Question

(a) Sketch the region \(R\) given by the problem. (b) Set up the iterated integrals, in both orders, that evaluate the given double integral for the described region \(R\) (c) Evaluate one of the iterated integrals to find the signed volume under the surface \(z=f(x, y)\) over the region \(R .\) \(\iint_{R}(4-3 y) d A,\) where \(R\) is bounded by \(y=0, y=x / e\) and \(y=\ln x\).

Short answer

The signed volume is the result of the evaluated integral for the given region \( R \).

Step-by-step solution

Understanding the Region R

First, identify the curves that bound the region \( R \). The bounds given are \( y = 0 \), \( y = \frac{x}{e} \), and \( y = \ln x \). The region is defined where all these inequalities are satisfied simultaneously. Thus, we need to sketch these curves on the xy-plane to find the region \( R \).

Sketching the Curves

Plot the three curves:1. \( y = 0 \): This is the x-axis.2. \( y = \frac{x}{e} \): This is a straight line passing through the origin with a slope of \( \frac{1}{e} \).3. \( y = \ln x \): This is a logarithmic curve starting from \( x = 1 \) and increasing to the right.Identify the enclosed region by these curves where all inequalities are satisfied.

Determining Intersection Points

To find intersection points, solve for where the curves intersect:1. \( y = \frac{x}{e} = \ln x \): Solve \( \frac{x}{e} = \ln x \) to find the intersection point.Using trial or numerical methods, find approximate intersection points.For the purpose of sketching, note that \( x = 1 \) is a rough intersection point for sketching.

Set Up Iterated Integrals: Order dy dx

For this setup:- Inner integral: Integrate with respect to \( y \) with limits from \( y = 0 \) to \( y = \frac{x}{e} \).- Outer integral: Integrate with respect to \( x \) over the range of possible values, from \( x_{min} \) to \( x_{max} \).\[ \int_{x_{min}}^{x_{max}} \int_{0}^{\frac{x}{e}} (4 - 3y) \, dy \, dx \]

Set Up Iterated Integrals: Order dx dy

For this setup:- Inner integral: Integrate with respect to \( x \) with limits from the solution of \( y = \ln x \) and \( x = e \cdot y \).- Outer integral: Integrate with respect to \( y \) from \( y = 0 \) to the intersection point.\[ \int_{0}^{y_{max}} \int_{e \cdot y}^{e} (4 - 3y) \, dx \, dy \]

Evaluate the Iterated Integral

We now evaluate one of the iterated integrals, let's choose the setup in Step 4.\[ \int_{x_{min}}^{x_{max}} \left[ \int_{0}^{\frac{x}{e}} (4 - 3y) \, dy \right] \, dx \]Evaluate the inner integral:\[ \int_{0}^{\frac{x}{e}} (4 - 3y) \, dy = \left[ 4y - \frac{3y^2}{2} \right]_{0}^{\frac{x}{e}} = 4\frac{x}{e} - \frac{3}{2} \left( \frac{x}{e} \right)^2 \]Replace in the outer integral and compute to find the signed volume.

Key concepts

Iterated Integrals

Understanding the concept of iterated integrals is crucial when dealing with double integrals. Essentially, an iterated integral is when you integrate a function first with respect to one variable and then with respect to the other. This approach is very useful for evaluating the double integral over a region, especially if the limits of integration are not constants but functions of the other variable.

• Order dy dx: First, you integrate with respect to y, keeping x constant, and then integrate the result with respect to x.
• Order dx dy: First, you integrate with respect to x, keeping y constant, and then integrate the result with respect to y.

Both orders will give the same result, although one may be easier to evaluate than the other, depending on the functions and limits involved.
Iterated integrals are set up by carefully analyzing the given region, which is bounded by the curves in the exercise. By correctly setting up these integrals, you accurately evaluate the signed volume over the region.

Region Sketching

The region sketching process begins with plotting the given boundaries that enclose the region. This typically involves graphing functions to see where they intersect and form a closed area in the plane. Let's walk through the steps needed to sketch the region in the exercise:

• Plot the line for \( y = 0 \), which is the x-axis along the plane.
• Plot the line for \( y = \frac{x}{e} \), a straight line through the origin with a slope of \( \frac{1}{e} \).
• Finally, plot the curve for \( y = \ln x \), which starts from \( x = 1 \) and increases gradually.

By analyzing these plots, you can determine the area enclosed between them, which is the region \( R \) for the double integral. The correct visualization helps in setting the right limits for integration.

Intersection Points

Finding the intersection points of the curves is a pivotal task when sketching region \( R \) and setting limits for integrals. These points occur where two curves meet. Identifying these points is crucial as they influence the boundaries for the integrals. In the exercise:

• Between \( y = \frac{x}{e} \) and \( y = \ln x \), you solve \( \frac{x}{e} = \ln x \). This equation may require numerical methods or graphical inspection to approximate or determine exact solutions if possible.

Such intersection points divide the region into smaller parts that might require integrating separately if the limits indeed change. Recognizing these points helps in correctly constructing the iterated integrals and ensures accuracy in your solutions.

Signed Volume

Signed volume under a surface with double integration often captures a physical notion of volume beneath a surface above the region \( R \) but respects the sign of the function being integrated. For the given exercise, calculate signed volume by evaluating the iterated integral.

• Setting up the integral: Deciding on which iterated integral to evaluate can depend on which one presents simpler computations. For example, evaluating in the order \( dy \, dx \) might simplify calculations.
• Evaluate the inner integral first, independently solving for one variable, and then plug this result into the outer integral which is then evaluated.

Understanding signed volume is important because, depending on the function \( f(x, y) \) over region \( R \), parts of the integral may be negative, representing actual subtraction of volume in a physical setting.