Problem 87 - Calculus Single Variable

Question

A flashlight reflector is made of an aluminum alloy whose mass density is \(3.74 \mathrm{~g} / \mathrm{cm}^{3} .\) The reflector occupies the solid region that is obtained when the region bounded by \(y=\) \(2.05 \sqrt{x}+0.496, y=2.05 \sqrt{x}+0.546, x=0,\) and \(x=2.80 \mathrm{~cm}\) is rotated about the \(x\) -axis. What is the mass of the reflector?

Step-by-Step Solution

Answer

The mass is determined by evaluating the volume integral and multiplying by the density, resulting in the mass of the reflector.

Step 1: Identify the Region of Rotation

The reflector is formed by rotating the region bounded by the curves \(y = 2.05 \sqrt{x}+0.496\) and \(y = 2.05 \sqrt{x}+0.546\) from \(x = 0\) to \(x = 2.80\) cm around the x-axis. The region between these two curves will form a solid when revolved around the x-axis.

Step 2: Determine the Radii of Rotation

The outer radius \(R\) and the inner radius \(r\) of the solid of revolution are given by the functions \(y = 2.05 \sqrt{x}+0.546\) and \(y = 2.05 \sqrt{x}+0.496\), respectively. These will be used to calculate the volume of the hollow cylinder formed by rotating the bounded region.

Step 3: Set Up the Volume Integral

The volume \(V\) of the solid can be calculated using the formula for the volume of a solid of revolution: \[ V = \pi \int_{a}^{b}(R^2 - r^2)\,dx \] where \(R = 2.05 \sqrt{x} + 0.546\) and \(r = 2.05 \sqrt{x} + 0.496\). Integrate from \(x = 0\) to \(x = 2.80\).

Step 4: Calculate the Volume Integral

Substitute \(R\) and \(r\) into the integral: \[ V = \pi \int_{0}^{2.80} \left[(2.05 \sqrt{x} + 0.546)^2 - (2.05 \sqrt{x} + 0.496)^2\right] dx \]. Simplify and evaluate the integral to get the volume of the solid.

Step 5: Compute the Mass of the Reflector

Calculate the mass by multiplying the volume of the reflector by the mass density of the aluminum alloy: \[ \text{Mass} = \text{Density} \times V = 3.74 \, \text{g/cm}^3 \times V \]. Substitute the calculated volume to find the mass.

Key Concepts

Volume Integral

When it comes to calculating volumes of complex objects, the volume integral is a powerful tool. In the context of a solid of revolution, such as the flashlight reflector in our exercise, the volume integral helps us find out how much space is enclosed by a 3D object formed by rotating a 2D shape around an axis.

For our flashlight reflector, we use two curves: \(y = 2.05 \sqrt{x}+0.496\) and \(y = 2.05 \sqrt{x}+0.546\). These curves define the top and bottom edges of the 2D shape that we will rotate. By revolving this area around the \(x\)-axis, we form the reflector.

Think of the volume integral as slicing the solid into infinitesimally thin disks or shells and adding up the volumes of all these tiny slices. The formula used for this calculation is:

• \( V = \pi \int_{a}^{b}(R^2 - r^2)\,dx \)

This integral sums up the area of each infinitesimally small disk, from the starting point \(x = a\) to the endpoint \(x = b\), multiplied by the thickness of the slices (\(dx\)). Here the integral calculates the difference between the outer radius \(R\) and the inner radius \(r\), resulting in the hollow area we are interested in.

Mass Density

Mass density is a measure of how much mass is contained in a certain volume. Think of it like packing objects tightly in a space. For our reflector, made from an aluminum alloy, the given mass density is \(3.74 \text{ g/cm}^3\). This density tells us how heavy one cubic centimeter of the material is.

To find the mass of a three-dimensional object like the reflector, we first need to find its volume, which we do with the volume integral. Once we have the volume, calculating the mass is straightforward.

The formula for mass using density is quite simple:

• \( \text{Mass} = \text{Density} \times V \)

This means, take the volume you calculated and multiply it by the object's density. The result is the mass of the object, which tells you how much it would weigh on a scale.

Rotating a Region

Creating a solid of revolution involves taking a region in the plane and spinning it around an axis. By doing this, we obtain a three-dimensional shape. In this particular exercise, the region is defined by two curves: \(y = 2.05 \sqrt{x}+0.496\) and \(y = 2.05 \sqrt{x}+0.546\). When this region is rotated about the \(x\)-axis, it forms the hollow reflector shape you would see in a flashlight.

Here's how it works: Imagine drawing the region between the two curves on a piece of paper, then pinning it so it can spin around the x-axis. As it rotates, every point in the region moves to form a circle around the axis, creating a "donut" shape but with the donut hole offset since the curves are not centered on the axis.

Key considerations when rotating a region include:

• Identify the curves that bound the region.
• Decide on the axis of rotation (in this case, the \(x\)-axis).
• Setting up the correct formula to calculate the volume or other properties of the resultant solid.
  

This concept is fundamental in calculus and helps engineers and scientists design and analyze objects with rotational symmetry.