Problem 29 - Calculus Early Transcendentals

Question

Evaluate the following integrals. A sketch is helpful. \(\iint_{R} y^{2} d A ; R\) is bounded by \(x=1, y=2 x+2,\) and \(y=-x-1\).

Step-by-Step Solution

Answer

Question: Evaluate the double integral $$\iint_R y^2 dA$$, where R is the region bounded by the curves \(x=1\), \(y=2x+2\), and \(y=-x-1\). Answer: The value of the given double integral over the region R is $$\iint_R y^2 dA = \frac{97}{27}$$

Step 1: Set up the limits of integration

Since the region is bounded by \(x=1\), \(y=2x+2\), and \(y=-x-1\), we need to determine how the region depends on these boundaries. The region is formed between the lines \(y=2x+2\) and \(y=-x-1\) considering \(x\) varies between \(x=-\frac{1}{3}\) and \(x=1\). Also, from the intersection points, \(y\) varies between \(\frac{4}{3}\) and 4 vertically. Thus, we have: $$\iint_R y^2 dA = \int_{-\frac{1}{3}}^{1}\int_{-x-1}^{2x+2} y^2 dy dx$$

Step 2: Integrate with respect to y

First, we need to find the integral with respect to y: $$\int_{-x-1}^{2x+2} y^2 dy = \left[\frac{1}{3}y^3\right]_{-x-1}^{2x+2} = \frac{1}{3}\left((2x+2)^3 - (-x-1)^3\right)$$

Step 3: Integrate with respect to x

Now, we integrate the result from step 2 with respect to x: $$\int_{-\frac{1}{3}}^{1} \frac{1}{3}\left((2x+2)^3 - (-x-1)^3\right) dx = \frac{1}{3}\left[\frac{1}{4}(2x+2)^4 - \frac{1}{4}(-x-1)^4\right]_{-\frac{1}{3}}^{1}$$

Step 4: Evaluate the integral

We can now evaluate the integral at the limits of integration: $$\frac{1}{3}\left[\frac{1}{4}(2(1)+2)^4 - \frac{1}{4}(-(1)-1)^4\right] - \frac{1}{3}\left[\frac{1}{4}(2(-\frac{1}{3})+2)^4 - \frac{1}{4}(-(-\frac{1}{3})-1)^4\right] = \frac{97}{27}$$ In conclusion, the value of the given double integral over the region R is: $$\iint_R y^2 dA = \frac{97}{27}$$

Key Concepts

Region of Integration

Understanding the region of integration is a vital step when solving double integrals. The region of integration is the area over which we evaluate the integral. In this exercise, we have a region, denoted as \( R \), bounded by the equations \( x = 1 \), \( y = 2x + 2 \), and \( y = -x - 1 \). These boundaries form a trapezoid-like area on the xy-plane.
The position and shape of this region are determined by the intersection points of these lines. By analyzing when \( y = 2x + 2 \) intersects with \( y = -x - 1 \), we establish points that outline the region.

• The line \( y = 2x + 2 \) represents a slanting upward line originating from the y-axis.
• The line \( y = -x - 1 \) is a downward-sloping line.
• Within the bounds of \( x = 1 \) and intersections with these lines, you have your region of integration.

This bounded region \( R \) is where the function \( y^2 \) will be integrated. Visualizing this region helps to make setting up our limits easier.

Limits of Integration

Setting the correct limits of integration is crucial as it determines the parts of the function being integrated. Double integrals generally require two sets of limits:

• One for the outer integral (x-axis).
• Another for the inner integral (y-axis).

The specific limits are derived from the bounds of the region of integration. Here, the limits for \( x \) are found from analyzing the intersections between \( y = 2x + 2 \) and \( y = -x - 1 \). For this problem, \( x \) varies from \( x = -\frac{1}{3} \) to \( x = 1 \).
Next, moving accordingly within these limits, \( y \) changes from \( y = -x - 1 \) to \( y = 2x + 2 \). Thus, the complete integral is structured as:
\[\int_{-\frac{1}{3}}^{1}\int_{-x-1}^{2x+2} y^2 \, dy \, dx\]
These limits ensure the function is evaluated over the entire specified region.

Integration Techniques

Various integration techniques simplify the evaluation of double integrals. This problem involves calculating a double integral by first integrating with respect to \( y \) and then \( x \).

• The first step involves the inner integral with \( y \), giving the antiderivative \( \frac{1}{3}y^3 \), calculated over its limits.
• The subtraction of these results, from the upper and lower limits, produces a new function expressed in terms of \( x \).

Once this is achieved, the new expression becomes the integrand for the outer integral. Here, integration is carried out with respect to \( x \).
After finding the antiderivative for \( x \), the next step is evaluating it at the given limits and calculating the final result. This two-step process simplifies complex areas using easier single-variable calculus. For this solution, these steps yield the value \( \frac{97}{27} \), presenting the integral's value over the defined region. This approach is systematic and highlights the importance of tackling double integrals incrementally.