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Chapter 8: Problem 73

For \(k>0,\) the volume of the solid created by rotating the region bounded by \(y=k x(x-2)\) and the \(x\) -axis between \(x=0\) and \(x=2\) around the \(x\) -axis is \(192 \pi / 5\) Find \(k\).

Short Answer

Expert verified
The value of \(k\) is 6.

Step by step solution

01

Set Up the Problem

We're rotating the region bounded by the curve \(y = kx(x-2)\) and the x-axis between \(x = 0\) and \(x = 2\) around the x-axis. This will create a solid whose volume can be calculated using the disk method. The volume \(V\) is given by the integral \(\pi \int_{a}^{b} [f(x)]^2 \, dx\). Here, \(f(x) = kx(x-2)\) and the bounds are \(a = 0\) and \(b = 2\).
02

Write the Volume Integral

The volume of the solid is expressed as: \[ V = \pi \int_{0}^{2} [kx(x-2)]^2 \, dx \] We need to evaluate this integral and set it equal to the given volume \(\frac{192\pi}{5}\) to find the value of \(k\).
03

Simplify the Integrand

First, simplify the function within the integral:\[ (kx(x-2))^2 = k^2x^2(x^2 - 4x + 4) = k^2(x^4 - 4x^3 + 4x^2) \]
04

Integrate the Function

Set up and integrate:\[ V = \pi k^2 \int_{0}^{2} (x^4 - 4x^3 + 4x^2) \, dx \]Calculate the antiderivative:\[ = \pi k^2 \left[ \frac{x^5}{5} - \frac{4x^4}{4} + \frac{4x^3}{3} \right]_{0}^{2} \]
05

Evaluate the Definite Integral

Evaluate the antiderivative from \(x = 0\) to \(x = 2\):\[ = \pi k^2 \left( \frac{32}{5} - 16 + \frac{32}{3} \right) = \pi k^2 \left( \frac{96}{15} - \frac{240}{15} + \frac{160}{15} \right) = \pi k^2 \left( \frac{16}{15} \right) \]Thus, the expression for the volume becomes:\[ V = \frac{16}{15} \pi k^2 \]
06

Solve for k

Set the derived volume expression equal to the given volume:\[ \frac{16}{15} \pi k^2 = \frac{192 \pi}{5} \]Cancel \(\pi\) from both sides:\[ \frac{16}{15} k^2 = \frac{192}{5} \]Multiply both sides by \(15\):\[ 16k^2 = 576 \]Divide by \(16\):\[ k^2 = 36 \]Take the square root:\[ k = 6 \]
07

Verify

Verify whether \(k = 6\) satisfies all conditions and calculations through reversing steps or plugging back into the integral itself to confirm the calculated volume matches \(\frac{192 \pi}{5}\).

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Key Concepts

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

Volume of Solids of Revolution
When we talk about solids of revolution, imagine rotating a 2D shape around an axis, creating a 3D shape. In this particular problem, we rotated the region under the curve of the polynomial function \( y = kx(x-2) \) around the x-axis. This results in a solid whose volume we want to calculate using the disk method.
  • **Disk Method**: This method considers the solid as a stack of thin circular disks. Each disk has a small thickness, typically represented as \( dx \).
  • The radius of each disk is the function value \( f(x) \). So, for our problem, the radius is \( kx(x-2) \).
  • The volume of each disk is the area of the circle \( \pi [f(x)]^2 \) times a small thickness \( dx \).
  • The total volume is found by integrating from the start to the end of the section along the x-axis, i.e., \( \pi \int_{0}^{2} [kx(x-2)]^2 \ dx \).
The solution uses this method to express the volume in terms of the variable \( k \) and perform calculations to find its exact value.
Definite Integral
The definite integral is a fundamental concept in calculus for finding areas under curves, but it also calculates the volume of solids, especially in problems like this one involving rotation. A definite integral from \( a \) to \( b \) accumulates values, effectively calculating the sum of infinitely many infinitesimally small quantities.
  • In our exercise, the definite integral is \( \int_{0}^{2} [kx(x-2)]^2 \ dx \), which represents the total volume of the solid formed.
  • The limits of integration \( 0 \) and \( 2 \) tell us that we are considering the region from \( x = 0 \) to \( x = 2 \).
  • Simplifying the integrand \( [kx(x-2)]^2 \) gives us a more straightforward polynomial to integrate: \( k^2(x^4 - 4x^3 + 4x^2) \).
  • The integration process involves finding the antiderivative of the polynomial within these bounds. Once integrated, it provides a numeric result once evaluated from \( x=0 \) to \( x=2 \).
This problem shows how definite integrals can be used to compute not just areas but volumes, especially for solids formed by rotation.
Polynomial Functions
Polynomial functions are equations based on terms of varying powers of variables, often providing curves that can represent real-world quantities. In the exercise, \( y = kx(x-2) \) is a quadratic polynomial function. Here鈥檚 how they fit in the context of the problem:
  • **Structure**: The given polynomial \( y = kx(x-2) \) is a product of \( k \), \( x \), and \( (x-2) \), simplifying to a quadratic form \( y = k(x^2 - 2x) \).
  • This polynomial intersects the x-axis at the solutions to \( kx(x-2) = 0 \), or \( x = 0 \) and \( x = 2 \), defining the region of interest for rotation.
  • By squaring the function as part of the disk method integral \( [kx(x-2)]^2 \), the component \( k^2(x^4 - 4x^3 + 4x^2) \) becomes a polynomial integration problem.
  • **Role in Integration**: Each term of the polynomial contributes separately to the integration, allowing for precise calculation of the volume.
Understanding polynomial functions is crucial for setting up and solving integrals, especially when they describe curves that generate volumes when rotated.

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Most popular questions from this chapter

A particle moves with velocity \(d x / d t\) in the \(x\) -direction and \(d y / d t\) in the \(y\) -direction at time \(t\) in seconds, where $$\frac{d x}{d t}=-\sin t \quad \text { and } \quad \frac{d y}{d t}=\cos t$$ (a) Find the change in position in the \(x\) and \(y\) coordinates between \(t=0\) and \(t=\pi\). (b) If the particle passes through (2,3) at \(t=0,\) find its position at \(t=\pi\). (c) Find the distance traveled by the particle from time \(t=0\) to \(t=\pi\).

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