Question

Evaluate the integral without using calculus: \(\int_{-3}^{3} \sqrt{9-x^{2}} d x\).

Short answer

The integral evaluates to \( \frac{9\pi}{2} \).

Step-by-step solution

Identify the Geometric Interpretation

The integral \( \int_{-3}^{3} \sqrt{9-x^2} \, dx \) can be interpreted geometrically. The function \( \sqrt{9-x^2} \) is the equation of a semicircle with radius 3 centered at the origin on the x-y plane. The limits \( -3 \) and \( 3 \) represent the ends of the diameter on the x-axis.

Determine the Area of the Geometric Shape

Since the function represents the upper half of a circle with radius 3, we need to determine the area of this semicircle. The area of a full circle is given by \( \pi r^2 \). For a semicircle, the area is half this value: \( \frac{1}{2} \pi r^2 \).

Calculate the Area using the Radius

Substitute the radius \( r = 3 \) into the area formula for a semicircle: \[ \frac{1}{2} \pi (3)^2 = \frac{1}{2} \pi \times 9 = \frac{9\pi}{2} \].

Conclude the Result

The calculated area of the semicircle is the value of the integral, as the integral represents the area under the curve \( \sqrt{9-x^2} \) from \( x = -3 \) to \( x = 3 \). Thus, the integral evaluates to \( \frac{9\pi}{2} \).

Key concepts

Area under a curve

When we talk about the area under a curve, we're referring to a key concept in calculus often represented by definite integrals. Consider a curve plotted on a coordinate plane. The area under this curve between two defined points on the x-axis is what we aim to find. This area gives us the integral of the function over that interval.

In the exercise we are looking at, the integral \( \int_{-3}^{3} \sqrt{9-x^2} \, dx \) measures the area under the semicircle formed by the equation \( \sqrt{9-x^2} \) between \( x = -3 \) and \( x = 3 \). Here, the definite integral doesn't just represent a number; it delivers the physical space under part of a circle.

Understanding how these areas are calculated using definite integrals helps connect algebraic expressions to geometric figures. It transforms a problem of numbers into one of shapes, providing a visual understanding of integrals.

Geometric Interpretation

The geometric interpretation of integrals allows us to visualize the integral as an area. In the problem at hand, the function \( \sqrt{9-x^2} \) is not a complex curve. Instead, it's a recognizable shape—the upper half of a circle, or semicircle. This function's geometric interpretation simplifies the problem, as it allows us to compute the integral by considering the shape's area rather than using traditional calculus methods.

Since the function forms a semicircle with a radius of 3 centered at the origin, its geometric interpretation provides a straightforward path to solving the integral. We avoid lengthy calculations by directly linking the function's shape to known area formulas. Thus, knowing about the semicircle's geometry and basic circle area concepts lets us find the solution much faster.

Being able to recognize these geometric shapes from algebraic functions is valuable. It lets us solve integrals by applying basic geometric principles rather than advanced calculus.

Semicircle Integration

Semicircle integration refers to finding the integral of a function representing a semicircle. The integral we look at here finds the area of a semicircle, which arises from the equation \( \sqrt{9-x^2} \). This equation is derived from the Pythagorean theorem and describes a circle's radius squared. For a semicircle, we only consider the x-axis's top half as seen between \( x = -3 \) and \( x = 3 \).

The process is simplified by knowing the circle's area formula \( \pi r^2 \). As the function gives us a semicircle, we take half of this area calculation. For our semicircle of radius 3, replacing into the formula \( \frac{1}{2} \pi \times 3^2 \) gives us \( \frac{9\pi}{2} \).

This value represents the area under the semicircle and thus the result of the integral. Semicircle integration combines geometry and calculus, using spatial awareness and mathematical principles to achieve the answer efficiently.