are \( y = -2x \) (lower bound) and \( y = 1 - x \) (upper bound). For the second integral, the bounding curves are \( y = -x/2 \) (lower bound) and \( y = 1 - x \) (upper bound). These curves intersect to form the area to be found.

Step 3: Sketch the Region

Sketch the region on the xy-plane by plotting the curves from both integrals. Start by sketching \( y = -2x \), which is a line passing through the origin with a slope of \(-2\), and \( y = -x/2 \), also passing through the origin but with a shallower slope. Sketch \( y = 1 - x \), a line that intercepts the y-axis at \( 1 \) with a slope of \(-1\). Shade the area between these curves over the given x-intervals. This includes the triangle formed between these lines.

Step 4: Find Intersection Points

Determine where the lines intersect. Solve \( -2x = 1-x \) to find the intersection for the first integral: \( -2x = 1 - x \) \( \Rightarrow x = 1 \). Solve \( -x/2 = 1 - x \) for the second integral: \( -x/2 = 1 - x \) \( \Rightarrow x = 2 \). Hence, the points of intersection are \((-1,3)\), \( (2,-2)\) and \((0,1)\)

Step 5: Evaluate the Integrals

Calculate each integral separately to find the area. For the first integral, evaluate \(\int_{-1}^{0} \int_{-2x}^{1-x} dy \, dx\). For the second integral, evaluate \(\int_{0}^{2} \int_{-x/2}^{1-x} dy \, dx\). Add these two results to find the total area of the region.

Step 6: Calculate Total Area

Calculate both double integrals to find the total area. The first integral evaluates to \( 1\), and the second integral evaluates to \( 2 \). Sum them to get the total area: \( 3 \).

Key Concepts

Bounding Curves

Bounding curves are crucial to understanding the area enclosed by double integrals. In this problem, they serve as the limits for the nested integrals that define specific regions in the xy-plane. Two sets of bounding curves need to be identified.

In the first double integral, the curves bounding the region are given as:

• \( y = -2x \): This is the lower boundary line and represents a slope steeply inclined downwards, crossing through the origin.
• \( y = 1 - x \): Serving as the upper boundary line, this curve crosses the y-axis at 1 and slopes downward, reaching zero at the point (1,0).

For the second integral, the bounding curves are:

• \( y = -x/2 \): This line slopes downwards more gently than the first, indicating a lesser rate of change for each step in \( x \).
• Again, \( y = 1 - x \): This upper boundary is shared between both integrals, thus contributing to forming a continuous boundary across both regions.

The intersection of these bounding curves defines the closed region whose area is to be computed.

Intersection Points

Intersection points are the coordinates where bounding curves meet, delineating the vertices of the enclosed region in the xy-plane. To find these, we solve equations for the pairs of curves:

1. **Intersection between \( y = -2x \) and \( y = 1-x \):** Solve \(-2x = 1 - x\). Reorganizing terms gives \( x = 1 \), which then gives \( y = -2(1) = -2 \). Thus, an intersection occurs at the point (1, -2).

2. **Intersection between \( y = -x/2 \) and \( y = 1-x \):** Solve \(-x/2 = 1 - x\). Simplifying provides \( x = 2 \), and thus \( y = -x/2 = -1 \). This intersection occurs at (2, -1).

The intersection points demarcate the joining segments and corners of the triangular regions bounded in part by the double integral.

Area in the XY-Plane

Calculating the area in the xy-plane involves understanding the terrains covered by the specified integrals. The goal is to find the cumulative area within the bounding curves for each integral section.

The process involves evaluating each integral separately and determining their sum. For the first integral: \[ \int_{-1}^{0} \int_{-2x}^{1-x} dy \, dx \]we calculate the "strip" of area for each x from -1 to 0, between the curves \( y = -2x \) and \( y = 1-x \). This gives an area of 1 unit.

The second integral: \[ \int_{0}^{2} \int_{-x/2}^{1-x} dy \, dx \]involves calculating another "strip" from x = 0 to 2 but this time between the curves \( y = -x/2 \) and \( y = 1-x \), yielding an area of 2 units. Summing the areas of each strip drawn under bounded curves provides the total area of the region, which is 3 square units.

Region of Integration

The region of integration refers to the actual space or area within the xy-plane defined by the specified bounds of the double integrals. For this exercise, it is essential to grasp how these regions are outlined.

This region of integration has been defined in two parts by the limits of the integrals, with each exhibiting a combination of vertical and horizontal boundaries provided by the bounding curves. When visualized, this often results in a composite shape or unified area that each double integral independently accounts for while combining their specific territories.

The first integral covers the territory from x = -1 to 0, with vertical slices extending from the line \( y = -2x \) up to \( y = 1-x \). Similarly, the second integral spans x from 0 to 2, but each slice runs vertically from \( y = -x/2 \) up to \( y = 1-x \). Together, they carve out a triangular region in the xy-plane, illustrating the effective range that our integral solution depends upon.