13 February 2012
Solids of Revolution
Graphs are often used to obtain a deeper understanding of a wide variety of topics. These can range from graphs of the speed of a vehicle over time to graphs of the rate of return on an investment through the course of its lifetime. In both of these cases, graphing a function of the situation would allow one to see how these variables change over time. Furthermore, the area “underneath the curve” can be calculated to determine the total distance traveled in the case of the vehicle and the total return on the investment in the second case. This area is referred to as the definite integral. Through the use of calculus, the definite integral can be determined, which can offer more insight into practically any scenario that can be portrayed using a function and a graph.
In order to determine the area under a curve, the function that represents that curve must be known. Once the function is determined, various methods may be employed to find the area underneath the curve, including Riemann Sums and Simpson’s Rule. However, the most accurate method of calculating the area is through the use of the definite integral. This determines the area by dividing it up into a practically infinite number of very thin rectangles, calculating the area of each rectangle, and then adding up all of these individual areas to obtain a total area, which is the area under the curve. The proper calculus notation for the definite integral is , where f(x) represents the function in question and the height of the rectangular sections, a represents the lower bound on the x-axis, b represents the upper bound on the x-axis, and dx stands for the change in x and the width of the rectangles.
Figure 1. Area Underneath a Curve
Figure 1 effectively demonstrates just what is meant by the definite integral. The blue curve is the function in question, f(x); the shaded region represents the area that is being calculated, ; and the two vertical lines on either side of this region are the lower and upper bounds of the region, a and b, respectively.
Figure 2. Area Between Two Curves
The definite integral can also be used to determine the area between two curves. Figure 2 shows the area underneath the graphs of two functions and the area between these two functions. As seen in the figure, the area between the two is essentially the difference between the area under the upper curve and the area under the lower curve. In order to determine the area between these two functions, the points at which the functions intersect must be determined because these will be the lower and upper bounds of the definite integral. This is done by setting the two equations equal to each other and solving for x. In this scenario, the lower and upper bounds come out to be 0 and 9, respectively. All that remains now is to determine what to plug in for f(x). It is known that f(x) is representative of the height of the rectangular sections that the overall area is comprised of. Thus, the difference between the upper curve, y=x1/2, and the lower curve, y=x/3, must be plugged in for f(x).
Figure 3. Calculations for Area Between y=x1/2 and y=x/3
Figure 3 shows how y=x1/2 was substituted for the upper curve and y=x/3 was substituted for the lower curve. The final area was found to be 4.5 units2.
Areas underneath and between functions, such as the aforementioned one, can be revolved around an axis to create a three dimensional solid of revolution. The definite integral can then be applied to calculate the volume of this solid. This can be accomplished using one of three different methods, shell integration, disk integration, and ring integration.
Figure 4. Visualization of Shell Integration
Figure 4 shows the first of these three, the shell method. This method involves “slicing” the solid into several “shells” of a cylinder. These shells can be compared to the label on a soup can. When the label is removed from the can, it resembles a rectangle, just as each shell in the figure would represent a rectangle once unwrapped. The area of each rectangle is determined by multiplying its width, the circumference of the circular face of the soup can, by its height, the height of the soup can. Since the circumference of a circle is equal to twice the radius times pi, the area of each shell comes out to be 2*π*r*h. The letter r stands for the variable x and the letter h represents the variable y, when revolving around the y-axis. However, when rotating around the x-axis, the letter r stands for the variable y and the letter h represents the variable x. The area of all of the shells are calculated and summed up to determine the volume of the solid. This is represented by the definite integral when the figure is revolved around the y-axis, because it is being integrated with respect to the variable x in this case. However, the definite integral is used when the figure is revolved around the x-axis, because it is being integrated with respect to the variable y.
Figure 5. Visualization of Disk Integration
The second method of determining the volume of a solid of revolution, the disk method, involves “slicing” the solid into multiple sections, which are referred to as slabs. As seen in Figure 5, these slabs are each in the shape of a circle; however, each of these circles has a different radius. The area of one of these circles is equal to the radius squared times pi. Since the radius corresponds to y when revolving about the y-axis and x when revolving around the x-axis, the area of each circle is π*y2 when revolving around the y-axis and π*x2 when revolving about the x-axis. The area of each circle is found using this formula, and then these individual areas are summated to obtain a total volume. This is represented by the definite integral when the figure is revolved around the x-axis, because it is being integrated with respect to the variable x in this case. However, the definite integral is used when the figure is revolved around the y-axis, because it is being integrated with respect to the variable y.
The third and final manner in which the volume of a solid of revolution may be calculated is the ring method, which is very similar to the disk method. The main difference here is that the disk method applies when there is a hole going through the solid of interest.
Figure 6. Visualization of Ring Integration
As seen above in Figure 6, the area that is being revolved is comparable to the area displayed in Figure 2; it is essentially the area between two different curves. Therefore, the integral used in the case of disk integration may be modified to reflect the hole in the figure. Since there are two radii that must be accounted for, the integral comes out to be , where R corresponds to the upper curve, y1, and r corresponds to the lower curve, y2. Thus, the integral may be expressed as when the figure is revolved around the x-axis, because it is being integrated with respect to the variable x in this case. Alternatively, it is written as is used when the figure is revolved around the y-axis, because it is being integrated with respect to the variable y. In conclusion, disk integration is to ring integration as the area under a curve is to the area between two curves.
An example of the ring integration method in use may be demonstrated when presented with the task of finding the volume of the solid generated when the region in the first quadrant bounded by the graphs of y=x1/2 and y=x/3 is rotated about the horizontal line y=-2. It may be recalled that this is the same region that was presented in Figure 2 and that a modified version of the definite integral presented in the last section may be used. The aforementioned integral used in the case of ring integration must be altered to reflect the change in the axis of revolution from the standard axis y=0 to the news axis y=-2. The modified integral is , where y1 is the upper curve, y2 is the lower curve, a is the lower bound, and b is the upper bound. The new axis of revolution, y=-2, is subtracted from both of the radii (the upper and lower curves). This results in the further modified definite integral , which equals 98.96 units3.
In addition to the case of solids of revolution, definite integrals may be used to find the volume of a three dimensional figure whose base is the area under a curve and whose cross-sections are some geometric shape.
Figure 7. Visualization of Cross-Section Method
An example of a situation where this method would need to be employed is seen above in Figure 7. A three dimensional figure is depicted whose base is the area under the graph of the function y. Each cross-section of this solid is in the shape of an equilateral triangle. Due to this, each side of the triangle is equal to y. The equation that gives the area of an equilateral triangle is , where s is the length of one side of the triangle. Therefore, the formula for the area of one of the cross-sections depicted in Figure 7 is . Due to the fact that the area of many cross-sections are being added, the integral that would give the volume of the solid is . This method does not just apply to solids whose cross-sections are equilateral triangles, but to various other shapes as well. These include, but are not limited to, squares, rectangles, semicircles and quarter-circles.
In order to find another case in which the cross-section method may be utilized, refer back to the area between the two curves shown in Figure 2. Let R represent this area. Suppose that the region R is the base of a solid, and for this solid, the cross-sections perpendicular to the x-axis are isosceles right triangles with one leg on the base. The volume of this shape can be easily found using the cross-section method. Since isosceles right triangles are being dealt with, it is known that the area of each cross section is equal to half of the length of one leg squared. According to Figure 2, the length of one leg is equal to the difference between the upper curve y1 and the lower curve y2, so this translates to for each isosceles right triangle whose leg is on the base of the solid. Applying the definite integral given in the previous section, the new integral that is obtained is . Note that the two intersections of the two curves were plugged in for the upper and lower bounds. Solving this definite integral gives the answer of 1.35 units3.
It is easily seen that the definite integral has a wide variety of applications which may offer deeper insight into many different situations. Calculus may seem like an abstract, purposeless subject, but it is actually very applicable. The field of mathematics has come very far and continues to expand. The possibilities are literally infinite.
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