Problem 55 - Thomas Calculus

Question

Sketch the region of integration, reverse the order of integration, and evaluate the integral. \begin{equation}\begin{array}{l}{\text { Square region } \iint_{R}\left(y-2 x^{2}\right) d A \text { where } R \text { is the region bounded }} \\ {\text { by the square }|x|+|y|=1}\end{array}\end{equation}

Step-by-Step Solution

Answer

The integral evaluates to 0 over the given square region.

Step 1: Understand the Region of Integration

The region is defined by the equation \(|x|+|y|=1\). This represents a square centered at the origin, with vertices at (1,0), (0,1), (-1,0), and (0,-1). This square can be broken down into four triangular regions based on which line segment you are on.

Step 2: Identify Limits for Initial Order of Integration

Initially, let's use vertical slices. For a given \(x\), the vertical limits for \(y\) depend on whether \(x\) is positive or negative due to the absolute values. For \(0 \leq x \leq 1\), the limits are \(-1+x\) to \(1-x\), and for \(-1 \leq x \leq 0\), the limits are \(-1-x\) to \(1+x\).

Step 3: Set Up the Integral with Original Order

We setup the double integral initially as two separate integrals covering each half of the square: \[\int_{0}^{1} \int_{-1+x}^{1-x} (y - 2x^2) \, dy \, dx + \int_{-1}^{0} \int_{-1-x}^{1+x} (y - 2x^2) \, dy \, dx\]

Step 4: Reverse the Order of Integration

To reverse the order of integration, we look at horizontal slices. For a given \(y\), the range for \(x\) is different. For \(0 \leq y \leq 1\), \(-1+y\) to \(1-y\) covers the range of \(x\), and similarly for \(-1 \leq y \leq 0\), we need \(-1-y\) to \(1+y\).

Step 5: Set Up the Double Integral with Reversed Order

Now, writing the integral with the reversed order from Step 4:\[\int_{0}^{1} \int_{-1+y}^{1-y} (y - 2x^2) \, dx \, dy + \int_{-1}^{0} \int_{-1-y}^{1+y} (y - 2x^2) \, dx \, dy\]

Step 6: Evaluate the Inner Integrals

In each case of the integral, focus on calculating the inner integral over \(x\). For example, consider: \[ \int_{-1+y}^{1-y} (y - 2x^2) \, dx \] Perform this integration for each region separately.

Step 7: Evaluate the Outer Integrals

After calculating the inner integrals in Step 6, evaluate the outer integrals for \(y\) in each case and combine results. Complete these calculations to arrive at the final numerical value of the integral.

Key Concepts

Double Integration

Double integration is a crucial topic in multivariable calculus, allowing us to find the volume under a surface over a region in the plane. Essentially, it involves adding up values of a function across a two-dimensional area.

In the given exercise, we are dealing with the integral of the function \( (y - 2x^2) \) over a region \( R \). This operation involves performing two integrations: one across the width of the region and the other across its height.

• The first integration is usually done with respect to one variable, keeping the other constant.
• Next, the resulting expression is integrated with respect to the second variable.

In more complex regions, the order of integration may need to be changed to make the calculations simpler, as we will explore in the next sections.

Changing the Order of Integration

Sometimes, changing the order of integration makes a problem much easier to solve. Originally, our integral involved integrating with respect to \( y \) first and then \( x \). However, we switched the order due to the symmetrical nature of our square region defined by \(|x| + |y| = 1\).

To reverse the order of integration:

• First, visualize the region into which we are integrating.
• Determine the new limits of integration for each variable by examining horizontal and vertical slices of the region.
• Adjust the integral expression to reflect the new order.

This method is effective here because it allows the integration to be performed more straightforwardly by adjusting how each coordinate is bounded within the region.

By changing the order, some integrals become solvable when they initially seemed too complex.

Absolute Value Functions

Absolute value functions, like \(|x|+|y| = 1\) in this problem, can define quite intriguing regions. They handle situations where both positive and negative values need to result in non-negative outcomes due to their nature of reflecting all negative inputs to positive values.

The presence of absolute values in limits often means you'll want to split your integral into pieces, one for each section or its boundaries:

• This allows for correct evaluation of the area covered by different linear behavior in each section.
• This typically creates multiple sub-integrals to solve, each with potentially simpler bounds.

Absolute values are a powerful tool to creatively define regions like the diamond-shaped square here. Understanding how they influence the definition of regions and boundaries is key in solving related integrals.

Region of Integration

Understanding the region of integration is crucial, especially when the integrals involve geometric shapes like our square region defined by \(|x|+|y|=1\). This describes a square centered at the origin, spanning from -1 to 1 on both the \(x\)- and \(y\)-axes.

Regions are determined by the constraints imposed by the absolute values, similar to how we identify slices or triangles:

• Visualizing these boundaries is the first step to recognizing the area over which integration occurs.
• Recognizing that this region can be divided into manageable shapes simplifies integration tasks.

By comprehending the topology of the region, we can set accurate integration limits, whether integrating vertically or horizontally, ensuring that all calculations precisely correspond to the set region.