Problem 44 - Thomas Calculus

Question

Sketch the region of integration and write an equivalent double integral with the order of integration reversed. \begin{equation} \int_{0}^{\pi / 6} \int_{\sin x}^{1 / 2} x y^{2} d y d x \end{equation}

Step-by-Step Solution

Answer

The reversed integral is \( \int_{0}^{1/2} \int_{0}^{\sin^{-1}(y)} x y^{2} \; dx \; dy \).

Step 1: Analyze the original integral's limits

The given double integral \( \int_{0}^{\pi / 6} \int_{\sin x}^{1 / 2} x y^{2} \; dy \; dx \) implies that for each fixed \( x \) in the interval \( [0, \pi/6] \), \( y \) ranges from \( \sin x \) to 1/2.

Step 2: Understand the region of integration

In the xy-plane, the region is the set of points (x, y) where \( 0 \leq x \leq \pi/6 \) and \( \sin x \leq y \leq 1/2 \). This region is bounded above by the line \( y = 1/2 \) and below by the curve \( y = \sin x \).

Step 3: Sketch the region in the xy-plane

First, sketch the curve \( y = \sin x \) from \( x = 0 \) to \( x = \pi/6 \). Draw the horizontal line \( y = 1/2 \). Note where these curves intersect and where they bound the region. The region lies between \( y = \sin x \) and \( y = 1/2 \) from \( x = 0 \) to \( x = \pi/6 \).

Step 4: Find the new limits for \( x \)

For reversing the order, note that \( y \) will vary from 0 to \( 1/2 \), and for each fixed \( y \), \( x \) will range from 0 to \( \sin^{-1}(y) \).

Step 5: Write the equivalent reversed double integral

The double integral with the order reversed is \( \int_{0}^{1/2} \int_{0}^{\sin^{-1}(y)} x y^{2} \; dx \; dy \). This integral iterates over \( x \) first, then \( y \).

Key Concepts

Order of Integration

When dealing with double integrals, the order of integration refers to the sequence in which the integrations are performed. In the given exercise, the originally provided order is integrating with respect to \( y \) first and then \( x \). This is indicated by the notation \( d y \; d x \) in the integral. Therefore, for a fixed \( x \), \( y \) is integrated over its given limits.
Reversing this order means that we will first integrate with respect to \( x \) and then with respect to \( y \), indicated by \( d x \; d y \). This change requires us to reorganize the integration limits.
Understanding the order of integration helps to simplify computations in some cases, especially when the region of integration is better suited to one order over the other. It can also aid in finding the region of integration based on real-world applications, as the order of integration reflects how these variables change across that region.

Region of Integration

The region of integration is essentially the area over which you integrate a function. In the xy-plane, it is defined by the limits of integration. For the given integral \( \int_{0}^{\pi / 6} \int_{\sin x}^{1 / 2} x y^{2} \; d y \; d x \), the region of integration is the area where \( 0 \leq x \leq \pi/6 \) and \( \sin x \leq y \leq 1/2 \).
Imagine plotting this on a graph: the curve \( y = \sin x \) from \( x = 0 \) to \( x = \pi/6 \) forms the lower boundary; the horizontal line \( y = 1/2 \) forms the upper boundary. These are the constraints in the xy-plane.
To visualize this, sketch the sine curve from 0 to \( \pi/6 \), add in the horizontal line at \( y = 1/2 \), and shade the area between these lines. This shaded area represents the region of integration. Understanding this region is crucial as it sets the domain for both integrals and dictates how to evaluate them.

Limits of Integration

Limits of integration tell you the bounds within which the integral is evaluated. In double integrals, these bounds define the region over two variables, typically given as intervals in the inner and outer integrals.
For the original integral \( \int_{0}^{\pi / 6} \int_{\sin x}^{1 / 2} x y^{2} \; d y \; d x \):

• \( x \) is restricted between 0 and \( \pi/6 \)
• \( y \) is limited between \( \sin x \) and \( 1/2 \)

When reversing the order of integration, the limits are adjusted accordingly. Now, \( y \) spans from 0 to 1/2, while for each \( y \), \( x \) covers from 0 to \( \sin^{-1}(y) \). This results in the equivalent integral \( \int_{0}^{1/2} \int_{0}^{\sin^{-1}(y)} x y^{2} \; d x \; d y \).
The limits of integration are not only essential for accurately setting up the problem but also critical for ensuring that both integrals actually represent the same area or volume.