Problem 17 - Thomas Calculus

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 area of the region is \(\frac{3}{2}\).

Step 1: Analyze the Expression

We have a double integral split into two different areas:\[\int _ { - 1 } ^ { 0 } \int _ { - 2 x } ^ { 1 - x } d y d x + \int _ { 0 } ^ { 2 } \int _ { - x / 2 } ^ { 1 - x } d y d x\]The first integral corresponds to the region between the lines \(y = -2x\) and \(y = 1-x\), from \(x = -1\) to \(x = 0\). The second integral corresponds to the region between the lines \(y = -\frac{x}{2}\) and \(y = 1-x\), from \(x = 0\) to \(x = 2\).

Step 2: Determine Intersections

To find where the curves intersect, set the equations equal to each other:1. \(-2x = 1-x\) solves to \(x = -1\) and substituting in gives \(y = 2\).2. \(-\frac{x}{2} = 1-x\) solves to \(x = 2\) and substituting in gives \(y = -1\).3. \(-2x = -\frac{x}{2}\) solves to \(x = 0\), \(y = 0\).4. \(1-x = -\frac{x}{2}\) solves to \(x = \frac{2}{3}\), \(y = \frac{1}{3}\).Intersection points are \((-1, 2)\), \((0, 0)\), \(\left(\frac{2}{3}, \frac{1}{3}\right)\), and \((2, -1)\).

Step 3: Sketch the Region

Draw the x-y plane and plot the lines from the bounds of the integrals. Sketch the region determined by the lines \(y = -2x\), \(y = 1-x\), and \(y = -\frac{x}{2}\). Mark the intersection points \((-1, 2)\), \((0, 0)\), \(\left(\frac{2}{3}, \frac{1}{3}\right)\), and \((2, -1)\) to see the shape of the area we are integrating over.

Step 4: Evaluate the Integrals

Find the area of each region by evaluating the integrals. For the first integral:\[\int _ { - 1 } ^ { 0 } \int _ { - 2 x } ^ { 1 - x } d y d x = \int _ { -1 } ^ { 0 } \left( (1-x) - (-2x) \right) dx = \int _ { -1 } ^ { 0 } (1+ x) dx = [x + \frac{x^2}{2}]_{-1}^{0} = \frac{1}{2}\]For the second integral:\[\int _ { 0 } ^ { 2 } \int _ { - x / 2 } ^ { 1 - x } d y d x = \int _ { 0 } ^ { 2 } \left( (1-x) - (-\frac{x}{2}) \right) dx = \int _ { 0 } ^ { 2 } (1 - \frac{x}{2}) dx = \left[x - \frac{x^2}{4}\right]_0^2 = 1\]The area is the sum of the two evaluated integrals: \(\frac{1}{2} + 1 = \frac{3}{2}.\)

Key Concepts

Area of Regions

In mathematics, finding the area of regions using double integrals involves integrating over a specific region in the x-y plane. The given exercise divides the area into two distinct parts that are evaluated separately and then summed. This showcases the ability of double integrals to capture complex areas involving different bounding conditions. The area of a region may exist between intertwining functions, and to calculate it, the boundaries formed by these functions need to be clearly understood. In our problem, the total area in question is the sum of the two regions defined by double integrals within specified bounds. Such computations are crucial for pinpointing areas in engineering and physics applications as they describe surface properties precisely. Double integrals help visualize and quantify the region's dimensions, allowing us to understand their real-world implications better.

Intersection Points

To solve problems involving areas bound by multiple curves, identifying the intersection points of these curves is imperative. They help divide the entire plane into regions or provide necessary boundaries for integration. For the task at hand, the curves of interest include:

• The line given by the equation \(y = -2x \).
• The line given by \(y = 1-x \).
• The line expressed by \(y = -\frac{x}{2} \).

Through step-by-step algebraic manipulation, we determine the points where these lines intersect. Solving equations for these intersections delivers points such as \((-1, 2)\), \((0, 0)\), \(\left(\frac{2}{3}, \frac{1}{3}\right)\), and \((2, -1)\). These intersections assist in setting up boundaries for calculating definite integrals that define specific regions, which are otherwise not intuitive.

Bounding Curves

Bounding curves are critical in determining the scope of integration within an x-y plane. They outline the limits within which the integration is performed. In double integrals, understanding these bounding curves translates into recognizing the lines where integration begins and stops. In our example, for both regions, the outermost limits are described by lines:

• For \(x = -1\) to \(x = 0\), bound by \(y = -2x\) and \(y = 1-x\).
• For \(x = 0\) to \(x = 2\), bound by \(y = -\frac{x}{2} \) and \(y = 1-x\).

By solving these conditions, the exact shape of the subject region is known, and integrating within these limits provides the area encompassed by these bonds. Bounding curves simplify complex regions into calculable segments by providing definite start and endpoints for integrations.

Evaluating Integrals

Finally, evaluating integrals is the execution phase of determining the area. Following the identification of bounds and intersections, utilizing integration enables the computation over defined limits. For regions described, we performed the integration in two parts.The first integral, evaluated from \(x = -1\) to \(x = 0\), has inner bounds between \(y = -2x\) and \(y = 1-x\). Correct calculation showed the area contribution as \(\frac{1}{2}\).The second integral involves \(x = 0\) to \(x = 2\) with boundaries \(y = -\frac{x}{2} \) to \(y = 1-x\). Solving gives an area of \(1\). Summing these results, the total area computed is \(\frac{3}{2}\). When faced with integration tasks, ensuring correct bounds and intersections will lead to accurate evaluations of the geometry in question, showcasing the power of calculus in deriving solutions to spatial problems.