Problem 213
Question
The area of the plane region bounded by the curves \(x+2 y^{2}=0\) and \(x+3 y^{2}=1\) is equal to \(\quad\) (A) \(\frac{5}{3}\) (B) \(\frac{1}{3}\) (C) \(\frac{2}{3}\) (D) \(\frac{4}{3}\)
Step-by-Step Solution
Verified Answer
The area is \( \frac{4}{3} \).
1Step 1: Identify the Curves Equations
First, isolate the variable \( x \) in each curve equation:\[ x = -2y^2 \] and \[ x = 1 - 3y^2 \]
2Step 2: Determine Intersection Points
Set the equations from Step 1 equal to find the \( y \)-values of intersection:\[ -2y^2 = 1 - 3y^2 \]\[ y^2 = 1 \]So the intersection points for \( y \) are \( y = 1 \) and \( y = -1 \).
3Step 3: Express Area as an Integral
The area between two curves is given by the integral of the difference between the top function and the bottom function. Thus, the limits of integration are from \( y = -1 \) to \( y = 1 \) with\[ \int_{-1}^{1} ((1 - 3y^2) - (-2y^2)) \, dy = \int_{-1}^{1} (1 - y^2) \, dy \].
4Step 4: Evaluate the Integral
Compute the integral from Step 3:\[ \int_{-1}^{1} (1 - y^2) \, dy = \left[ y - \frac{y^3}{3} \right]_{-1}^{1} \].Calculating this, we have:At \( y = 1 \):\[ (1 - \frac{1}{3}) = \frac{2}{3} \].At \( y = -1 \):\[ (-1 + \frac{1}{3}) = -\frac{2}{3} \].Taking the difference, \[ \frac{2}{3} - (-\frac{2}{3}) = \frac{4}{3} \].
Key Concepts
Definite IntegralsIntersection PointsIntegral Calculus
Definite Integrals
Definite integrals are a powerful tool in calculus, primarily used to calculate areas under curves or between curves. When working with multiple functions, we can find the area between them by integrating the difference of the two functions. These integrals have specific limits, known as the bounds, which dictate the region of interest.
In this particular problem, we computed the area between two curves, and since they are given in terms of different equations of function of y, our integrals go from -1 to 1. When you calculate the integral, you actually accumulate the total area between the curves from the lower limit to the upper limit on the specified interval.
Here, setting up the integral as \[ \int_{-1}^{1} (1 - y^2) \, dy \] helps achieve the desired outcome. The results lead to the exact area across the bounded region by interpreting the geometric shapes formed by the curves. The fact that the answer \( \frac{4}{3} \) cannot be negative indicates a successful computation of the area between these curves.
In this particular problem, we computed the area between two curves, and since they are given in terms of different equations of function of y, our integrals go from -1 to 1. When you calculate the integral, you actually accumulate the total area between the curves from the lower limit to the upper limit on the specified interval.
Here, setting up the integral as \[ \int_{-1}^{1} (1 - y^2) \, dy \] helps achieve the desired outcome. The results lead to the exact area across the bounded region by interpreting the geometric shapes formed by the curves. The fact that the answer \( \frac{4}{3} \) cannot be negative indicates a successful computation of the area between these curves.
Intersection Points
Finding intersection points is vital to determining the exact interval over which you will integrate to find the area between curves. To find these points, you set the equations of the two curves equal to each other and solve for the variable.
In this exercise, let's look at the steps: Given the modified equations \( x = -2y^2 \) and \( x = 1 - 3y^2 \), setting them equal (\(-2y^2 = 1 - 3y^2\)) allows for finding that \(y^2 = 1\), producing \(y = 1\) and \(y = -1\). These values are where the two curves intersect, and thus determine the bounds of integration.
This process is significant since it establishes the limits of integration, enabling us to capture the entirety of the area of interest. Without these precise outputs, the actual region captured by the integral would be unknown or inaccurate.
In this exercise, let's look at the steps: Given the modified equations \( x = -2y^2 \) and \( x = 1 - 3y^2 \), setting them equal (\(-2y^2 = 1 - 3y^2\)) allows for finding that \(y^2 = 1\), producing \(y = 1\) and \(y = -1\). These values are where the two curves intersect, and thus determine the bounds of integration.
This process is significant since it establishes the limits of integration, enabling us to capture the entirety of the area of interest. Without these precise outputs, the actual region captured by the integral would be unknown or inaccurate.
Integral Calculus
Integral calculus is all about accumulation and measuring quantities. In this context, it deals with finding areas, volumes, and other quantities when the variable changes continuously. When calculating the area between curves, integral calculus comes into action, allowing you to condensely package and solve complex measures.
Here’s how it worked in the problem case: The area was determined by integrating the function that describes the distance between the curves at each point, which is \(1 - y^2\) in our instance. After simplifying and setting up your definite integral based on the intersection points, the next step entails doing the integration itself.
It is wrapped up by performing the fundamental theorem of calculus which involves evaluating the integral's antiderivative at the upper limit minus that at the lower limit. This approach is insightful for students learning how to manipulate differentials and apply them in real-world scenarios.
Here’s how it worked in the problem case: The area was determined by integrating the function that describes the distance between the curves at each point, which is \(1 - y^2\) in our instance. After simplifying and setting up your definite integral based on the intersection points, the next step entails doing the integration itself.
It is wrapped up by performing the fundamental theorem of calculus which involves evaluating the integral's antiderivative at the upper limit minus that at the lower limit. This approach is insightful for students learning how to manipulate differentials and apply them in real-world scenarios.
Other exercises in this chapter
Problem 211
The area enclosed between the curves \(y^{2}=x\) and \(y=|x|\) is [2007] (A) \(2 / 3\) (B) 1 (C) \(1 / 6\) (D) \(1 / 3\)
View solution Problem 212
Let \(I=\int_{0}^{1} \frac{\sin x}{\sqrt{x}} d x\) and \(J=\int_{0}^{1} \frac{\cos x}{\sqrt{x}} d x .\) Then which one of the following is true? (A) \(I>\frac{2
View solution Problem 214
\(\int_{0}^{\pi}[\cot x] d x,[.]\) denotes the greatest integer function, is equal to (A) \(\frac{\pi}{2}\) (B) 1 (C) \(-1\) (D) \(-\frac{\pi}{2}\)
View solution Problem 215
The area of the region bounded by the parabola \((y-2)^{2}=x-1\), the tangent to the parabola at the point \((2,3)\) and the \(x\)-axis is (A) 3 (B) 6 (C) 9 (D)
View solution