Problem 46 - Thomas Calculus

Question

Sketch the region of integration and write an equivalent double integral with the order of integration reversed. \begin{equation} \int_{0}^{\sqrt{3}} \int_{0}^{\tan ^{-1} y} \sqrt{x y} d x d y \end{equation}

Step-by-Step Solution

Answer

Equivalent double integral: \( \int_{0}^{\frac{\pi}{3}} \int_{0}^{\tan x} \sqrt{x y} \, dy \, dx \).

Step 1: Understand the given region

The original integral is \( \int_{0}^{\sqrt{3}} \int_{0}^{\tan^{-1} y} \sqrt{x y} \, dx \, dy \). Here, \( y \) ranges from 0 to \( \sqrt{3} \) and for each fixed \( y \), \( x \) ranges from 0 to \( \tan^{-1} y \). We need to sketch this region in the \( xy \)-plane.

Step 2: Sketch the region

On the \( xy \)-plane, draw the line \( y = \sqrt{3} \) which is a horizontal line segment starting from \( (0, 0) \) to \( (\sqrt{3}, \sqrt{3}) \). Now, for each \( y \) from 0 to \( \sqrt{3} \), \( x \) goes from 0 to \( \tan^{-1} y \), which indicates an area under the curve \( x = \tan^{-1} y \). The curve starts from the origin and increases as \( y \) increases.

Step 3: Determine limits after reversing the order

To reverse the order of integration, express the region in terms of \( x \) first. The original region has upper bounds dependent on \( y = \tan(x) \) and \( y \) from 0 to \( \sqrt{3} \). The reversed region requires us to find allowable bounds of \( x \) first, which are 0 to \( \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \). For each \( x \), \( y \) will range from 0 to \( \tan x \).

Step 4: Write the equivalent double integral

Using the region description above, rewrite the double integral as \( \int_{0}^{\frac{\pi}{3}} \int_{0}^{\tan x} \sqrt{x y} \, dy \, dx \). This reverses the order of integration correctly.

Key Concepts

Order of Integration

In double integrals, the order of integration determines the sequence in which integration is carried out. The order is indicated by the differential elements and their corresponding limits.

• The first integration is performed with respect to the innermost variable, followed by the outer variable.
• In the original problem, the order is first integrating with respect to \(x\) and then \(y\).

Always carefully consider the given limits as they govern the integration region in the coordinate plane, affecting your approach.
To change the order of integration, you need to assess the region of integration in the graph to determine the new limits for the reversed order.
This means you’d integrate with respect to \(y\) first and then \(x\), as reflected in the rewritten integral for the given exercise.

Region of Integration

The region of integration refers to the area over which we integrate. It’s dictated by the limits in a double integral, which form a shape in the plane.
For the provided exercise, the original region is defined by:

• \( y \) ranging from 0 to \( \sqrt{3} \)
• \( x \) ranging from 0 to \( \tan^{-1} y \)

This forms a region under the curve \( x = \tan^{-1} y \) in the \(xy\)-plane, representing the dependency between \( x \) and \( y \).
When reversing the integration order, the region’s description changes, needing reassessment of bounds and new limits.

Sketching Regions

Sketching the region of integration is a critical step in understanding a given double integral. It provides a visual representation of the area over which the integration is performed.

• Start by plotting the limits of integration on the \(xy\)-plane.
• Identify lines, curves, or points that represent bounds. In this case, draw a horizontal line at \( y = \sqrt{3} \), and curve \( x = \tan^{-1} y \).

Connect these elements to form a closed shape which represents the area of integration.
Re-examine the sketch when changing the order of integration. This allows us to reimagine limits to see how the original shape can be described differently with respect to swapping \(x\) and \(y\), as done in this exercise.