Chapter 5: Problem 7

Evaluate the double integrals over the areas described. To find the limits, sketch the area and compare $$ \iint_{A}(2 x-3 y) d x d y \text {, where } A \text { is the triangle with vertices }(0,0),(2,1),(2,0) \text {. } $$

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Step by step solution

Sketch the Area

Sketch the triangle with vertices (0,0), (2,1), and (2,0). This triangle lies in the x-y plane.

Determine the Limits of Integration

Observe the triangular area: for each value of x ranging from 0 to 2, y ranges from the line y = 0 up to the line y = 0.5x. Thus, the limits are 0 ≤ x ≤ 2 and 0 ≤ y ≤ 0.5x.

Set Up the Double Integral

Use the established limits to set up the integral:

Integrate with Respect to y

Evaluate the inner integral:

Integrate with Respect to x

Evaluate the outer integral:

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Key Concepts

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

Sketching Regions

To start understanding the concept of double integrals, it's essential to visualize the region over which the integration occurs. In our problem, we need to sketch the triangular region with vertices at $(0,0)$, $(2,1)$, and $(2,0)$.

Begin by plotting these points on the x-y plane.
Next, connect the points to form the triangle.
Note that this triangle lies entirely in the first quadrant.

Sketching helps you see the area you're working with, making it easier to understand the integration limits and setup later.

Limits of Integration

After sketching the region, the next step is to determine the limits of integration. For each x, we need to see how y values range within the triangular area.

First, observe that the base of the triangle runs from x = 0 to x = 2.
For each x in this interval, y starts at 0 and goes up to the line described by varies depending on the triangle's boundaries.
By observing the triangle, we can see y changes from 0 to $\frac{1}{2}x$.

This gives us the range of 0 ≤ x ≤ 2 and 0 ≤ y ≤ $\frac{1}{2}x$. Setting these limits correctly is crucial for accurately evaluating the integral.

Evaluating Integrals

Now, we set up and solve the double integral using our limits.

First, write the integral as $\begin{aligned} \iint_{A}(2x-3y)dA &= \int_{0}^{2} \int_{0}^{0.5x} (2x - 3y) dy dx \end{aligned}$.
Start by integrating with respect to y: $\begin{aligned} \int_{0}^{0.5x}(2x-3y)dy &= [2xy - \frac{3}{2}y^{2}]_{0}^{0.5x} \ &= \(2x \times 0.5x - \frac{3}{2}(0.5x)^{2}$ - 0 \ &= x^{2} - \frac{3}{8}x^{2} \ &= \frac{5}{8}x^{2}.onumber \end{aligned}\)
Next, integrate with respect to x: $\begin{aligned} \int_{0}^{2}\frac{5}{8}x^{2}dx &= \frac{5}{8} \int_{0}^{2}x^{2}dx \end{aligned}$.
The integral of $x^{2}$ is $\frac{x^{3}}{3}$, so we evaluate: $\begin{aligned} \frac{5}{8}[\frac{2^{3}}{3}-\frac{0^{3}}{3}] &= \frac{5}{8}\times \frac{8}{3} = \frac{5}{3}.onumber \end{aligned}$

So, $\frac{5}{3}$ is the final value of the double integral over the triangular region. Practicing these steps will enhance your ability to solve similar problems with confidence.