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Chapter 13: Problem 80

Use double integrals to compute the area of the following regions. Make a sketch of the region. The region bounded by the lines \(x=0, x=4, y=x\), and \(y=2 x+1\)

Short Answer

Expert verified
In the given problem, we are asked to find the area of the region bounded by the lines \(x=0\), \(x=4\), \(y=x\), and \(y=2x+1\). Following the steps: 1. Sketch the region and find intersection points. 2. Set up the double integral to represent the area. 3. Evaluate the double integral. After completing the steps, we found that the area of the region bounded by the given lines is \(12\) square units.

Step by step solution

01

Sketch the region and find intersection points

First, let's draw the lines on the coordinate plane and find their intersection points: 1. \(x=0\): Vertical line passing through the origin. 2. \(x=4\): Vertical line passing through \((4,0)\). 3. \(y=x\): Diagonal line passing through the origin with a slope of \(1\). 4. \(y=2x+1\): Diagonal line passing through \((0,1)\) with a slope of \(2\). Intersection points: - \(x=0\) and \(y=x\): Point \((0,0)\). - \(x=0\) and \(y=2x+1\): Point \((0,1)\). - \(x=4\) and \(y=x\): Point \((4,4)\). - \(x=4\) and \(y=2x+1\): Point \((4,9)\). - \(y=x\) and \(y=2x+1\): Set \(x=y\) and \(2x+1=y\), solving for x, we get \(x=1\) and \(y=1\) (Intersection Point is \((1,1)\)). Now, we have the intersection points, and we can sketch the region.
02

Set up the double integral

Since the expressions given are in terms of \(x\) and \(y\), we can use either \(dy dx\) or \(dx dy\). In this case, it is easier to integrate over the region with respect to \(y\) first and then \(x\). We will calculate the area using the following double integral: $$ Area = \int_{a}^{b} \int_{c}^{d} dA, $$ where \(a\) and \(b\) are the limits of integration with respect to \(x\), and \(c\) and \(d\) are the limits of integration with respect to \(y\). The limits of integration can be calculated from the intersection points and the equations of the lines: $$ a=x=0, \quad b=x=4, \\ c=y=x, \quad d=y=2x+1. $$ Now, we set up the double integral: $$ Area = \int_{0}^{4} \int_{x}^{2x+1} dy dx. $$
03

Evaluate the double integral

Now, evaluate the double integral by first integrating with respect to \(y\) and then with respect to \(x\): $$ Area = \int_{0}^{4} \left[ \int_{x}^{2x+1} (1) dy \right] dx = \int_{0}^{4} (2x+1 - x) dx. $$ Evaluate the integral with respect to \(x\): $$ Area = \int_{0}^{4} (x+1) dx = \left[\frac{1}{2}x^2 + x\right]_{0}^{4} = \frac{1}{2}(4^2) + 4 - 0 = 12. $$ So, the area of the region bounded by the given lines is \(12\) square units.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Area Calculation
Calculating the area of a region defined by several boundaries is a fundamental application of double integrals in multivariable calculus. The technique involves creating a visual representation of the region, commonly referred to as sketching, which aids in visualizing the space that will be analyzed.

For the given exercise, the region of interest is a polygon bounded by the lines described by the equations:
  • \( x=0 \: \) which is a vertical line passing through the origin.
  • \( x=4 \: \) which is another vertical line but passing through the point (4,0).
  • \( y=x \: \) is a diagonal line indicating a 45-degree angle with the x-axis, passing through the origin.
  • \( y=2x+1 \: \) is also a diagonal line, but with a steeper slope than the line \( y = x \), passing through the point (0,1).
Once these lines are plotted, and the intersection points determined, we have a well-defined closed region whose area can be computed. The integral is set up over the region, and the sketch assists in the identification of the variable limits and points of intersection.
Integration Limits
Integration limits are crucial for performing double integrals; they define the range over which the integration will occur. To correctly calculate the area of the region in question, it is essential to determine the correct integration limits for both variables鈥攐ften denoted as \( x \) and \( y \)鈥攖hat outline the region.

In the exercise, after the region is sketched and the intersection points identified, the limits can be deduced: For \( x \) the range is from 0 to 4, and for \( y \) the limits vary with \( x \) and are bound between \( y = x \) and \( y = 2x+1 \) for each respective value of \( x \) within its range. These limits, once established, are applied to the double integral鈥擻( c \) and \( d \) are the moving vertical boundaries (respecting \( y \) values) while \( a \) and \( b \) are the fixed horizontal boundaries (respecting \( x \) values). Understanding and correctly setting these limits are key steps in the process.
Double Integral Evaluation
Once the area is sketched and integration limits are set, the next step is evaluating the double integral, which computes the total area of the region. The process involves integrating with respect to one variable first, generally the one with variable limits, followed by the other.

In the given problem, the evaluation begins with integrating with respect to \( y \) across the boundary lines from \( y=x \) to \( y=2x+1 \) for each value of \( x \) between 0 and 4. The integral is calculated in a nested fashion鈥攆irst the inside integral with respect to \( y \) is solved, leading to an expression in terms of \( x \) alone. Subsequently, we integrate this resulting function with respect to \( x \) to find the area.

The evaluation of the inner integral with respect to \( y \) is relatively straightforward since the integral of 1 with respect to \( y \) is just \( y \) itself. Once we subtract the bottom \( y \) limit from the top, the resulting function is then integrated with respect to \( x \), yielding the area of the defined region as the final answer.

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