Problem 17
Question
Graph the integrands and use known area formulas to evaluate the integrals. $$\int_{-3}^{3} \sqrt{9-x^{2}} d x$$
Step-by-Step Solution
Verified Answer
The integral evaluates to \( \frac{9\pi}{2} \).
1Step 1: Recognize the Function
The integrand, \( \sqrt{9-x^2} \), is the equation of a semicircle. It represents the top half of a circle centered at the origin with radius 3. This is because the equation of a circle is given by \( x^2 + y^2 = r^2 \), so \( y = \sqrt{9-x^2} \) represents the top half.
2Step 2: Graph the Function
Sketch the function \( y = \sqrt{9-x^2} \) from \( x = -3 \) to \( x = 3 \). This function will form a semicircle with a radius of 3, above the x-axis, and symmetric about the y-axis from \( x = -3 \) to \( x = 3 \).
3Step 3: Calculate the Area of the Semicircle
Calculate the area of the semicircle using the formula for the area of a circle, \( A = \pi r^2 \), where \( r \) is the radius of the circle. In this case, the radius \( r = 3 \). Therefore, the area of the full circle would be \( \pi \times 3^2 = 9\pi \). The semicircle is half of this area, so the area is \( \frac{9\pi}{2} \).
4Step 4: Evaluate the Integral
The integral \( \int_{-3}^{3} \sqrt{9-x^2} \, dx \) represents the area under the curve of the semicircle from \( x = -3 \) to \( x = 3 \). Therefore, the value of the integral is \( \frac{9\pi}{2} \).
Key Concepts
Understanding SemicirclesCalculating the Area Under a CurveCircle Equations and Their Derivatives
Understanding Semicircles
A semicircle is essentially half of a circle. It is created by cutting a whole circle along its diameter into two equal halves. Each half is a semicircle. A common way to represent a semicircle in mathematics, particularly when graphed, is to consider a function like \( y = \sqrt{r^2 - x^2} \) for the top half of the semicircle, where \( r \) is the radius and \( x \) ranges over the interval which defines the semicircle's width.
Recognize that this equation stems from the general circle equation \( x^2 + y^2 = r^2 \), but rewritten to solve for \( y \). This specific function expresses just the positive values of \( y \), which corresponds to the top half of the circle. Thus, the function \( \sqrt{9 - x^2} \) represents a semicircle with a radius of 3 above the x-axis covering the range from \( x = -3 \) to \( x = 3 \).
The semicircle concept is fundamental to understanding processes where symmetry or half-circle areas are considered, such as certain physical phenomena and mathematical integrations.
Recognize that this equation stems from the general circle equation \( x^2 + y^2 = r^2 \), but rewritten to solve for \( y \). This specific function expresses just the positive values of \( y \), which corresponds to the top half of the circle. Thus, the function \( \sqrt{9 - x^2} \) represents a semicircle with a radius of 3 above the x-axis covering the range from \( x = -3 \) to \( x = 3 \).
The semicircle concept is fundamental to understanding processes where symmetry or half-circle areas are considered, such as certain physical phenomena and mathematical integrations.
Calculating the Area Under a Curve
Finding the area under a curve is a critical element of integral calculus. When you graph a function on a coordinate plane, the integral of that function over an interval gives you the area under the curve between two points, typically on the x-axis.
To compute the area under a curve like our semicircle, use known geometrical area formulas. For instance, the area of a full circle is \( \pi r^2 \), where \( r \) is the radius. For a semicircle, you take half of that, resulting in \( \frac{\pi r^2}{2} \).
When you evaluate definite integrals, like \( \int_{-3}^{3} \sqrt{9-x^2} \, dx \), you're calculating the total area under the semicircle from \( x = -3 \) to \( x = 3 \). This integral specifically utilizes the semicircle's symmetry to compute its area efficiently.
To compute the area under a curve like our semicircle, use known geometrical area formulas. For instance, the area of a full circle is \( \pi r^2 \), where \( r \) is the radius. For a semicircle, you take half of that, resulting in \( \frac{\pi r^2}{2} \).
When you evaluate definite integrals, like \( \int_{-3}^{3} \sqrt{9-x^2} \, dx \), you're calculating the total area under the semicircle from \( x = -3 \) to \( x = 3 \). This integral specifically utilizes the semicircle's symmetry to compute its area efficiently.
Circle Equations and Their Derivatives
Circle equations follow the general form \( x^2 + y^2 = r^2 \), defining all the points on the circumference of a circle with radius \( r \). When graphing or solving parts of a circle, like a semicircle, it's useful to rearrange this formula to solve for \( y \).
In the context of semicircles, we derive the function \( y = \sqrt{r^2 - x^2} \) to represent the upper half of the circle, especially when \( y \) must remain non-negative within a certain range of \( x \). This format allows for intuitive integration practices and area calculations.
Understanding how to manipulate and apply circle equations is crucial. It aids not only in graphing but also in transitioning to integrals, which often require specific forms for simplification. Recognizing how these transformations work is key to solving real-world and purely mathematical problems alike.
In the context of semicircles, we derive the function \( y = \sqrt{r^2 - x^2} \) to represent the upper half of the circle, especially when \( y \) must remain non-negative within a certain range of \( x \). This format allows for intuitive integration practices and area calculations.
Understanding how to manipulate and apply circle equations is crucial. It aids not only in graphing but also in transitioning to integrals, which often require specific forms for simplification. Recognizing how these transformations work is key to solving real-world and purely mathematical problems alike.
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