Problem 35 - Thomas Calculus

Question

Sketch the region of integration and write an equivalent double integral with the order of integration reversed. \begin{equation} \int_{0}^{1} \int_{y}^{\sqrt{y}} d x d y \end{equation}

Step-by-Step Solution

Answer

Reversed integral: \( \int_{0}^{1} \int_{x^2}^{x} d y d x \).

Step 1: Understand the Original Limits of Integration

The given double integral is \( \int_{0}^{1} \int_{y}^{\sqrt{y}} d x d y \). This means the outer integral \( dy \) runs from \( y=0 \) to \( y=1 \), while the inner integral \( dx \) runs from \( x=y \) to \( x=\sqrt{y} \). This describes the region bounded by \( y \leq x \leq \sqrt{y} \) for \( 0 \leq y \leq 1 \).

Step 2: Sketch the Region of Integration

To sketch the region of integration, plot the curves \( x = y \) and \( x = \sqrt{y} \). The line \( x = y \) is a straight line through the origin with slope 1, while \( x = \sqrt{y} \) is a curve that starts from the origin and increases more gradually. The region of integration is between these two curves for \( y \) ranging from 0 to 1.

Step 3: Interpret the Region

For a better understanding, recognize that the region is the area where \( y \leq x \leq \sqrt{y} \). This is a wedge-like area starting from the point (0,0) bounded by two curves and the line \( y=1 \).

Step 4: Determine New Limits for y

To reverse the order of integration, first find the vertical span of the region. The smallest possible \( x \) is 0 (where \( x = y \, \text{and} \, x = \sqrt{y} \) intersect), and the largest is 1 (where \( y = 1 \)). For a given \( x \), \( y \) goes from \( y = x^2 \) to \( y = x \) since solving \( x = \sqrt{y} \) gives \( y = x^2 \) and using the original \( x = y \) gives \( y = x \).

Step 5: Setup the Reversed Integral

With the new limits, express the integral as \( \int_{0}^{1} \int_{x^2}^{x} d y d x \). Here, \( x \) varies from 0 to 1, and \( y \) varies from \( y = x^2 \) to \( y = x \).

Key Concepts

Region of Integration

In the world of double integrals, the region of integration is a vital concept that ensures we correctly evaluate the bounds or limits over which we integrate. For the given problem, we have a double integral represented as \( \int_{0}^{1} \int_{y}^{\sqrt{y}} d x d y \). This specific integral is defined over a region in the xy-plane that is formed between two curves: \( y=x \) and \( x=\sqrt{y} \).
To define this region of integration, we examine the intervals and boundaries:

• The outer integral \( dy \) runs from \( y=0 \) to \( y=1 \).
• The inner integral \( dx \) runs from \( x=y \) to \( x=\sqrt{y} \).

These describe a wedge-like area bounded within these limits. Understanding this bounded region is key to visualizing and solving the problem effectively.

Order of Integration

The order of integration in a double integral dictates the sequence in which the variables are integrated. For many problems, choosing the correct order of integration simplifies the solution. Here, the original order is \( dx \) first and then \( dy \), which means integrating in the \( x \)-direction first and then across \( y \).
Changing the order of integration is sometimes necessary, particularly when it helps in simplifying the integral or making it possible to integrate analytically. For this exercise, reversing the order of integration involves integrating with respect to \( y \) first (hence, \( dy \) becomes the inner integral) and then \( x \) (making \( dx \) the outer integral). This requires finding a new set of limits that reflect the new integration order.

Limits of Integration

In double integrals, the limits of integration define the bounds for each variable. These bounds can be numbers or expressions and dictate the area of the xy-plane we are integrating over.
The original integral's limits are as follows:

• Outer limit for \( y \): 0 to 1.
• Inner limit for \( x \): from \( x=y \) to \( x=\sqrt{y} \).

When we reverse the order, we need a fresh look at how \( y \) and \( x \) vary with each other. After sketching the region, we find:

• \( x \) ranges from 0 to 1, as given by left-most intersection and endpoint.
• For each \( x \), \( y \) ranges between \( y = x^2 \) and \( y = x \). These were obtained from evaluating where the curves \( x = y \) and \( x = \sqrt{y} \) intersect the x-axis and each other.

Thus, for the reversed integral: \( \int_{0}^{1} \int_{x^2}^{x} dy dx \).

Sketching Regions

Drawing the region of integration is an essential step in solving double integrals, especially when changing the order of integration. It visualizes the area you are considering, which can guide you in adapting limits correctly.
To sketch the region for the given problem:
1. **Identify the curves**: We have \( x=y \) and \( x=\sqrt{y} \).
2. **Plot the curves**:

• \( x = y \) is a straight line with a 45-degree angle, passing through the origin.
• \( x = \sqrt{y} \) is a curve starting from the origin that climbs more steeply as \( y \) increases.

3. **Identify intersections and bounds**:

• The curves intersect at the origin \((0,0)\).
• Both curves touch the line \( y=1 \), giving a clearer idea of where the region stops.

Understanding these intersections and how the graphs interact helps you determine new limits when reversing the order of integration. Drawing it accurately enables more precise adjustments in the limits when you change from \( dx \,dy \) to \( dy \,dx \).