Question

In Exercises 23-28, find the volume of the solid generated by revolving the region about the y-axis. the region enclosed by the triangle with vertices \((1,0),(2,1),\) and \((1,1)\)

Short answer

The volume of the solid generated by revolving the region about the y-axis is \(\pi/12\).

Step-by-step solution

Sketch the Region and Setup the Integrals

Firstly, it is crucial to sketch the triangular region based on given vertices which are (1,0),(2,1), and (1,1). We note that the lines connecting these points are \(y=x-1\) and \(x=1\). This shape will be revolved about the y-axis to create the solid, giving two separate solids from \(y=0\) to \(y=1\) and \(y=1\) to \(y=2\). Thus, the integral limit should be set from 0 to 1 for the first solid and from 1 to 2 for the second solid.

Calculate the Radius for Each Solid

The radius of each disk in the solid of revolution is simply the x value of the function since it is revolved around the y-axis. As such, the radius for the first solid is \(1-y\), and the radius for the second solid is \(y\).

Apply the Disk Method to Find Volume

The disk method states that the volume of a solid of revolution is given by the formula \(\pi \int_{a}^{b} [f(y)]^2 dy\), where \(f(y)\) is the radius as a function of y. Plug in the radius function and the limits from step 1 and 2 into the disk method formula. Thus, we have two separate volume calculations: one for the solid formed by the line segment from (1,0) to (1,1), and another for the line segment from (2,1) to (1,1) due to the different radius of each part. The total volume is the sum of these two separate volume calculations. This gives us the final integral to solve: \(V = \pi \int_{0}^{1} (1-y)^2dy + \pi \int_{1}^{2} y^2dy\).

Evaluate the Integrals

Evaluate the integral to find the volume of the solid. This results in \(V = [1/3\pi - \pi/4] + [\pi/2 - 1/3\pi] = \pi/12\).

Key concepts

Disk Method

The disk method is a key technique in integral calculus for finding the volumes of solids of revolution. When a region in the plane is revolved around an axis, it creates a three-dimensional shape. Imagine slicing this shape into thin circular disks. The volume of each disk can be approximated as \( V_{\text{disk}} = \pi r^2 h \), where \( r \) is the radius of the disk and \( h \) is its thickness.

To construct these disks when rotating around the y-axis, as in our exercise, the radius is determined by the x-coordinate of the function. Let's dissect the process: you first identify the function that determines the radius of rotation, then you square this function (since the disk area formula includes \( r^2 \) ), and lastly, you multiply it by \( \pi \) to account for the circular area. Integration across the specified bounds yields the total volume of the solid.

For our triangular volume problem, the radius changes depending on the y-value. By integrating these radius functions squared along the specified y-interval, we capture the volume of the resultant solid. The disk method simplifies what seems complex into a manageable calculation.

Solid Geometry Calculus

Solid geometry calculus is the study of shapes with three dimensions—length, width, and height—using the principles of calculus. While traditional solid geometry involves calculating volumes and areas of 3D shapes using established formulas, solid geometry calculus allows us to determine these measures for shapes that are not standard, often involving curves or irregular boundaries.

In the classroom exercise, we have a triangular region creating a solid when revolved around an axis. Instead of using a predetermined formula, we employ calculus to deduce the volume of this non-standard shape. The disk method falls under this branch of calculus, serving as a strategic approach to handle such complex volumes by breaking the solid into infinitesimally thin slices—disks in this case—and summing their individual contributions through integration.

Integral Calculus

Integral calculus is the branch of calculus focused on the concept of an integral, which allows us to calculate things like areas under curves and, by extension, the volumes of solids. Essentially, an integral sums an infinite number of infinitesimally small quantities to determine a total.

Connection to Solids:

When applied to solid geometry, integral calculus lets us add up the countless small elements that contribute to a solid's volume. This provides a powerful tool for capturing the entirety of three-dimensional space occupied by complex shapes. In our exercise, integral calculus pairs with the disk method to perform the summation of volumes of infinite, thin disks, each corresponding to a segment of the function's domain, detailing the total volume of the solid of revolution.

Volumes by Integration

Calculating volumes by integration is a direct application of integral calculus to three-dimensional space. This technique is particularly useful for finding the volume of a solid that is difficult to describe using simple geometric formulas.

In practice, this involves setting up an integral or a series of integrals that represent the volume of infinitesimally thin cross-sections of the solid. When we rotate a region around an axis, the cross-sections become disks or washers, and integrating these across the bounds of the region yields the total volume.

The approach used in our exercise showcases this method well. By integrating the squares of the radius functions from each portion of the triangle, considering it separately due to the difference in boundaries, we originate a comprehensive understanding of how calculus enables us to decode the complexities of three-dimensional shapes and their volumes.