Problem 24 - Thomas Calculus

Question

In Exercises \(21-30,\) sketch the region of integration and write an equivalent double integral with the order of integration reversed. $$ \int_{0}^{1} \int_{1-x}^{1-x^{2}} d y d x $$

Step-by-Step Solution

Answer

Reverse the order to: \(\int_{0}^{1} \int_{1-y}^{\sqrt{1-y}} d x d y\).

Step 1: Understanding the Region

First, recognize the limits of the inner integral, which are from \(y = 1-x\) to \(y = 1-x^2\), and the outer integral goes from \(x=0\) to \(x=1\). Thus, the region of integration is bounded by these curves which need to be sketched.

Step 2: Sketch the Curves

Consider the curves \(y = 1 - x\) and \(y = 1 - x^2\). Plot these on the coordinate plane for \(x\) ranging from \(0\) to \(1\). The curve \(y = 1-x\) is a straight line, while \(y = 1-x^2\) is a parabola opening downwards. These intersect between \(x = 0\) and \(x = 1\).

Step 3: Determine Intersection Points

Set \(1-x = 1-x^2\) to find intersection points: \[1-x = 1-x^2\] \[x^2 - x = 0\]\[x(x-1) = 0\]This gives \(x = 0\) and \(x = 1\). At these points, the value of \(y\) is the same thus confirming \(y = 1\) at both intersections.

Step 4: Find Region Bounded in y-direction

To reverse the order of integration, determined by the lower boundary \(y = 0\) (horizontal line) and the upper boundaries which are defined by \(y = 1\). Therefore as \(x\) moves from these intersections along \(y\), the region is vertically bounded by the curve \(y = 1-x^2\) above and \(y = 1-x\) below.

Step 5: Reverse the Order of Integration

Set up a new integral that uses \(y\) as the outer integral ranging from \(y = 0\) to \(y = 1\). For a given \(y\), \(x\) spans from the right boundary of \(x = 1-y\) to the left boundary of \(x = \sqrt{1-y}\). Therefore, rearrange the integral as follows:\[\int_{0}^{1} \int_{1-y}^{\sqrt{1-y}} d x d y\]

Key Concepts

Region of Integration

In double integration, the region of integration is the area over which the integration is performed. This region must be defined clearly to set up the integral correctly. In this exercise, our region of integration is bounded by two curves:

• The line: \(y = 1-x\)
• The parabola: \(y = 1-x^2\)

The values of \(x\) range from 0 to 1, defining the horizontal limits, while the vertical limits are governed by the bounds of the inner integral, which are the two curves mentioned above. To understand where exactly this region lies, it's helpful to plot these curves on a coordinate plane. The area found between these curves as \(x\) moves from 0 to 1 is the region of integration. Sketching is an essential step as it visually confirms the boundaries and helps in reversing the integration order.

Order of Integration

The order of integration determines which variable is integrated first. In a double integral, either the variable \(x\) or \(y\) can be integrated first. In the original set up for this exercise, the inner integral is computed along \(y\) first, while the outer integral is over \(x\) values ranging from 0 to 1.

• Inner integral: \( \int_{1-x}^{1-x^2} dy \)
• Outer integral: \( \int_{0}^{1} dx \)

This order is chosen based on how the expressions are set and depends on the specific boundaries the problem defines. Different situations or problems might require reversing this order for simplification or necessity, which involves re-evaluating the limits of integration.

Reversing Integration

Reversing the order of integration in a double integral involves flipping which variable is initially integrated. This often simplifies the integral or accommodates new limits of integration. In this exercise, the original order of integration is inner \(y\) first, then \(x\). To reverse, we integrate with respect to \(x\) first.

To find the new limits, observe the plotted region. Now, for any selected \(y\), \(x\) spans from the curve \(x = 1-y\) to \(x = \sqrt{1-y}\). The outer integral, therefore, changes to a range along \(y\), from \(y = 0\) to \(y = 1\).

• Original integral: \( \int_{0}^{1} \int_{1-x}^{1-x^2} dy \, dx \)
• Reversed integral: \( \int_{0}^{1} \int_{1-y}^{\sqrt{1-y}} dx \, dy \)

By reversing, it reveals another perspective of the same region, often helping with more straightforward integration computation.

Boundary Curves

Boundary curves are the mathematical expressions that define the edges of the region of integration. In double integrals, these curves play pivotal roles in determining integration limits for each variable. For this problem:

• \(y = 1-x\) is a straight line that serves as a lower bound in the original setup.
• \(y = 1-x^2\) is a downward-opening parabola, narrowing the integration range in the initial setup.

Both curves intersect precisely at points \(x = 0\) and \(x = 1\), which helps confirm that the value of \(y\) is consistently 1 at these points. Understanding these boundary curves helps adjust your integration, especially when reversing, so that you establish the new limits properly. Recognizing and plotting these curves first articulates the sketch of the integration region and ensures both the accuracy and success of the integral resolution.