Problem 48 - Essential Calculus

Question

The integral represents the volume of a solid of revolution. Identify (a) the plane region that is revolved and (b) the axis of revolution. $$ 2 \pi \int_{0}^{1}\left(y-y^{3 / 2}\right) d y $$

Step-by-Step Solution

Answer

The plane region that is being revolved is the area bound by the functions \(y\) and \(y^{3 / 2}\) from 0 to 1 on the y-axis. The axis of revolution is the y-axis.

Step 1: Understanding the integral

First, interpret the integral. The integral should be viewed as an accumulation of infinitesimal volumetric disc elements along the y-axis. The limits of the integral, 0 and 1, designate the interval on the y-axis where the shape lies.

Step 2: Identifying the plane region

Generate the plane region. The plane region is the area bound by the functions \(y\) and \(y^{3 / 2}\) from 0 to 1 on the y-axis. To visualize this, it may be helpful to plot these two functions on the same graph. The function \(y\) is just a straight line from the origin, and the function \(y^{3 / 2}\) is a curve.

Step 3: Identifying the axis of revolution

The quantity under the integral shows the radius of the discs as a function of y, indicating that the axis of revolution lies along the y-axis. The axis of revolution is therefore the y-axis.

Key Concepts

Calculus

Calculus is a branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series. It’s divided into two major areas: differential calculus, concerning rates of change and slopes of curves, and integral calculus, which focuses on the accumulation of quantities and the areas under and between curves.

For students learning calculus, it’s essential to understand how these concepts are applied to solve problems. In our exercise, integral calculus is used to determine the volume of a solid of revolution. This calculation involves taking infinitesimally small discs or rings, and stacking them together to find the solid's total volume.

Integral Calculus

Integral calculus is the study of integrals and their properties. It deals with the summation of continuous quantities, which can be areas, volumes, or other quantities described by a function. When you integrate a function over a given interval, you're effectively computing the total accumulated value between two points.

The integral given in the exercise, \( 2 \pi \int_{0}^{1}\left(y-y^{3 / 2}\right) dy \), is an application of integral calculus to calculate the volume of a solid that has been created by revolving a plane region around an axis. Each infinitesimal part of the accumulated value represents a tiny volume of the solid.

Volumes of Solids

In calculus, one common application is finding the volumes of solids with known cross sections. These cross sections can be squares, rectangles, triangles, or other shapes. A solid of revolution, which is a solid obtained by rotating a plane curve about an axis, is a special case where the cross sections are circular discs or washers.

The integral given represents the volume of a solid of revolution where the cross-sectional area varies with the radius described by the function \( y - y^{3 / 2} \). Using the method of discs or washers, each circular slice's volume is added up over the interval to get the total volume of the solid.

Axis of Revolution

The axis of revolution in a solid of revolution is the line around which the plane region is rotated. This axis can be horizontal, vertical, or any axis in three-dimensional space. It's crucial to identify correctly because it influences the shape and volume of the solid produced.

In our exercise example, the axis of revolution is along the y-axis since the radius of the discs depends on the value of y, which is established through the integral. The resulting solid, therefore, has a varying radius that changes according to the given functions as you move along the y-axis.