Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 6: Problem 59
Find the area of the region described in the following exercises. The region bounded by \(x=y(y-1)\) and \(y=\frac{x}{3}\)
Short Answer
Expert verified
Answer: The area of the region is \(\frac{32}{3}\) square units.
Step by step solution
01
Find the intersection points
First, we need to find the intersection points between the given curves. To do this, we set \(x=y(y-1)\) equal to \(x=3y\). In order to find y values of the intersection points, we need to solve the equation \(y(y-1)=3y\):
$$
y(y-1)=3y \\
y^2-y=3y \\
y^2-4y=0 \\
y(y-4)=0
$$
The solutons are \(y=0\) and \(y=4\). Now that we have found the y values of intersection points, we can find the corresponding x values using either of the equations:
\(x=y(y-1)=0\)
\(x=y(y-1)=12\)
So, the intersection points are \((0,0)\) and \((12,4)\).
2.
02
Find the difference between the functions
We need to find the difference between the given functions, which will be the integrand we'll use to find the area of the region. We express \(x\) as a function of \(y\) for both equations:
$$
x_1 = y(y-1) \\
x_2 = 3y
$$
The difference between these functions will be the integrand of our integral:
$$
x_2 - x_1 = 3y - y(y-1) = 3y - y^2 + y = 4y - y^2
$$
3.
03
Integrate the difference between the functions
We now integrate the difference between the functions over the interval of intersection \(\left[0,4\right]\):
$$
A = \int_0^4 (4y - y^2) dy = \int_0^4 (4y - y^2) dy = \left[2y^2 - \frac{1}{3}y^3\right]_0^4 \\
A = \left(2\cdot 4^2 - \frac{1}{3}\cdot 4^3\right) - \left(2\cdot 0^2 - \frac{1}{3}\cdot 0^3\right) = 32 - \frac{64}{3} \\
A = \frac{32}{3}
$$
So, the area of the region bounded by the curves \(x=y(y-1)\) and \(y=\frac{x}{3}\) is \(\frac{32}{3}\) square units.
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.
Calculus
Calculus is a fascinating and comprehensive field of mathematics focused on the study of change, in the same way that geometry is about shape and algebra is about operations and their application to solving equations. It is divided into two major branches: differential calculus and integral calculus. Differential calculus concerns the concept of a derivative, which allows for the calculation of how a function changes at any given point. Integral calculus, on the other hand, deals with the integral, which provides a way to determine the total accumulation of quantities. In our context, we're using integral calculus to find the area between two curves, which is a classic application of the definite integral. This calculation requires identifying the region of interest, finding intersection points, and evaluating the integral across this region to accumulate the total area.
Integration
Integration, one of the principal operations in calculus, is used to compute areas, volumes, and many other significant values. It is essentially the process of finding the integral of a function, which represents the accumulation of a quantity over an interval. When you integrate a function over an interval, you're adding up an infinite number of infinitesimally small quantities represented by the function. This process is crucial when determining the area between curves. The integral provides a way to calculate this total area by considering the difference between the upper and lower bounds of the region at every point. The act of integration can be visualized as calculating the area under each curve and then determining the difference to find the area trapped between them.
Intersection Points
Intersection points are where two mathematical objects鈥攕uch as lines, curves, or surfaces鈥攎eet. Finding the intersection points of two curves is vital for calculating the area between them because it establishes the boundaries for integration. Once we know where the curves intersect, we can set the limits of our definite integral. These points ensure that we consider only the region of interest for which we want to find the area. In the given exercise, we found the intersection points by setting the equations of the curves equal to each other and solving for the variables. This gave us precise locations which serve as the starting and ending points for our integral.
Definite Integral
The definite integral is a central concept in calculus used to calculate the exact area under a curve bounded by an interval on the x-axis. Unlike indefinite integrals, which represent a family of functions, a definite integral gives us a specific numerical result. It represents the total accumulation of the function's values over a certain interval, which can be thought of as the sum of an infinite number of infinitesimally thin rectangles under the curve. This creates a precise method to calculate areas, especially between curves. In our example, once the intersection points were determined, the definite integral of the difference of the two functions over the interval defined by these points yielded the area of the region between the two curves.
