Chapter 5: Problem 12
Evaluate the double integrals over the areas described. To find the limits, sketch the area and compare \(\iint y d x d y\) over the triangle with vertices \((-1,0),(0,2)\), and \((2,0)\).
Short Answer
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Step by step solution
01
Understand the Region
First, sketch the triangle with vertices at \((-1,0)\), \(0,2)\) , and \((2,0)\). This helps visualize the region over which to integrate.
02
Determine the Equations of the Triangle's Boundaries
1. For the line connecting \((-1,0)\) and \((0,2))\), use the points to find the slope and intercept: The equation is \(y = 2(x + 1)\).
03
Find the Equation of the Other Boundaries
2. For the line connecting the points \((0,2)\) and \((2,0))\), the equation is \(y = -x + 2\).3. For the line connecting \((-1,0)\) and \((2,0))\), the equation is simply y = 0.
04
Set Up the Limits for the Integral
To integrate, one needs to split the integral into two parts or properly set up the limits. The limits in y are from 0 to 2. The x-limits depend on y: from \(-\frac{y}{2}-1\) to \(-y+2\).
05
Set Up the Double Integral
Convert the double integral limits into the region found earlier: \[\int_{0}^{2} \int_{-\frac{y}{2}-1}^{-y+2} y \, dx \, dy\].
06
Integrate with Respect to x
Integrate the inner integral first: \( \int_{- \frac{y}{2}-1}^{-y + 2} y \, dx = y [x]_{- \frac{y}{2}-1}^{-y + 2}\) which simplifies to \( = y[ - y + 2 -(-\frac{y}{2} - 1)] = y(-y + 2 + \frac{y}{2} + 1) = y(-\frac{y}{2} + 3). \)
07
Integrate with Respect to y
Now, integrate the resulting expression over y: \( \int_{0}^{2} y(-\frac{y}{2} + 3) \, dy \). This gives \( \int_{0}^{2} ( -\frac{y^2}{2} + 3y ) \, dy\).
08
Evaluate the Integral
Calculate the integral \( \int_{0}^{2} ( -\frac{y^2}{2} + 3y ) \, dy =\ \left[\frac{-y^3}{6} + \frac{3y^2}{2} \right]_{0}^{2} \). This simplifies to \left\( \left[ \frac{ - 8}{6} + 6 \right] - 0 \right)= 5 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Double Integrals
Double integrals are a powerful tool for calculating areas, volumes, and other quantities across two-dimensional regions. They extend the concept of a single integral, which measures area under a curve, to two dimensions. Think of a double integral as summing up tiny pieces of area in a region. When you see a double integral, it means you're working in a plane. For example, if you see something like \( \iint_R f(x, y) \, dA \), it means you're summing the function \( f(x, y) \) over a region \( R \). The double integral notation \( \iint \, dx \, dy \) means you'll integrate first with respect to \( x \), then \( y \). This order can also be reversed. Each order might simplify the problem differently, depending on the region of integration.
Limits of Integration
Understanding the limits of integration is crucial for evaluating double integrals. The limits define the region over which you're summing. For our triangle with vertices at \((-1,0)\), \((0,2)\), and \( (2,0) \), the region is bounded by lines. To set the limits:
- First, identify the range for \( y \) which, here, is from 0 to 2.
- Next, find how \( x \) changes for a given \( y \). For our triangle, \( x \) ranges from the line \( x = -\frac{y}{2} - 1 \) to \( x = -y + 2 \).
Triangle Area Calculation
To successfully compute a double integral over a triangular region, understanding the equations of the triangle's boundaries is essential:
- For the line \((-1,0)\) to \((0,2)\), we found: \( y = 2(x + 1) \).
- For the line \((0,2)\) to \((2,0)\), the equation is: \( y = -x + 2 \).
- The base is simply the x-axis: \( y = 0 \).
Integration Technique
Evaluating double integrals involves integrating one variable at a time. Here's how you solve our example:
- Set up the integral: \( \int_{0}^{2} \int_{-\frac{y}{2} - 1}^{-y +2} y \, dx \, dy \).
- First, integrate with respect to \( x \), treating \( y \) as a constant:
\( \int_{-\frac{y}{2}-1}^{-y+2} y \, dx \). This simplifies to:
\( y \left[ x \right]_{-\frac{y}{2}-1}^{-y+2} = y \left( -y + 2 + \frac{y}{2} + 1 \right) = y \left( -\frac{y}{2} + 3 \right) \). - Next, integrate the resulting expression with respect to \( y \):
\( \int_{0}^{2} y \left( -\frac{y}{2} + 3 \right) \, dy = \int_{0}^{2} \left( -\frac{y^2}{2} + 3y \right) \, dy \). - Finally, calculate the integral:
\( \int_{0}^{2} \left( -\frac{y^2}{2} + 3y \right) \, dy = \left[ \frac{-y^3}{6} + \frac{3y^2}{2} \right]_{0}^{2} \), which simplifies to: \( \left[ \frac{-8}{6} + 6 \right] - 0 = 5 \).