Question

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

5

Step-by-step solution

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.

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 \(y = 2(x + 1)\).

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.

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 \(-y+2\).

Set Up the Double Integral

Convert the double integral limits into the region found earlier: \[\int_{0}^{2} \int_{-\frac{y}{2}-1}^{-y+2} y \, dx \, dy\].

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 \( = y[ - y + 2 -(-\frac{y}{2} - 1)] = y(-y + 2 + \frac{y}{2} + 1) = y(-\frac{y}{2} + 3). \)

Integrate with Respect to y

Now, integrate the resulting expression over y: \( \int_{0}^{2} y(-\frac{y}{2} + 3) \, dy \). This gives \( \int_{0}^{2} ( -\frac{y^2}{2} + 3y ) \, dy\).

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\( \left[ \frac{ - 8}{6} + 6 \right] - 0 \right)= 5 \).

Key concepts

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 \( x = -y + 2 \).

These boundaries split the region correctly and allow the double integral to compute the desired area or function over the addressed region. Proper sketching helps in visualizing these limits.

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: \( y = 0 \).

These equations help in setting up the accurate limits for integrating. Sketching the triangle is beneficial as it provides a clearer view of how the limits and the region look.

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: \( \left[ \frac{-8}{6} + 6 \right] - 0 = 5 \).

Breaking complex integrals into step-by-step parts makes them manageable and much easier to solve!