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Chapter 6: Problem 42

For the following regions \(R\), determine which is greater- the volume of the solid generated when \(R\) is revolved about the x-axis or about the y-axis. \(R\) is bounded by \(y=4-2 x,\) the \(x\) -axis, and the \(y\) -axis.

Short Answer

Expert verified
Answer: The volume of the solid generated when the region R is revolved around the x-axis is greater than the volume of the solid generated when the region R is revolved around the y-axis.

Step by step solution

01

Identifying Boundaries

We are given the function \(y=4-2 x\). The region \(R\) is bounded by this function, the x-axis, and the y-axis. Let's find the boundaries for x and y by setting y = 0 and x = 0 respectively. For x-axis: \(y = 4 - 2x\) \(0 = 4 - 2x\) \(x = 2\) For y-axis: No calculation needed as the boundary is exactly the y-axis, x = 0. So, the boundaries for x are \(0\le x \le 2\), and for y, the function \(y=4-2x\) represents the upper boundary and the y-axis, \(y=0\) represents the lower boundary. Now, let's find the volume when this region \(R\) is revolved around the x-axis and the y-axis.
02

Volume around the x-axis

Using the disk method, the volume of the solid generated by revolving the region around the x-axis can be found using the following formula: $$V_{x}=\pi\int_{0}^{2} [y(x)]^2 dx $$ Here, \(y(x) = 4-2x\). So we have, $$V_{x}=\pi\int_{0}^{2} (4-2x)^2 dx $$ Now, calculate the integral.
03

Evaluating the integral for the volume around the x-axis

Integrate and evaluate the definite integral: $$V_{x} = \pi\int_{0}^{2} (4-2x)^2 dx$$ $$= \pi\int_{0}^{2} (16-16x+4x^2) dx$$ $$= \pi\left[16x-\frac{16}{2}x^2+\frac{4}{3}x^3\right]_{0}^{2}$$ $$= \pi(32-16\cdot 2 + \frac{4}{3} \cdot 8)$$ $$= \frac{64}{3}\pi$$ The volume of the solid generated when the region is revolved about the x-axis is \(\frac{64}{3}\pi\).
04

Volume around the y-axis

Now let's find the volume of the solid generated by revolving the region around the y-axis. We must express x in terms of y for this case. Given the function \(y = 4 - 2x\), we have \(x = \frac{4-y}{2}\). Now using the disk method, the volume of the solid generated by revolving the region around the y-axis can be found using the following formula: $$V_{y}=\pi\int_{0}^{4}\left [x(y)\right] ^2 dy$$ Here, \(x(y) = \frac{4-y}{2}\). So we have, $$V_{y}=\pi\int_{0}^{4}\left(\frac{4-y}{2}\right)^2 dy $$
05

Evaluating the integral for the volume around the y-axis

Integrate and evaluate the definite integral: $$V_{y} = \pi\int_{0}^{4} \left(\frac{4-y}{2}\right)^2 dy$$ $$= \pi\int_{0}^{4} \frac{(4-y)^2}{4} dy$$ $$= \frac{\pi}{4}\int_{0}^{4} (16-8y+y^2) dy$$ $$= \frac{\pi}{4}(16y-4y^2+\frac{1}{3}y^3)\Big|_{0}^{4}$$ $$= \frac{\pi}{4}(64 - 64 + \frac{64}{3})$$ $$= \frac{16}{3}\pi$$ The volume of the solid generated when the region is revolved about the y-axis is \(\frac{16}{3}\pi\).
06

Comparing the volumes

Now, let's compare \(V_{x}\) and \(V_{y}\): $$V_{x} = \frac{64}{3}\pi$$ $$V_{y} = \frac{16}{3}\pi$$ Since \(\frac{64}{3}\pi > \frac{16}{3}\pi\), we can conclude that the volume of the solid generated when the region \(R\) is revolved around the x-axis is greater than the volume of the solid generated when the region \(R\) is revolved around the y-axis.

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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 a solid of revolution is a fascinating concept in calculus. It involves rotating a region or shape around a specified axis to create a three-dimensional object.
This method is widely used to solve various engineering and physics problems where solid objects with rotational symmetry need to be analyzed.
To determine the volume, we must first define the shape or region to be rotated, along with the axis of rotation. The shape could be a simple curve, such as a line or parabola, which can create complex volumes when rotated.
  • Identify the function or curve representing the boundary of the region.
  • Determine the limits or bounds of the region, usually defined by intersections with other curves or axes.
  • Select the axis of rotation: the x-axis, y-axis, or even a vertical or horizontal line different from the coordinate axes.
The fundamental idea is that by rotating a flat, two-dimensional area around an axis, we create a volume."},{
Disk Method
The disk method is a practical technique in calculus for calculating the volume of solids of revolution. This method derives its name from the way it slices the solid into a stack of disks or washers.
Each disk is treated as a tiny cylinder, and by taking the infinite sum of these small volumes, we obtain the total volume.
To use the disk method, you need to convert everything into terms that can relate to the axis of rotation. For this, you will:
  • Start by finding the radius of each disk. For rotation around the x-axis, the radius comes from the function value, i.e., the vertical distance from the curve to the axis of rotation.
  • The thickness of each disk is a small value along the axis of rotation, either \(dx\) or \(dy\), depending on the rotation.
  • The volume of each disk is then \( \pi [\text{radius}]^2 \cdot [\text{thickness}] \).
The summation of these disk volumes from the lower to the upper bounds gives the total volume. This integral setup captures the idea that integrations of tiny slices approximate the full volume. The method is versatile and can be used for solids rotated about either the x-axis or the y-axis.
Definite Integral
A definite integral is an essential tool in calculus used to calculate exact quantities, such as areas under a curve, total distance traveled, or, in this case, volumes of solids.
The power of the definite integral lies in its ability to provide the summation of varying quantities continuously over an interval. When computing volumes, a definite integral adds up infinitesimally small slices to approximate a total. These slices can be incredibly thin, allowing for precise measurements.
Here's how it works:
  • The integral is set up with limits that indicate where the process of summation begins and ends for a given variable. These are the bounds.
  • For volume, typically, the integrand is based on a function that defines the shape of the solid being rotated.
  • The integral of the area of each slice, multiplied by the thickness, provides the exact volume without approximation.
The definite integral elegantly combines the precision of calculus with practical application, making it indispensable for solving real-world problems where dimensions and measures must be exact.

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