Problem 29 - Thomas Calculus

Question

In Exercises \(21-30,\) sketch the region of integration and write an equivalent double integral with the order of integration reversed. $$ \int_{0}^{1} \int_{-\sqrt{1-y^{2}}}^{\sqrt{1-y^{2}}} 3 y d x d y $$

Step-by-Step Solution

Answer

Reversed integral: \( \int_{0}^{1} \int_{0}^{\sqrt{1-x^2}} 3y \, dy \, dx \).

Step 1: Revisiting the Region of Integration

First, let's understand the given integral. We have the integral \( \int_{0}^{1} \int_{-\sqrt{1-y^{2}}}^{\sqrt{1-y^{2}}} 3y \, dx \, dy \). In this setup, \( x \) varies from \(-\sqrt{1-y^2}\) to \(\sqrt{1-y^2}\), indicating we are dealing with a circle of radius 1. Additionally, \( y \) varies from 0 to 1, restricting us to the upper semicircle in the first quadrant.

Step 2: Understanding the Boundaries

The integral is defined over the region that can be described by the equations: \( x = -\sqrt{1-y^2} \), \( x = \sqrt{1-y^2} \), and \( 0 \leq y \leq 1 \). This corresponds to the right side of the circle \( x^2 + y^2 = 1 \), above the x-axis.

Step 3: Sketching the Region

Draw a unit circle centered at (0,0). Highlight the region in the first quadrant between \( y = 0 \) and \( y = 1 \), bounded by \( x = -\sqrt{1-y^2} \) and \( x = \sqrt{1-y^2} \). This is the top right quarter of the circle.

Step 4: Reversing the Order of Integration

We need to rewrite the integral with the order of integration reversed (i.e., integrating with respect to \( y \) first). For the given region, \( x \) ranges from 0 to \( \sqrt{1-x^2} \), and for each \( x \), \( y \) must range from 0 to 1.

Step 5: Writing the Reversed Double Integral

With the reversed order, the integral becomes \( \int_{0}^{1} \int_{0}^{\sqrt{1-x^2}} 3y \, dy \, dx \). First, \( y \) ranges from 0 to \( \sqrt{1-x^2} \) for each \( x \) between 0 and 1.

Key Concepts

Order of Integration

When dealing with double integrals, the order of integration refers to the sequence in which we integrate with respect to each variable. In our problem, the original order of integration is \(\int_{0}^{1} \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} 3y \, dx \, dy\). This means that for a given value of \(y\), we integrate over \(x\) first (from \(-\sqrt{1-y^2}\) to \(\sqrt{1-y^2}\)), and then with respect to \(y\) (from 0 to 1).
To reverse the order of integration, we find a new double integral expression where we swap the order of integration. The key here is determining the new bounds for \(y\) when \(x\) is given. After drawing the region of integration, we notice that \(x\) ranges from 0 to 1 and \(y\) from 0 to \(\sqrt{1-x^2}\). Thus, the reversed double integral becomes \(\int_{0}^{1} \int_{0}^{\sqrt{1-x^2}} 3y \, dy \, dx\).
The choice of order can affect both the difficulty of solving the integral and the result of the integration, especially when the limits of integration aren't constants. Switching the order can simplify the computation in many cases.

Region of Integration

The region of integration is the area over which the double integral is evaluated. In this exercise, the region is dictated by the boundaries of the integral: \(x = -\sqrt{1-y^2}\), \(x = \sqrt{1-y^2}\), and \(0 \leq y \leq 1\).
This describes the top half of a circle of radius 1 centered at the origin, specifically in the first quadrant. The boundaries indicate that for each value of \(y\), \(x\) varies symmetrically from \(-\sqrt{1-y^2}\) to \(\sqrt{1-y^2}\).
This semicircular region is significant because it cuts the circle in such a way that it fits perfectly when swapping the order of integration. By drawing this region, it becomes easier to visualize and understand the integration limits when writing the double integral in terms of different variables.

Coordinate Transformation

In the context of double integrals, coordinate transformation can simplify evaluation and provide a clearer geometrical interpretation of the problem. Common transformations include changing from Cartesian to polar coordinates when dealing with circular regions.
For this particular exercise, we are dealing with a semicircular region in the first quadrant of a circle. If needed, a coordinate transformation to polar coordinates could simplify the integration process. In polar coordinates, the transformation is \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\), with \(r\) representing the radius and \(\theta\) the angle.
If applied, the integral over the semicircular region can be expressed as \(\int_0^{\frac{\pi}{2}} \int_0^1 3r\sin(\theta) \, r \, dr \, d\theta\). Although the transformation is not used directly in the exercise solution, it serves as an alternative method to handle complex boundaries and can simplify the integration process in practice.