https://r2.erweima.ai/mathbot/image/1726855678704upie9lac.jpg

The problem asks to compute the integral \( \iint_A 3 \, dx \, dy \), where the function is \( f(x, y) = 3 \) and the region \( A \) is defined by a set of inequalities:

- \( y \geq -1 \)
- \( y \leq 0 \)
- \( y \geq -x + 2 \)
- \( y \leq \frac{x}{3} + \frac{5}{3} \)

### Step 1: Understand the region \( A \)
The region \( A \) is bounded by the lines described in the inequalities. Let's examine these inequalities one by one:

- \( y \geq -1 \): This is a horizontal line, where \( y = -1 \).
- \( y \leq 0 \): This is another horizontal line, where \( y = 0 \).
- \( y \geq -x + 2 \): This is a line with slope \( -1 \) and intercept \( 2 \).
- \( y \leq \frac{x}{3} + \frac{5}{3} \): This is a line with slope \( \frac{1}{3} \) and intercept \( \frac{5}{3} \).

### Step 2: Determine the bounds of integration
Now, to set up the double integral, we need to determine the bounds for \( x \) and \( y \). To do this, we need to find the points where the lines intersect and determine the limits for \( y \) in terms of \( x \).

#### Intersection points:
- Intersection of \( y = -x + 2 \) and \( y = 0 \): Set \( -x + 2 = 0 \), which gives \( x = 2 \).
- Intersection of \( y = \frac{x}{3} + \frac{5}{3} \) and \( y = 0 \): Set \( \frac{x}{3} + \frac{5}{3} = 0 \), which gives \( x = -5 \).

So the region \( A \) is bounded by \( x = -5 \) and \( x = 2 \).

### Step 3: Set up the double integral
The function \( f(x, y) = 3 \) is constant, so the integral becomes:

\[
\iint_A 3 \, dx \, dy = 3 \cdot \text{Area}(A)
\]

We now need to find the area of the region \( A \). The bounds for \( x \) are from \( -5 \) to \( 2 \), and for each \( x \), the bounds for \( y \) are from the line \( y = -x + 2 \) to the line \( y = \frac{x}{3} + \frac{5}{3} \).

Thus, the integral becomes:

\[
3 \int_{-5}^{2} \left( \frac{x}{3} + \frac{5}{3} - (-x + 2) \right) dx
\]

### Step 4: Compute the integral
Now we can compute the integral:

\[
\int_{-5}^{2} \left( \frac{x}{3} + \frac{5}{3} + x - 2 \right) dx
\]
Simplify the integrand:

\[
\frac{x}{3} + x + \frac{5}{3} - 2 = \frac{4x}{3} - \frac{1}{3}
\]

Now integrate:

\[
\int_{-5}^{2} \left( \frac{4x}{3} - \frac{1}{3} \right) dx = \frac{4}{3} \int_{-5}^{2} x \, dx - \frac{1}{3} \int_{-5}^{2} 1 \, dx
\]

Let's compute each part:

1. \( \frac{4}{3} \int_{-5}^{2} x \, dx = \frac{4}{3} \left[ \frac{x^2}{2} \right]_{-5}^{2} = \frac{4}{3} \left( \frac{2^2}{2} - \frac{(-5)^2}{2} \right) = \frac{4}{3} \left( \frac{4}{2} - \frac{25}{2} \right) = \frac{4}{3} \cdot \frac{-21}{2} = \frac{-84}{6} = -14 \)
2. \( \frac{1}{3} \int_{-5}^{2} 1 \, dx = \frac{1}{3} \left[ x \right]_{-5}^{2} = \frac{1}{3} \left( 2 - (-5) \right) = \frac{1}{3} \cdot 7 = \frac{7}{3} \)

Now subtract:

\[
-14 - \frac{7}{3} = \frac{-42}{3} - \frac{7}{3} = \frac{-49}{3}
\]

Thus, the area of \( A \) is \( \frac{-49}{3} \).

### Step 5: Multiply by 3
The integral is:

\[
3 \cdot \frac{-49}{3} = -49
\]

### Final Answer:
The value of the integral is \( \boxed{-49} \).

Would you like more details, or do you have any questions? Here are some related questions you could explore:

1. How do we handle non-constant functions in double integrals?
2. What happens when the region \( A \) is more complex?
3. Can we use polar coordinates for different types of regions?
4. How do we interpret the result of negative area or integral values?
5. What are the properties of definite integrals in multivariable calculus?

**Tip**: When solving a double integral, always sketch the region of integration to ensure you're setting the bounds correctly.

Given the set A = {(x, y) ∈ R | y ≥ -1, y ≤ 0, y ≥ -x + 2, y ≤ x/3 + 5/3} and the function f(x, y) = 3, find the integral ∫∫_A 3 dx dy.

Learn how to evaluate a double integral over a region bounded by linear inequalities. This problem demonstrates how to find the area between lines and compute the integral over the defined region. The function used in the example is f(x, y) = 3, and the solution involves setting up proper integration limits.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation