Question

In Exercises 35-38, use the cylindrical shell method to find the volume of the solid generated by revolving the region bounded by the curves about the y-axis. $$y=x, \quad y=-x / 2, \quad x=2$$

Short answer

The volume of the solid is \(\frac{61\pi}{18} cubic units\).

Step-by-step solution

Identify the Region

Firstly, the region bounded by the curves \(y=x\), \(y=-x/2\), and \(x=2\) should be identified. This region is a triangle with vertices at (0,0), (2,2) and (2,-1). This region will be revolved around the y-axis to form the solid.

Set Up the Integral

Using the formula for the volume of a cylinder \(\text{volume} = \pi* r^2*h\), we can express the volume of the solid as an integral. Since we're using the cylindrical shell method and revolving about the y-axis, the height of the shell is \(x\) and the radius of the shell is \(y\), yielding a representative volume \(2\pi*y* f(y)\,dy\). Here \(f(y)\) denotes difference in x-coordinates which serve as our height. For the region above the x-axis, the height is given by \(f(y)= x - y = 2 - y\) (isotropic line minus equation \(y=x\)). For the region below x-axis, the height is given by \(f(y) = 2 + (1/2)y\), where \(y<=0\).

Calculate the Integral

We have two parts to calculate, for the part \(y>=0\) we have from 0 to 2; and for part \(y<=0\) we go from 0 to -1. Therefore, we calculate the integral: \(\int_{0}^{2} 2\pi*y * (2 - y) dy + \int_{-1}^{0} 2\pi*y * (2 + (1/2)y) dy\)

Evaluate the Integral

Evaluating the integral, we get \(2 \pi * [\frac{y^2}{2}-\frac{y^3}{3}]_{0}^{2} - \pi *[\frac{-y^2}{2} - \frac{y^3}{6} ]_{-1}^{0}\) which simplifies to \( \frac{8\pi}{3} + \frac{5\pi}{6} = \frac{61\pi}{18}\)

Key concepts

Volume of Solids of Revolution

Understanding the concept of solids of revolution is pivotal for students tackling integral calculus problems involving volume. When a two-dimensional shape is rotated around an axis, the resulting three-dimensional object is called a solid of revolution. Imagine taking a paper cut-out of a shape and spinning it around a pencil; the shape sweeps out a volume in space.
To calculate the volume mathematically, calculus comes into play, allowing us to sum an infinite number of infinitesimally thin slices of the solid. In our exercise, the region is a triangle, and when it is revolved around the y-axis, it forms a solid that has different radii at different points along its height, which is why the cylindrical shell method is so effective here.
Furthermore, the exercise can be improved by visualizing the problem. Drawing the region and the resulting solid can provide a clearer understanding of what the volume actually represents. Moreover, ensuring that the formula for the volume of a cylinder \( \text{volume} = \text{π} r^2 h \) is well understood will facilitate the comprehension of how that translates into the integrals used in the solution.

Integral Calculus

Integral calculus is one of the two principal branches of calculus, the other being differential calculus. It focuses on the accumulation of quantities, such as areas under curves, and in our context, volumes of solids. The integration process involves finding a function whose derivative matches the original function and can be thought of as the inverse operation of taking a derivative.
In our example, evaluating the integral is essential to find the volume. When we set up the integral, what we're essentially doing is summing up all the infinitesimally small shells to find the total volume. Practicing setting up integral expressions from geometric situations is a critical skill for mastering this part of calculus. Breaking down the problem into understandable parts, such as identifying the radius and height in terms of variables, can make the process of setting up the integral more intuitive.

Volume by Cylindrical Shells

The cylindrical shell method is particularly handy when the axis of rotation is perpendicular to the axis we're integrating along. Visualizing the solid of revolution in terms of concentric cylindrical shells, instead of disks or washers, can simplify the process dramatically. The volume of each cylindrical shell is given by the formula \( 2\text{π} y f(y) dy \) where \( y \) is the radius of the shell, and \( f(y) \) is the height.
In our exercise, we differentiate between the region above and below the x-axis, which is crucial since it affects how the height function \( f(y) \) is defined. An essential piece of advice is to ensure that students can identify how to set up these height functions correctly. This involves understanding the geometry of the region and how it changes with respect to \( y \)—a step that requires solid visualization skills and a good grasp on how to reflect those visualizations in the integral setup.