Problem 10 - College Algebra and Calculus: An Applied Approach

Question

Set up the integral for both orders of integration and use the more convenient order to evaluate the integral over the region \(R\). $$ \begin{aligned} &\int_{R} \int x d A\\\ &R: \text { semicircle bounded by } y=\sqrt{25-x^{2}} \text { and } y=0 \end{aligned} $$

Step-by-Step Solution

Answer

The value of the integral over the region \(R\) is 0.

Step 1: Understand the region of integration

The region \(R\) is a semicircle bounded by \(y=\sqrt{25-x^{2}}\) (the upper semicircle with radius 5 centered at the origin) and \(y=0\) (the x-axis). This semicircle is in the xy-plane with the flat part of the semicircle on the x-axis. We'll define this region in two different ways to set up the double integral in both orders of integration.

Step 2: Set up the integral in dy dx order

The limits of integration for y go from the lower boundary \(y=0\) to the upper boundary \(y=\sqrt{25-x^{2}}\). The limits of integration for x go from the leftmost point of the semicircle \(-5\) to the rightmost point \(5\). So the double integral over \(R\) is given by \[\int_{-5}^{5}\int_{0}^{\sqrt{25-x^{2}}} x dy dx\].

Step 3: Set up the integral in dx dy order

The limits of integration for x go from the left boundary \(-\sqrt{25-y^{2}}\) to the right boundary \(\sqrt{25-y^{2}}\). The limits of integration for y go from \(0\) to \(5\). So the double integral over \(R\) in dx dy order is \[\int_{0}^{5}\int_{-\sqrt{25-y^{2}}}^{\sqrt{25-y^{2}}} x dx dy\].

Step 4: Choose the more convenient order

The dy dx order seems to be the more convenient way to evaluate the integral because we do not need to integrate the square root function, which is a more complex function.

Step 5: Evaluate the Integral

First, integrate \(x\) with respect to \(y\)\[\int_{0}^{\sqrt{25-x^2}} x dy = xy \Big |_0^{\sqrt{25-x^2}} = x\sqrt{25-x^2}\]Then, integrate this result from \(-5\) to \(5\) with respect to \(x\) yields \[\int_{-5}^{5}x\sqrt{25-x^2} dx = 0\]This result is due to the fact that \(x\sqrt{25-x^2}\) is an odd function and the interval \([-5, 5]\) is symmetric about the origin.

Key Concepts

Understanding the Region of Integration

To understand a double integral, it's crucial to first visualize the region you're integrating over. In this exercise, the region of integration \(R\) is related to a geometric shape—specifically, a semicircle in the xy-plane. The curve \(y=\sqrt{25-x^{2}}\) forms the top part of the semicircle, while the line \(y=0\) simply represents the x-axis. This semicircle has a radius of 5, centered at the origin \(0, 0\), and stretches horizontally from -5 to 5 along the x-axis.

Understanding the region of integration also involves recognizing the bounds of this region. For example, the boundaries \(y=0\) and \(y=\sqrt{25-x^{2}}\) create a flat-bottom semicircle. This step is vital because it sets the stage for how you set up the double integral's limits.

Exploring Orders of Integration

When dealing with double integrals, you have two choices for the order of integration: \(dy \, dx\) and \(dx \, dy\). Choosing the right order can simplify the problem and make the integral easier to evaluate. In this exercise, these orders of integration refer to which variable you integrate first.

• The \(dy \, dx\) order means you first integrate with respect to \(y\) while keeping \(x\) fixed, then integrate the resulting expression with respect to \(x\).
• The \(dx \, dy\) order works in the opposite way: first with respect to \(x\), then with respect to \(y\).

In this scenario, \(dy \, dx\) order is more convenient because it leads us to solving a simpler integral first. By recognizing and choosing this order, you bypass the need to handle more complex functions early on.

The Process of Evaluating Integrals

Evaluating double integrals involves a sequential process that takes each function of the integral in turn. Start by solving the inner integral, then move on to the outer one. In this example, once the order \(dy \, dx\) is selected, you first integrate \(x\) with respect to \(y\) from \(y=0\) to \(y=\sqrt{25-x^{2}}\).

This results in the expression \(x\sqrt{25-x^2}\). The next step is to tackle the outer integral by integrating this expression with respect to \(x\) from \(x=-5\) to \(x=5\). Fortunately, \(x\sqrt{25-x^2}\) is an odd function over a symmetric interval, causing the integral to evaluate to zero.

Recognizing such properties, like symmetry or odd/even functions, can significantly simplify your integral evaluation.

The Role of the Semicircle in Integration

A semicircle often appears in integrals related to problems in geometry and calculus because of its distinct shape characterized by its flat side along one axis. In this exercise, only the upper semicircle is considered. The special equation \(y=\sqrt{25-x^2}\) represents the boundary of this semicircle, derived from the circle equation \(x^2 + y^2 = 25\). Here, the semicircle limits \(y\) values between 0 and \(\sqrt{25\−x^2}\), and the \(x\) values between -5 and 5.

Understanding the semicircle's role offers insight into setting up your integral limits, determining symmetry, and simplifying calculations. Semicircles are often involved in elegant simplifications due to their symmetrical properties.