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Chapter 6: Problem 24

Shell method Let R be the region bounded by the following curves. Use the shell method to find the volume of the solid generated when \(R\) is revolved about indicated axis. \(y=\sqrt{4-2 x^{2}}, y=0,\) and \(x=0,\) in the first quadrant; about the \(y\) -axis

Short Answer

Expert verified
Based on the provided step-by-step solution, the short answer for this question is: The volume of the solid generated when the region R is revolved about the y-axis is \(\frac{8}{3}\pi\) cubic units.

Step by step solution

01

1. Find the Intersection Points of the Given Curves

The region R is bounded in part by the curve \(y = \sqrt{4 - 2x^2}\). Since this is in the first quadrant, the intersection point will be with the y-axis and the x-axis. To find where this curve intersects with the y-axis \(x=0\), insert the value of \(x\) in the equation: \(y = \sqrt{4 - 2(0)^2} \\ y = \sqrt{4} \\ y = 2\) To find where the curve intersects the x-axis \(y = 0\): \(0 = \sqrt{4 - 2x^2} \\ 0 = 4 - 2x^2 \\ x^2 = 2 \\ x = \sqrt{2}\) The region R has intersection points at (0,0), (0,2), and \((\sqrt{2}, 0)\).
02

2. Set up the Shell Method Integral for Volume

We are using the shell method with the y-axis as the axis of rotation. In this case, we can represent the volume as: \(V = 2\pi \int_{a}^{b} radius \cdot height \: dx \) Here, the limits of integration (a and b) will be the x-values of the intersection points: \(a = 0 \\ b = \sqrt{2}\) The radius of each shell will be the distance from the y-axis, which is just x, so: \(radius = x\) The height of each shell is the difference between the curve and the x-axis: \(height = \sqrt{4 - 2x^2} \) Now, we can plug these values into the integral to represent the volume of the solid: \(V = 2\pi \int_{0}^{\sqrt{2}} x(\sqrt{4 - 2x^2}) \: dx\)
03

3. Evaluate the Integral

In order to evaluate the integral, we will use substitution: Let \(u = 4 - 2x^2\) \(du = -4x \: dx\) Rearrange for x and dx: \(x \: dx = -\frac{1}{4} du\) With new limits of integration: \(u(0) = 4 - 2(0)^2 = 4\) \(u(\sqrt{2}) = 4 - 2(\sqrt{2})^2 = 0\) The integral is now transformed to: \(V = 2\pi (-\frac{1}{4}) \int_{4}^{0} u^\frac{1}{2} \: du\) Now we can evaluate the integral: \(V = -\frac{1}{2}\pi \int_{4}^{0} u^\frac{1}{2} \: du\) \(V = -\frac{1}{2}\pi [(\frac{2}{3})u^\frac{3}{2}]_{4}^{0}\) \(V = -\frac{1}{3}\pi [(0) - (8)]\) \(V = \frac{8}{3}\pi\) cubic units. The volume of the solid generated when the region R is revolved about the y-axis is \(\frac{8}{3}\pi\) cubic units.

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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
The volume of solids of revolution involves calculating the 3D space occupied by a shape obtained by rotating a 2D region around an axis. This is common in integral calculus where certain curves are revolved around an axis to form solid shapes, often for practical applications like creating symmetrical objects by rotation.
When determining these volumes, two primary methods are used: the Disk Method and the Shell Method.
  • Disk Method: This is used when slices perpendicular to the axis of rotation are made. These slices form disks or washers when revolved.
  • Shell Method: This is typically used for regions rotated around the y-axis. Cylindrical shells are formed by slicing parallel to the axis. It's especially useful when the bounds of the function are easier to integrate with respect to other variables, such as the given problem.
The premise with the Shell Method is to use cylinders as approximations of volume. In our example, we consider the distance from the y-axis (radius) and the height of the shell obtained from the function, leading to the integral formula used.
Integral Calculus
Integral calculus forms the backbone for many of the computations involved in calculating volumes, such as those of solids of revolution. This branch of calculus deals with the accumulation of quantities, which in this context is volume accumulated by slices or shells around an axis.
The crux of these problems is setting up an appropriate integral that represents the problem's geometric situation. We determined the integral for the Shell Method as follows:
\[ V = 2\pi \int_{a}^{b} (radius \times height) \, dx \]In this setup:
  • The radius corresponds to the distance from the axis of rotation, which is expressed as the x-value of the shell.
  • The height refers to the function value, i.e., the curve's distance from the x-axis or other boundaries.
Successfully solving the integral involves transforming and integrating using techniques like substitution, allowing us to find exact values. In our problem, this led to the final computed volume of the solid after evaluating the integral.
Curve Intersection Points
Identifying and understanding curve intersection points is crucial in any problem involving bounded regions. These points determine the limits or bounds of integration and help in sketching the region accurately.
Considering the given curves, we can solve their equations to find intersection points that define boundaries of region R. This process involves substituting values like finding:
  • Intersection with the y-axis by setting \(x=0\).
  • Intersection with the x-axis by setting \(y=0\).
For the given curves, this resulted in intersection points of \((0, 0)\), \((0, 2)\), and \((\sqrt{2}, 0)\). These points are essential in establishing limits for the Shell Method, guiding us from the lower bound of \(0\) to an upper bound of \(\sqrt{2}\). Consequently, correctly identifying these helps in proper integral setup and, ultimately, correct volume calculation.

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Most popular questions from this chapter

Emptying a cylindrical tank A cylindrical water tank has height \(8 \mathrm{m}\) and radius \(2 \mathrm{m}\) (see figure). a. If the tank is full of water, how much work is required to pump the water to the level of the top of the tank and out of the tank? b. Is it true that it takes half as much work to pump the water out of the tank when it is half full as when it is full? Explain.

Leaky Bucket A \(1-\mathrm{kg}\) bucket resting on the ground contains \(3 \mathrm{kg}\) of water. How much work is required to raise the bucket vertically a distance of \(10 \mathrm{m}\) if water leaks out of the bucket at a constant rate of \(\frac{1}{5} \mathrm{kg} / \mathrm{m} ?\) Assume the weight of the rope used to raise the bucket is negligible. (Hint: Use the definition of work, \(W=\int_{a}^{b} F(y) d y,\) where \(F\) is the variable force required to lift an object along a vertical line from \(y=a\) to \(y=b\).)

A data collection probe is dropped from a stationary balloon, and it falls with a velocity (in \(\mathrm{m} / \mathrm{s}\) ) given by \(v(t)=9.8 t\) neglecting air resistance. After \(10 \mathrm{s}\), a chute deploys and the probe immediately slows to a constant speed of \(10 \mathrm{m} / \mathrm{s}\), which it maintains until it enters the ocean. a. Graph the velocity function. b.how far does the probe fall in the first 30 s after it is released? c. If the probe was released from an altitude of \(3 \mathrm{km}\), when does it enter the ocean?

Coiling a rope \(A\) 60-m-long, 9.4 -mm-diameter rope hangs freely from a ledge. The density of the rope is \(55 \mathrm{g} / \mathrm{m}\). How much work is needed to pull the entire rope to the ledge?

Shell method Use the shell method to find the volume of the following solids. The solid formed when a hole of radius 3 is drilled symmetrically along the axis of a right circular cone of radius 6 and height 9
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