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 \(R .\) \(\iint_{R}(6-3 x-2 y) d A,\) where \(R\) is bounded by \(x=0, y=0\) and \(3 x+2 y=6\)

Short answer

The signed volume under the surface is 9.

Step-by-step solution

Identify the boundaries of the region

To sketch the region \( R \), note that it is bounded by \( x = 0 \), \( y = 0 \), and \( 3x + 2y = 6 \). Start by identifying the intersections of these boundaries:- Intersection of \( x = 0 \) and \( 3x + 2y = 6 \): Set \( x = 0 \), then \( 3(0) + 2y = 6 \) gives \( y = 3 \).- Intersection of \( y = 0 \) and \( 3x + 2y = 6 \): Set \( y = 0 \), then \( 3x = 6 \) gives \( x = 2 \).- The line \( 3x + 2y = 6 \) itself also intersects the axes at these points.Thus, the vertices of the region are \((0, 0)\), \((2, 0)\), and \((0, 3)\), forming a right triangle.

Sketch the region R

Plot the points \((0, 0)\), \((2, 0)\), and \((0, 3)\) on the coordinate plane. Connect these points to form the triangular region bounded by the x-axis, y-axis, and the line \( 3x + 2y = 6 \). This triangular region is the area \( R \) over which you will integrate.

Set up the iterated integral in dydx order

For \( \, dydx \, \) order, the limits for \( y \) range from \( 0 \) to \( 3 - \frac{3}{2}x \), and for \( x \) range from \( 0 \) to \( 2 \). Thus, the iterated integral is:\[\int_{0}^{2} \int_{0}^{3 - \frac{3}{2}x} (6 - 3x - 2y) \, dy \, dx\]

Set up the iterated integral in dxdy order

For \( \, dxdy \, \) order, the limits for \( x \) range from \( 0 \) to \( 2 - \frac{2}{3}y \), and for \( y \) range from \( 0 \) to \( 3 \). Thus, the iterated integral is:\[\int_{0}^{3} \int_{0}^{2 - \frac{2}{3}y} (6 - 3x - 2y) \, dx \, dy\]

Evaluate the iterated integral in dydx order

Choose to evaluate the integral in \( \, dydx \, \) order:1. Evaluate the inner integral with respect to \( y \): \[ \int_{0}^{3 - \frac{3}{2}x} (6 - 3x - 2y) \, dy = \left[ 6y - 3xy - y^2 \right]_{0}^{3 - \frac{3}{2}x} \] Substituting the limits: \[ \left( 6(3 - \frac{3}{2}x) - 3x(3 - \frac{3}{2}x) - (3 - \frac{3}{2}x)^2 \right) \]2. Simplify and integrate with respect to \( x \) from \( 0 \) to \( 2 \): \[ \int_{0}^{2} \left( 18 - 9x - 2(3 - \frac{3}{2}x)^2 \right) dx \] 3. Solving this integral gives the final answer: After simplification and integration, you find that the value is 9.

Key concepts

Iterated Integrals

Iterated integrals are a method used for evaluating double integrals over a specified region. In the context of this exercise, you start with a double integral that involves integrating a function over a two-dimensional region. Typically, this process involves two steps:

• Integrate with respect to one variable while holding the other constant.
• Afterwards, integrate with respect to the second variable.

The order in which you perform these integrations can vary, resulting in different forms of iterated integrals, such as (dy dx) or (dx dy). In this example, we set up integrals in both orders: - For the (dy dx) order: This means you first integrate with respect to y and then with respect to x. - For the (dx dy) order: First, you integrate with respect to x and then with respect to y. These different orders might seem confusing, but they essentially boil down to the boundaries of your region of integration and how it's defined. Paying attention to these limits ensures that you correctly include all parts of the region.

Region Sketch

Sketching the region over which you’re integrating is crucial for solving double integrals. In this exercise, the region is defined by boundaries including the x-axis, y-axis, and a line given by the equation. Begin by identifying the points where these boundaries intersect. - Find intersection points, such as where the line intersects the axes. - In our case, these points are (0,0), (2,0) and (0,3). These intersection points outline the vertices of a right triangle in the coordinate plane. Once these vertices are plotted, connect them to sketch the triangular region. This visualization helpfully illustrates the finite space where you’re calculating the integral, providing clear insight into the limits of integration applied in the iterated integral setup.

Signed Volume

A double integral over a two-dimensional region can be interpreted as finding the signed volume under a surface. Here, the surface is defined by the function \( z = f(x, y) \).When evaluating the iterated integral:
- You're determining how the function's value at each point in the region affects the overall volume.- Think of it as "summing up" countless infinitesimally small blocks under the surface, each contributing to the total volume.The notion of signed volume comes into play when the function \( f(x, y) \) has both positive and negative values over the region.- Positive parts of this surface will add to the volume.- Negative parts will subtract from it.In this solution, after integrating, you found a signed volume of 9, meaning that over the defined region, more volume is accumulated under the positive parts of the surface compared to any negative parts.

Boundaries Identification

Identifying boundaries is a preliminary step in evaluating double integrals. Clear boundary identification ensures that you correctly define the limits of integration in your iterated integral.Each boundary serves as one of the constraints for the region you are working with:
- Understand each equation representing a boundary. For instance, the line \( 3x + 2y = 6 \) is a critical boundary in this problem.- Check how these lines interact with the axes, giving intersection points like (0,3) and (2,0)- These intersections allow you to effectively sketch and delimit the region of interest.Boundaries set the scope within which you're evaluating your integral.Paying careful attention to these boundaries guarantees accurate integral evaluation.Misidentifying any limits can lead to incorrect outcomes, which is why meticulously sketching and calculating intersections is key.