/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 58 Use a geometric formula to compu... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 58

Use a geometric formula to compute the integral. $$\int_{-3}^{0} \sqrt{9-x^{2}} d x$$

Short Answer

Expert verified
By replacing the integral with the semicircle area formula, the area, which represents the value of the integral, is \(\frac{1}{2} \pi (3)^2 = \frac{9}{2} \pi\).

Step by step solution

01

Identifying the Geometric Figure

The given integrand, \(\sqrt{9 - x^{2}}\), represents a semicircle. So, instead of thinking of it as a traditional integral, it could be approached as a geometric problem. The definite integral will be equal to the area of this semicircle.
02

Calculating the Area of a Semicircle

The formula to find the area of a semicircle is \(\frac{1}{2} \pi r^2\), where r represents the radius of the circle. In this case, the radius of the semicircle is 3 (since the integrand yields a circle with radius 3 when the square root is not applied). Substituting the radius into the formula, we can calculate the area of the semicircle.
03

Evaluating the Area

Substitute the radius value into the semicircle area formula: \(\frac{1}{2} \pi (3)^2\). Simplify this expression to obtain the area.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Semicircle
The concept of a semicircle is crucial when discussing geometric integration. A semicircle is exactly half of a full circle, meaning its area is half of the area of a complete circle. In problems involving functions like \( \sqrt{9 - x^2} \), the shape of the graph is a semicircle. This is because the equation \( x^2 + y^2 = r^2 \) describes a circle, and when solved for \( y \) as \( y = \sqrt{r^2 - x^2} \), it represents the upper half鈥攆orming a semicircle. Recognizing the connection between the integrand and a geometric shape simplifies the approach to solving integrals by viewing them as area calculations.
Area Calculation
Calculating the area of geometric shapes, such as a semicircle, can often replace more complex integral evaluation. The area of a semicircle can be determined using the formula \( \frac{1}{2} \pi r^2 \). This formula arises because the area of a full circle is \( \pi r^2 \). Since a semicircle is half of a circle, you simply take half of that area. By substituting the radius into this formula, you directly compute the area without needing further calculations. This approach is particularly useful in geometric integration when the function represents part of a simple geometric figure.
Definite Integrals
Definite integrals can often calculate the area under a curve between two specific points. In our case, integrating \( \sqrt{9 - x^2} \) from \(-3\) to \(0\) calculates the area under a semicircular arc on a Cartesian plane. While typically the indefinite integral of a function would provide a general solution, definite integrals include limits to provide a specific area. By understanding the geometric shape, recognizing the semicircle allows you to apply geometric formulas directly instead, simplifying the process significantly and avoiding unnecessary complexity often associated with integral calculus.
Radius of a Circle
The radius of a circle is the distance from its center to any point on its perimeter. Understanding the role of the radius helps in solving problems involving circles and semicircles. Specifically, for the function \( \sqrt{9 - x^2} \), recognizing that the equation originates from \( x^2 + y^2 = 9 \) implies a circle with a radius of \(3\). This understanding is crucial because the radius directly substitutes into the area formula \( \frac{1}{2} \pi r^2 \) for semicircle calculations. Comprehending how the radius fits into these equations is essential for solving geometric integrals effectively.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The number of items that consumers are willing to buy depends on the price of the item. Let \(p=D(q)\) represent the price (in dollars) at which \(q\) items can be sold. The integral \(\int_{0}^{Q} D(q) d q\) is interpreted as the total number of dollars that consumers would be willing to spend on \(Q\) items. If the price is fixed at \(P=D(Q)\) dollars, then the actual amount of money spent is \(P Q .\) The consumer surplus is defined by \(C S=\int_{0}^{Q} D(q) d q-P Q .\) Compute the consumer surplus for \(D(q)=150-2 q-3 q^{2}\) at \(Q=4\) and at \(Q=6 .\) What does the difference in \(C S\) values tell you about how many items to produce?

Find the average value of the function on the given interval. \(f(x)=x^{2}-1,[1,3]\)

Starting with \(\quad e^{x}=\lim _{n \rightarrow \infty}\left(1+\frac{x}{n}\right)^{n}, \quad\) show \(\quad\) that \(\ln x=\lim _{n \rightarrow \infty}\left[n\left(x^{1 / n}-1\right)\right] .\) Assume that if \(\lim _{n \rightarrow \infty} x_{n}=x,\) then \(\lim _{n \rightarrow \infty}\left[n\left(x_{n}^{1 / n}-1\right)\right]=\lim _{n \rightarrow \infty}\left[n\left(x^{1 / n}-1\right)\right].\)

Evaluate the definite integral. $$\int_{1}^{4} \frac{x-1}{\sqrt{x}} d x$$

Generalize exercise 51 to \(I=\int_{0}^{d} \frac{f(x)}{f(x)+f(a-x)} d x\) for any positive, continuous function \(f\) and then quickly evaluate \(\int_{0}^{\pi / 2} \frac{\sin x}{\sin x+\cos x} d x\)
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Applied Mathematics

Read Explanation

Logic and Functions

Read Explanation

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.