Problem 41 - Thomas Calculus in SI Units

Question

Find the volume of the solid generated by revolving each region about the \(y\) -axis. The region enclosed by the triangle with vertices (1,0),(2,1) and (1,1)

Step-by-Step Solution

Answer

The volume is \(\frac{7\pi}{3}\).

Step 1: Understand the Problem

We need to find the volume of the solid formed by revolving the given triangle about the y-axis. The vertices of the triangle are (1,0), (2,1), and (1,1). This forms a right triangle in the xy-plane.

Step 2: Identify the Function and Region

The triangle is bounded by the vertical line x=1, the horizontal line y=1, and the line connecting points (1,0) and (2,1). We can express this line as the linear equation derived from these two points.

Step 3: Find the Equation of the Line

Calculate the slope between (1,0) and (2,1): \(m = \frac{1-0}{2-1} = 1\). Using the point-slope form \(y - y_1 = m(x - x_1)\) with point (1,0), we get \(y = x - 1\).

Step 4: Set Up the Integral for Volume

Use the disk method for rotation around the y-axis. The radius of each disk is \(x\), so we solve for \(x\) from \(y = x - 1\): \(x = y + 1\). The volume \(V\) is given by \( \pi \int_{0}^{1} (y + 1)^2 \, dy \).

Step 5: Integrate to Find the Volume

Calculate the integral: \( \int_{0}^{1} (y + 1)^2 \, dy = \int_{0}^{1} (y^2 + 2y + 1) \, dy = \left[ \frac{y^3}{3} + y^2 + y \right]_{0}^{1} \).

Step 6: Evaluate the Definite Integral

Evaluate the expression: \( \left[ \frac{1^3}{3} + 1^2 + 1 \right] - \left[ \frac{0^3}{3} + 0^2 + 0 \right] = \frac{1}{3} + 1 + 1 = \frac{7}{3} \).

Step 7: Find the Final Volume

Multiply by \(\pi\) to get the volume: \(V = \frac{7\pi}{3}\).

Key Concepts

Disk Method

The disk method is a powerful technique to find the volume of a solid of revolution. It involves slicing the solid into thin, circular disks and summing their volumes. Each disk is formed by revolving a region around an axis. In this method, knowing the radius of these disks is crucial. When revolving around the y-axis, the radius is determined by the x-value at a given y. In this exercise, the function defining our curve was derived as \(y = x - 1\), which we rearranged to express \(x\) in terms of \(y\): \(x = y + 1\). This expression \(x = y + 1\) gives us the radius of the disks. The volume \(V\) is calculated using the integral formula:

• \( V = \pi \int_{a}^{b} [radius]^2 \, dy \)
• In our specific example, the radius became \(y + 1\) when revolving around the y-axis.

To find the complete volume, we integrate over the intersection region from \(y = 0\) to \(y = 1\). This computes the total volume of these infinite disks, stacked side by side.

Definite Integrals

Definite integrals are used to calculate the accumulated sum of quantities along an interval. They provide us with a finite answer, which is golden when determining areas or volumes. In the context of this exercise, the definite integral aids in summing the infinite volume slices from the disks. In our problem, we wanted the integral of \((y + 1)^2\) over \(y\) from 0 to 1. This specific integral represents the collective volume of all disks within our defined region:

• \( \int_{0}^{1} (y + 1)^2 \, dy = \int_{0}^{1} (y^2 + 2y + 1) \, dy \)
• The integrand \((y^2 + 2y + 1)\) arises from the squared radius, \((y + 1)^2\).

This indicates that you calculate the volume across the entire y-interval. Evaluating the integral, we find account for each portion of the volume generated by deriving and computing each integral term. This thorough computation is essential for acquiring an exact volume for our solid.

Revolving Around the y-axis

Revolving shapes around the y-axis forms striking three-dimensional solids. It's a trusty method for creating structures like cones, cylinders, and even doughnut-shaped entities. Understanding how to depict a radius in terms of \(y\) for the rotational axis is essential here.In this exercise, the region bounded by the triangle was rotated around the y-axis, transforming flat boundaries into circular rims. To achieve this, each \(x\)-value became the radius of a disk:

• The line \(y = x - 1\) was transformed to \(x = y + 1\), establishing the radius for each y value.
• Each radius \(x = y + 1\) rotated around the central axis to describe a circular face.

Hence, our triangle, through revolution, was sculpted into a conic-like, elegant shape, anchored around the y-axis. Such insight underscores the conversion of 2D figures into volumetric solids via proper geometric transformations.