91Ó°ÊÓ

The concept of finding the area of a geometric shape is fundamental, especially when dealing with integrals of certain functions. In our example, the integrand \( \sqrt{9-x^{2}} \) represents a portion of a circle. Specifically, this is the equation for the upper half, or semicircle, of a circle with radius 3.

To understand why this is the case, you start with the equation of a full circle, \( x^2 + y^2 = 9 \). Solving for \( y \) gives you \( y = \sqrt{9-x^2} \), which describes a semicircle centered at the origin, extending from \( x = -3 \) to \( x = 3 \).

The full circle has an area calculated by the formula \( \pi r^2 \). For a semicircle, you simply take half of this area. So, for our semicircle with a radius of 3:

• The area of a full circle would be \( \pi (3)^2 = 9\pi \).
• Thus, the area of the semicircle is \( \frac{1}{2} \pi (3)^2 = \frac{9\pi}{2} \).

This area represents the definite integral of the function from \(-3\) to \(3\).

Integrals don't only exist in isolation on paper; they have real-world meanings and applications, one of which is representing areas under curves. In this exercise, the integral of \( \sqrt{9 - x^2} \) from \(-3\) to \(3\) represents the area of a semicircle.

When interpreting integrals geometrically, the integral provides the total area between the curve of the function and the x-axis over a specified interval. Here, the function corresponds to a semicircle where the integral tells us about the area coverage by this upper semicircle, specifically between \( x = -3 \) and \( x = 3 \).

This geometric interpretation strengthens your understanding by visualizing the problem:

• The integrand \( \sqrt{9 - x^2} \) is the curve, described by half a circle.
• The interval from \(-3\) to \(3\) effectively spans the diameter of the semicircle.
• Thus, the integral gives us \( \frac{9\pi}{2} \), the area of the semicircle.

Graphing functions is a crucial skill in understanding integrals. To solve such problems, it is often helpful to first sketch the function. Here, the function \( y = \sqrt{9 - x^2} \) corresponds to the top half of a circle with radius 3.

When graphing \( \sqrt{9 - x^2} \), pay close attention to the nature of the function:

• It is defined for \( x \) between \(-3\) and \(3\), forming the semicircle along the x-axis.
• The function peaks at \( y = 3 \) when \( x = 0 \), which is the top of the semicircle.
• At \( x = \pm 3 \), the function reaches the x-axis, completing the half-circle.

By graphing, you see the geometrical object that corresponds to the definite integral, helping you visualize how the bounded area looks, with limits referring directly to the diametrical points. Converting functions into visual graphs aids in better appreciating mathematical concepts and solutions.