Problem 60 - Calculus Early Transcendentals

Question

Find the volume of the following solids using the method of your choice. The solid formed when the region bounded by \(y=2, y=2 x+2\), and \(x=6\) is revolved about the \(y\) -axis

Step-by-Step Solution

Answer

The final volume of the resulting solid is \(\frac{8924\pi}{12}\).

Step 1: Sketch the graph and identify the region

First, we need to sketch the graph and identify the region that we're revolving around the \(y\)-axis. This will help us visualize the problem and set up the integral properly. We can easily see that the line \(y=2\) is a horizontal line, the line \(y=2x+2\) is a line that intersects the \(y\)-axis at the point \((0,2)\) and has a positive slope of 2, and \(x=6\) is a vertical line. When we sketch these lines and shade the region they enclose, we can see that it's a triangle.

Step 2: Set up the volume integral using the disk method

Since we're revolving the region around the \(y\)-axis, we want to use the disk method with respect to \(y\). The volume of each infinitesimally small disk can be represented by the formula \(dV = \pi r^2 dy\). The radius of each disk can be found by noticing that it is equal to the \(x\)-coordinate along the line \(y=2x+2\), which can be rewritten as \(x = \frac{y-2}{2}\). Based on the bounding lines, our region falls within \(y\) values of 2 and 14. So, setup the integral to find the volume like this: \(V = \int_2^{14} \pi (\frac{y-2}{2})^2 dy\)

Step 3: Solve the volume integral

Now, we need to solve the integral to find the volume. We start by expanding the integrand and factoring out any constants: \(V = \pi \int_2^{14} (\frac{y^2-4y+4}{4}) dy = \frac{\pi}{4} \int_2^{14} (y^2-4y+4) dy\) Next, we integrate the polynomial term by term: \(V = \frac{\pi}{4} [\frac{y^3}{3} - 2y^2 + 4y] |_2^{14}\) Now, we plug in the upper and lower limits of integration and subtract: \(V = \frac{\pi}{4} [( \frac{14^3}{3} - 2(14)^2 + 4(14)) - (\frac{2^3}{3} - 2(2)^2 + 4(2))]\) \(V = \frac{\pi}{4} [(4312 - 392 + 56) - (\frac{8}{3} - 8 + 8)]\) \(V = \frac{\pi}{4} [2976 - \frac{8}{3}]\)

Step 4: Calculate the final volume

Finally, multiply the volume by \(\pi / 4\): \(V = \pi ( \frac{2976 - \frac{8}{3}}{4})\) \(V = \frac{8924\pi}{12}\) So the volume of the solid formed when the region bounded by the lines \(y=2, y=2x+2\), and \(x=6\) is revolved about the \(y\)-axis is \(\frac{8924\pi}{12}\).

Key Concepts

Disk Method

The disk method is a popular technique for calculating the volume of solids generated by revolving a region around an axis. It's particularly useful when the solid resembles a stack of disks or washers. Imagine slicing a loaf of bread; each slice or disk has a thickness (which is infinitesimally small) and a circular shape. The disk method uses this principle.

• The idea is to sum up the volume of each disk to get the total volume of the solid.
• When revolving around the y-axis, the radius of each disk is determined by the function defining the curve.
• The thickness of each disk is a small measure along the y-axis, represented by the differential dy.

This method involves setting up an integral, as shown in the original problem, where the formula for the volume of an individual disk is used: \[ dV = \pi r^2 dy \]. This formula arises because the area of a circle is \( \pi r^2 \), and multiplying by an infinitesimal thickness gives the volume.
Applying this method involves integrating the formula over the limits of the region, such as from \( y = 2 \) to \( y = 14 \) in our exercise. It's crucial to express the radius in terms of y, which usually requires solving the function for x in terms of y, like \( x = \frac{y-2}{2} \) in this case.

Integrals

Integrals are fundamental in calculus and are essential when calculating volumes, areas, and more. In this context, we use integration to sum up infinitely small quantities to determine the total volume of a solid.

• The integral \( \int \) symbol represents this summation process.
• By integrating a radius function squared across the limits of the region, we can calculate the volume of the solid of revolution.

For instance, our integral \( \int_2^{14} \frac{(y-2)^2}{4} dy \) runs from lower limit 2 to upper limit 14. This represents using all y-values between these points to fully account for the entire volume of the solid.
Solving an integral often involves factoring out constants and expanding terms, simplifying the calculation. It's a matter of integrating each term (like \( y^2, y \)) individually and then substituting the boundary values. Integration basics conclude with evaluating the integral at the upper limit and subtracting the evaluation at the lower limit, which gives you the exact volume as in our example.

Revolving a Region

Revolving a region is a geometric concept where a two-dimensional shape or region on a plane is rotated about a line (often an axis) to produce a three-dimensional solid. The revolution is a simple yet powerful way to create volume and is commonly used to design objects with rotational symmetry.

• The original region is typically defined by curves or lines and bounded by specific limits.
• As the region rotates, every point traces out a circle, creating a solid.

In the original exercise, the region bounded by \(y=2, y=2x+2\), and \(x=6\) revolves around the y-axis. The line creates a triangle-like shape, which, when rotated, produces a solid that can be analyzed using calculus techniques like the disk method.
Starting by sketching the region helps visualize the set-up and ensure the correct application of methods and integration limits. Always pinpoint where the shape is bounded and what variable curves determine these boundaries. This influence when and where the function is solved for x (e.g., \( x = \frac{y-2}{2} \)) and how the integral's limits are defined, such as from \( y=2 \) to \( y=14 \). Revolving a region transforms the original 2D shape into a full 3D object, allowing its volume to be calculated accurately.