/*! 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 38 Solve the given problems. Find... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Solve the given problems. Find the positive value of \(c\) such that the region bounded by \(y=x^{2}-c^{2}\) and \(y=c^{2}-x^{2}\) has an area of 576

Short Answer

Expert verified
The positive value of \(c\) is 6.

Step by step solution

01

Set Up the Equations

The given equations are \( y = x^2 - c^2 \) and \( y = c^2 - x^2 \). These are both parabolas. The first one is an upward opening parabola shifted down by \(c^2\), and the second is a downward opening parabola shifted up by \(c^2\). We need to determine where they intersect to find the bounded area.
02

Find Points of Intersection

To find the points of intersection, set the equations equal to each other:\[x^2 - c^2 = c^2 - x^2\]Simplify the equation:\[2x^2 = 2c^2 \Rightarrow x^2 = c^2 \Rightarrow x = \pm c\]Thus, the parabolas intersect at points \((c, 0)\) and \((-c, 0)\).
03

Set Up the Integral for Area

To find the area between the curves from \(-c\) to \(c\), set up the integral of the difference of the functions:\[A = \int_{-c}^{c} \left((c^2 - x^2) - (x^2 - c^2)\right) \, dx = \int_{-c}^{c} (2c^2 - 2x^2) \, dx\]
04

Integrate and Solve for Area

Integrate the expression:\[A = \int_{-c}^{c} (2c^2 - 2x^2) \, dx = 2 \left[ c^2x - \frac{x^3}{3} \right]_{-c}^{c}\]Now evaluate the integral from \(-c\) to \(c\):\[A = 2 \left[ (c^3 - \frac{c^3}{3}) - (-c^3 + \frac{c^3}{3}) \right]\]This simplifies to:\[A = 2 \left[ \frac{2c^3}{3} + \frac{2c^3}{3} \right] = 2 \times \frac{4c^3}{3} = \frac{8c^3}{3}\]
05

Solve for \(c\) Given Area

We know that the area is 576, so we set up the equation:\[\frac{8c^3}{3} = 576\]To solve for \(c\), multiply both sides by 3:\[8c^3 = 1728\]Divide by 8:\[c^3 = 216\]Take the cube root of both sides:\[c = 6\]
06

Conclusion

The positive value of \(c\) such that the area of the region bounded by the given curves is 576 is \(c = 6\).

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 91Ó°ÊÓ!

Key Concepts

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

Integral Calculus
Integral Calculus is a branch of calculus focused on the concept of integration. Integration can be thought of as the reverse process of differentiation. While derivatives measure the rate of change, integrals measure the total accumulation of a quantity or the area under a curve.

When we talk about integration, we're often discussing finding the integral of a function. The definite integral, which we'll cover in detail later, helps find the total area under a curve between two specific points on the x-axis. This ability to measure area is key in solving problems like the one in the original exercise, where the region between two curves needs to be quantified.

In our exercise, the integral helps determine the area between two parabolic curves over a given interval, from \(-c\) to \(c\). The basic concept involves recognizing the function to integrate and applying the integral across the predefined limits.
Area Between Curves
The area between curves focuses on finding the region that lies between two functions plotted on the same graph. In the example provided, we are trying to calculate the area enclosed by two parabolas: \(y = x^2 - c^2\) and \(y = c^2 - x^2\).

This calculation involves:
  • Identifying the intersection points of the curves. These represent where the curves meet, and our area is confined within these points.
  • Setting up the integral using the difference between the two functions over the determined interval. This requires subtracting the lower function from the higher function at every point along the interval.
By following these steps, we use the definite integral formula to account for the area that lies between these two curves. This is the process we applied in the given exercise to calculate the desired area of 576.
Parabolas
Parabolas are U-shaped curves that are defined by quadratic equations. In mathematics, the standard form of a parabola is given by \(y = ax^2 + bx + c\).

In our exercise, we are dealing with two parabolas:
  • \( y = x^2 - c^2 \), which opens upwards.
  • \( y = c^2 - x^2 \), which opens downwards.
These particular parabolas are symmetrical about the y-axis.

To solve the problem, we identify where these parabolas intersect. By setting the equations equal, we find the x-values where their graphs touch. These values also help define the limits \(-c\) and \(c\) for the definite integral used to calculate the enclosed area. Understanding the nature of parabolas and how they interact is crucial for problems involving areas between curves.
Definite Integrals
Definite Integrals are a fundamental concept in calculus used to calculate the accumulation of quantities, such as areas under a curve, between specified limits. When we talk about \( ext{the definite integral from } -c ext{ to } c\), it refers to the sum of the infinitesimally small areas under the curve between these points.

In calculus notation, it is expressed as \(\int_{-c}^{c}f(x) \, dx\), which tells us how to accumulate the areas from \(-c\) to \(c\) based on the function \(f(x)\).

In our solution, we used the concept of a definite integral to compute the exact area between the two parabolas. The function evaluated was the difference \((c^2 - x^2)-(x^2 - c^2)\), which expresses the height of the region at any x-value within the limits \(-c\) and \(c\).

This method allowed us to conclusively solve for the specific area, meeting the problem's requirement to equal 576, all by using the properties of definite integrals.

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

Find the coordinates of the centroids of the given figures. Each region is covered by a thin, flat plate. The region bounded by \(y=x^{2 / 3}, x=8,\) and \(y=0\)

Find the work done in pumping the water out of the top of a cylindrical tank 3.00 ft in radius and 10.0 ft high, given that the tank is initially full and water weighs \(62.4 \mathrm{lb} / \mathrm{ft}^{3}\). [Hint: If horizontal slices \(d x\) ft thick are used, each element weighs \(62.4(\pi)\left(3.00^{2}\right) d x\) lb, and each element must be raised \(10-x \mathrm{ft}\), if \(x\) is the distance from the base to the element (see Fig. 26.66 ). In this way, the force, which is the weight of the slice, and the distance through which the force acts are determined. Thus, the products of force and distance are summed by integration.]

Find the coordinates of the centroids of the given figures. Each region is covered by a thin, flat plate. The region above the \(x\) -axis, bounded by the ellipse with vertices (a, 0) and \((-a, 0),\) and minor axis \(2 b\). (The area of an ellipse is \(\pi a b\).

A current \(i=t / \sqrt{t^{2}+1}(\text { in } \mathrm{A})\) is sent through an electric dryer circuit containing a previously uncharged (zero voltage) \(2.0-\mu\) F capacitor. How long does it take for the capacitor voltage to reach \(120 \mathrm{V} ?\)

The angular acceleration \(\alpha\) is the time rate of change of angular velocity \(\omega\) of a rotating object. See Fig. \(26.3 .\) When starting up, the angular acceleration of a helicopter blade is \(\alpha=\sqrt{8 t+1} .\) Find the expression for \(\theta\) if \(\omega=0\) and \(\theta=0\) for \(t=0\)

See all solutions

Recommended explanations on Math Textbooks

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.