91Ó°ÊÓ

When you have a double integral, one of the first tasks is to identify the region of integration. This region tells you over which area you are integrating. In our example, we deal with a triangular region where the boundaries are defined by the lines \(y=1\), \(y=1-x\), and \(y=x-1\).

To sketch this region, it's essential to understand how these lines intersect, forming corners or vertices of the region. With vertices at \((0,1), (2,1)\), and \((1,0)\), you can see that these points form a triangle when plotted on a coordinate plane. This visual representation helps you understand the scope of your integration and the limits it imposes on the variables \(x\) and \(y\). Recognizing the shape of the region, whether it's a triangle, rectangle, or another shape, is critical for setting up the integral correctly.

Finding intersection points is a crucial step in defining your region of integration. These are the points where the lines that bound your region cross each other. In our problem, you find these points by solving systems of equations.

Start with each pair of lines:

• The intersection of \(y=1\) and \(y=1-x\) gives the point \((0,1)\).
• The intersection of \(y=1\) and \(y=x-1\) gives the point \((2,1)\).
• The intersection of \(y=1-x\) and \(y=x-1\) yields the point \((1,0)\).

These points outline the corners of your region, which in this example turns out to be a triangle. Once you have these intersection points, you can easily sketch the region and visually verify its shape.

Choosing the right order of integration can simplify solving the integral. In double integrals, you have two choices for the integration order: doing \(dx\) first or \(dy\). Sometimes, one order may lead to easier computations compared to the other.

In this exercise, we select \(dy \ dx\) because of the triangular region's orientation. Why is this? Along each vertical line (constant \(x\)), \(y\) runs from one boundary line to another. The variable \(x\) then spans across the entire base of the triangle. This order matches the natural orientation of the region, making it easier to set up limits and perform calculations.

Determining the limits of integration correctly is vital to evaluate the integral accurately. In the case of the integral \(\int \int_{R} y^2 \, dA\) with the region \(R\) defined by the given lines, we need suitable limits for both \(x\) and \(y\).

The horizontal span of the region is from \(x=0\) to \(x=2\). For a fixed \(x\) in this range, \(y\) starts on the line \(y=x-1\) and ends on the line \(y=1-x\).

• The outer integral's limits are simple: \(x\) ranges from 0 to 2.
• The inner integral considers \(y\)'s variation, starting from the lower line \(y=x-1\) and ending at the upper line \(y=1-x\).

These limits ensure you account for all points within the triangular region. Setting them up correctly lets you solve the integral step-by-step without missing any part of the region.