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Chapter 6: Problem 40

Find the volumes of the solids generated by revolving the regions bounded by the lines and curves about the \(x\)-axis. \(y=4-x^{2}, \quad y=2-x\)

Short Answer

Expert verified
The volume of the solid is \(\frac{117\pi}{5}\).

Step by step solution

01

Identify the area to be revolved

First, you need to find the points of intersection of the curves to determine the region to revolve. Set the equations equal to each other: \(4-x^2 = 2-x\). Solve for \(x\) to find the points of intersection, which will give you the limits for integration.
02

Solve for points of intersection

Rearrange the equation \(4 - x^2 = 2 - x\) to form a standard quadratic equation: \(-x^2 + x + 2 = 0\). Simplify to get \(x^2 - x - 2 = 0\). Factor the quadratic to find \((x-2)(x+1) = 0\). Thus, the solutions are \(x = 2\) and \(x = -1\).
03

Set up the integral for the volume

The volume \(V\) of the solid formed by revolving the region around the x-axis is found using the disk method. Use the formula: \[ V = \pi \int_{a}^{b} \left( R(x)^2 - r(x)^2 \right) \, dx \]where \(R(x)\) is the outer radius and \(r(x)\) is the inner radius. Here, \(R(x) = 4-x^2\) and \(r(x) = 2-x\).
04

Calculate the volume integral

Substitute \(R(x)\) and \(r(x)\) into the volume formula:\[ V = \pi \int_{-1}^{2} \left((4-x^2)^2 - (2-x)^2 \right) \, dx \]. Expand each term and simplify the integrand, then perform the integration.
05

Evaluate the integral

Simplify the expression \((4 - x^2)^2 - (2 - x)^2 = 16 - 8x^2 + x^4 - (4 - 4x + x^2) = x^4 - 9x^2 + 4x + 12\). Integrate term by term over the interval \([-1, 2]\):\[ V = \pi \left[ \frac{x^5}{5} - 3x^3 + 2x^2 + 12x \right]_{-1}^{2} \].
06

Solve for numeric value of integral

Calculate the definite integral:\[ V = \pi \left[ \left(\frac{32}{5} - 24 + 8 + 24\right) - \left(-\frac{1}{5} + 3 + 2 - 12\right) \right] \]. Simplify each term inside the parenthesis to find that \(V = \pi \times \frac{117}{5} = \frac{117\pi}{5}\).

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Key Concepts

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

Disk Method
The disk method is a technique used to find the volume of a solid of revolution. This method revolves a region around an axis. Imagine a stack of very thin disks (or circles) that together form the entire solid. The volume of each disk can be found using its radius and thickness. Here's how it works:
  • Identify the region to be revolved around the axis.
  • Determine the radii of the disks, often expressed as functions of x or y.
  • Set up the integral over the interval that represents the region being revolved.
The general formula when revolving around the x-axis is:\[ V = \pi \int_{a}^{b} [R(x)^2 - r(x)^2] \, dx \]where \(R(x)\) and \(r(x)\) are the outer and inner radii of the disk, respectively. For this exercise, there is no inner radius, so focus on \(R(x)^2\). To understand why this method is useful, visualize slices of the solid, each adding up to form the whole.
Integration
Integration is a mathematical process used to calculate the area under a curve. In the context of finding volumes of revolution, integration helps sum up the volumes of infinitely thin disks to get the total volume of the solid. When working with the disk method, integration helps formalize the summation into an integral. Here's a breakdown of how it's used:
  • The limits of integration are determined by the points of intersection of the curves or lines that form the boundary of the region.
  • The integrand (the expression inside the integral) represents the squared radius of the disks, capturing all points from the outer edge of the solid to the axis of revolution.
  • Once set up, the integral is evaluated to get the exact volume of the solid.
Think of integration as building the entire solid one tiny piece at a time. Each infinitesimally small disk volume contributes to the grand total.
Quadratic Equations
Quadratic equations are essential when solving for the points of intersection between curves. A general quadratic equation is in the form \(ax^2 + bx + c = 0\). Solving this type of equation gives us the values that allow us to find intersections.In this particular problem, the process involves:
  • Setting the equations equal to each other to find common points.
  • Rearranging and simplifying the equation to a standard quadratic form.
  • Finding the roots of the equation using methods like factoring.
For the exercise given, the equation \(-x^2 + x + 2 = 0\) is transformed into \(x^2 - x - 2 = 0\) and factored into \((x-2)(x+1) = 0\). Solving for \(x\) gives \(x = 2\) and \(x = -1\), which are critical for defining the limits of integration.
Points of Intersection
Finding points of intersection involves determining where two or more curves meet. These points are crucial as they define the boundaries of the region we revolve.To find these points:
  • Set the equations of the curves equal to each other.
  • Solve this equation for the variable, typically \(x\).
  • Use these solutions as the limits of integration when setting up the integral for volume.
In our example, the intersection points are where \(y = 4-x^2\) intersects with \(y = 2-x\). Solving for \(x\) through the quadratic equation reveals the intersection points: \(x = 2\) and \(x = -1\). These values are integral in carving out the region of interest, paving the way to understanding the physical scope of any problems involving revolved solids.

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Most popular questions from this chapter

Find the lateral (side) surface area of the cone generated by revolving the line segment \(y = x / 2,0 \leq x \leq 4 ,\) about the \(x\) -axis. Check your answer with the geometry formula Lateral surface area \(= \frac { 1 } { 2 } \times\) base circumference \(\times\) slant height.

Tilted plate Calculate the fluid force on one side of a 5 \(\mathrm{ft}\) by 5 \(\mathrm{ft}\) square plate if the plate is at the bottom of a pool filled with water to a depth of 8 \(\mathrm{ft}\) and \begin{equation}\begin{array}{l}{\text { a. lying flat on its } 5 \text { ft by } 5 \text { ft face. }} \\ {\text { b. resting vertically on a } 5 \text { -ft edge. }} \\ {\text { c. resting on a } 5 \text { -ft edge and tilted at } 45^{\circ} \text { to the bottom of the pool. }}\end{array}\end{equation}

For some regions, both the washer and shell methods work well for the solid generated by revolving the region about the coordinate axes, but this is not always the case. When a region is revolved about the \(y\) -axis, for example, and washers are used, we must integrate with respect to \(y .\) It may not be possible, however, to express the integrand in terms of \(y .\) In such a case, the shell method allows us to integrate with respect to \(x\) instead. Exercises 29 and 30 provide some insight. $$ \begin{array}{l}{\text { Compute the volume of the solid generated by revolving the triangular }} \\ {\text { region bounded by the lines } 2 y=x+4, y=x, \text { and } x=0} \\ {\text { about }}\end{array} $$ $$ \begin{array}{l}{\text { a. the } x \text { -axis using the washer method. }} \\\ {\text { b. the } y \text { -axis using the shell method. }} \\ {\text { c. the line } x=4 \text { using the shell method. }} \\ {\text { d. the line } y=8 \text { using the washer method. }}\end{array} $$

The region in the first quadrant that is bounded above by the curve \(y=1 / \sqrt{x},\) on the left by the line \(x=1 / 4,\) and below by the line \(y=1\) is revolved about the \(y\) -axis to generate a solid. Find the volume of the solid by $$\text{a. the washer method}. \quad \text{b. the shell method}.$$

The region between the curve \(y=2 / x\) and the \(x\) -axis from \(x=1\) to \(x=4\) is revolved about the \(x\) -axis to generate a solid. \begin{equation}\begin{array}{l}{\text { a. Find the volume of the solid. }} \\\ {\text { b. Find the center of mass of a thin plate covering the region if }} \\ {\text { the plate's density at the point }(x, y) \text { is } \delta(x)=\sqrt{x} \text { . }} \\ {\text { c. Sketch the plate and show the center of mass in your sketch. }}\end{array}\end{equation}
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