Question

Think About It In Exercises 39 and \(40,\) determine which value best approximates the volume of the solid generated by revolving the region bounded by the graphs of the equations about the \(y\) -axis. (Make your selection on the basis of a sketch of the solid and not by performing any calculations.) \(y=2 e^{-x}, y=0, x=0, x=2\) (a) \(\frac{3}{2}\) (b) -2 (c) 4 (d) 7.5 (e) 15

Short answer

The volume of the solid is best approximated by (a) \(\frac{3}{2}\).

Step-by-step solution

Understand the given equations

The given equations include \(y=2 e^{-x}\), which is an exponential decay function, \(y=0\) which is the x-axis, and \(x=0\) and \(x=2\) which are vertical lines along the y-axis. These equations bound our region.

Sketch the bounded region and the solid

First plot the primary function \(y=2 e^{-x}\). It starts at \(y=2\) when \(x=0\) and approaches \(y=0\) as \(x\) increases. After identifying the boundary lines \(y=0, x=0\), and \(x=2\), shade the bounded region. Now, imagine this region revolved around the y-axis, forming a three-dimensional solid.

Approximate the volume of the solid

The solid looks like a right circular cone that is constructed by rotating the region around the y-axis. Its dimensions are approximately a height of 2 and a radius of 2. The volume \(V\) of a cone is given by \(V= \frac{1}{3} \pi r^2 h\). Since it appears that the radius and height are equal (\(r=h=2)\), we could guess that the volume would be around \(\frac{1}{3} \pi 2^2 \times 2 = \frac{8}{3} \pi\), which is roughly 8.377. However, the profile of the solid is not exactly a triangle due to the curvature of the \(y=2 e^{-x}\) part. The volume is therefore likely to be less than that of the cone with equivalent dimensions. Looking at the answer choices, option (a) \(\frac{3}{2}\) seems to be the best fit.

Key concepts

Exponential Decay Function

One of the primary concepts in our exercise is the exponential decay function, expressed as y=2e^{-x}. This type of function is pervasive in fields such as physics, chemistry, and economics, and it describes a quantity that decreases over time at a rate proportional to its current value. For our purposes, it's important to note that as x increases, the value of y will decrease rapidly at first and then more slowly, approaching but never quite reaching zero. This behavior is the 'decay' part, where the function is continually halved over a certain period – this period is called the 'half-life'.

In a graphical representation, the exponential decay function forms a curve that starts from a certain y-value (the initial value) and curves downwards towards the x-axis as x increases. Understanding the nature of this function helps students imagine how the solid of revolution will look when the area under this curve is rotated around an axis.

Solid Generated by Revolving a Region

The second key concept is the solid generated by revolving a region. In geometric terms, when we revolve a two-dimensional region around an axis, we generate a three-dimensional solid known as a solid of revolution. This is akin to spinning a shape around a spindle on a lathe to create a symmetrical object. In the case of our exercise, the region bounded by the exponential decay function, the x-axis, and the given vertical lines x=0 and x=2, form the two-dimensional area that, when revolved around the y-axis, creates a three-dimensional object.

Visualizing the Solid

To visualize this solid, imagine a piece of paper with a graph drawn on it. If you roll this paper into a tube such that the two vertical edges touched (without overlapping), you'd have a rough idea of what the solid would look like: the curves in the graph become the 'walls' of the solid, giving it its unique shape. It is often helpful for students to sketch this bounded region before imagining the revolution, as it aids in visualizing the final solid shape and makes the subsequent steps of the volume approximation more intuitive.

Volume Approximation

Lastly, the concept of volume approximation is used to find an estimate for the volume of irregularly shaped objects, such as the solid generated in our problem. There are various methods to approximate the volume of a solid of revolution, with one of the simpler ways being to compare it to a solid with a known formula, such as a cylinder or cone, as done in the exercise solution.

Approaching the Approximation

For objects with a relatively simple structure, we can sometimes use formulas for regular shapes as a starting point for approximation. For instance, if a solid looks similar to a cone, we could estimate its volume using the formula for a cone's volume, V = (1/3)πr^2h, where r is the radius of the base and h is the height. By comparing the actual shape to the regular one, adjustments can be made to better suit its dimensions. It is crucial to remember that approximation by its nature implies some level of inaccuracy, and we are aiming for the best possible estimate with our chosen method. For complex shapes, other methods such as the disk method or the shell method may be more suitable to achieve a closer approximation.