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Chapter 15: Problem 82

How would you evaluate the double integral of a continuous function \(f(x, y)\) over the region \(R\) in the \(x y\) -plane enclosed by the triangle with vertices \((0,1),(2,0),\) and \((1,2) ?\) Give reasons for your answer.

Short Answer

Expert verified
Integrate over the triangular region by setting up a double integral with appropriate boundaries from vertex equations.

Step by step solution

01

Define the region of integration

First, identify the region of integration, which is the triangle with vertices \( (0, 1), (2, 0), \) and \( (1, 2) \). These vertices form a triangular region in the \( xy\)-plane.
02

Determine the boundaries

Determine the lines that form the boundaries of the triangular region. The sides of the triangle can be described by the lines connecting the vertices: between \((0, 1) \) and \((2, 0) \), between \((2, 0) \) and \((1, 2) \), and between \((1, 2) \) and \( (0, 1) \).
03

Find equations of the boundary lines

Find the equations for each side of the triangle: 1. The line from \((0, 1)\) to \((2, 0) \): \( y = 1 - 0.5x \).2. The line from \((2, 0)\) to \((1, 2) \): \( y = -2x + 4 \).3. The line from \((1, 2)\) to \((0, 1) \): \( y = x + 1 \).These equations will help set the limits of integration.
04

Set up the double integral

Decide the order of integration. Here, it's simpler to integrate with respect to \( x \) first and then \( y \). The integration limits for \( x \) are functions of \( y \). The integral becomes: \[\int_{y=0}^{2} \int_{x=g_1(y)}^{g_2(y)} f(x, y) \, dx \, dy\]Where \( g_1(y) = \frac{y - 1}{-0.5} \) and \( g_2(y) = \min(2-y, \frac{y-4}{-2}) \).
05

Evaluate the inner integral

For the inner integral, integrate \( f(x, y) \) with respect to \( x \) from \( x=g_1(y) \) to \( x=g_2(y) \). This will yield a function of \( y \).
06

Evaluate the outer integral

Substitute the result of the inner integral into the outer integral and integrate with respect to \( y \) from \( y=0 \) to \( y=2 \). This will provide the result of the double integral over the triangular region.

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Key Concepts

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

Triangular Region
When dealing with double integrals, understanding the shape of the region of integration is crucial. In this case, the region is a triangle formed by the vertices
  • (0,1)
  • (2,0)
  • (1,2)
These points are plotted in the xy-plane to create a triangular region.
To comprehend the triangular region better, visualize it as a polygon. This shape will serve as the domain over which we integrate the continuous function.
To find the boundaries that define this triangle, connect the given points with straight lines. These lines help us establish the limits of integration necessary for the double integral.
Continuous Function
In calculus, a continuous function is key when evaluating integrals. A function, \(f(x, y)\), is continuous if there are no breaks, jumps, or holes in its graph over the region we are interested in.
For double integrals over a region like our triangle, having a continuous function ensures that the area under the surface represented by \(f(x, y)\) can be accurately calculated.
The continuity of the function simplifies evaluating the integral, as theoretical properties like the Fundamental Theorem of Calculus hold.
  • Makes computations easier.
  • Results are more reliable.
  • Integral methods assume continuity for simplicity.
Region of Integration
The region of integration is the area over which the double integral is evaluated. In our exercise, this is the triangular region we've defined.
To set up the integral, we need to express this region clearly using its boundaries. These are the lines deduced from:
  • The line between (0,1) and (2,0).
  • The line between (2,0) and (1,2).
  • The line between (1,2) and (0,1).
Defining the correct region is crucial. It determines where the function \(f(x, y)\) is integrated over.
By clearly defining these boundaries, we get precise limits for our integrals, ensuring we calculate the integral correctly within the triangle.
Limits of Integration
Setting the correct limits of integration is crucial when evaluating a double integral over the triangular region. The problem requires setting up the integral with respect to both \(x\) and \(y\).
For this exercise, we integrate first with respect to \(x\). We determine the limits for \(x\) as functions of \(y\), based on our boundary lines:
  • Lower limit: This corresponds to the line from (0,1) to (2,0) with equation \( y = 1 - 0.5x \).
  • Upper limit: This is determined by taking the minimum of the other lines.
Once the x-limits are set, we integrate with respect to \(y\) over its range from 0 to 2. The correct application of these limits ensures that we evaluate the integral over the desired triangular area.

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Most popular questions from this chapter

Sketch the region of integration and evaluate the integral. \begin{equation} \int_{0}^{1} \int_{0}^{y^{2}} 3 y^{3} e^{x y} d x d y \end{equation}

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Use a CAS double-integral evaluator to estimate the values of the integrals. \begin{equation}\int_{0}^{1} \int_{0}^{1} e^{-\left(x^{2}+y^{2}\right)} d y d x\end{equation}
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