Jordan 1 Sydni Jordan Mr. Acre AP Calculus 23 February 2015 Solids of Integration Integration is just one of the few central topics that make up the branch of mathematics we call Calculus. It is a very broad term considering that there are numerous applications of topic โ such as reverse differentiation and summation. Although integration is used in many contexts and math problems, this discussion will focus primarily on the use of integration in the calculations of area and volume of graphed regions and solids. Definite integration is a straightforward method that is used to calculate the area under a curve on an interval. To begin, it must be understood that when the equation of a function is graphed, there exists a space between the x-axis and said function, f(x). It is the area of a portion of this region that the definite integral calculates. The shaded area in Figure 1 shows this region under the curve of a sample equation, ๐ฆ = โ๐ฅ. Jordan 2 Figure 1. The area under the curve of the sample graph. When calculating the integral of a function, an interval is often included to place upper and lower boundaries on the designated portion of the area. The x-values that serve as this boundary are notated as b and a respectively. In the sample graph from Figure 1, these lower and upper boundaries are set at approximately 0 and 4, indicating that an integral would evaluate all of the x values within this closed interval of the sample function. The calculus notation of the definite integral function is as follows: ๐ โซ ๐(๐ฅ) ๐๐ฅ ๐ When the equation of the function and its lower and upper x-values are inserted into the placeholders of the integral function, the area that is bound by the function and the xaxis on that designated interval is computed. However, there is more to it than simply plugging an equation into a formula and coming up with a numerical value. The symbol, dx, emphasizes the change in x along the designated interval. To better understand, imagine slicing the graph into vertical strips that are equal in width. The symbol, dx, Jordan 3 represents the width of the slices and, therefore, signifies the change in the x values along the x-axis (see Figure 2). Figure 2. The changes in x labeled on the sample graph Because the definite integral function includes the dx and is evaluated on an interval, it is the sum of all of the areas between a and b. Furthermore, the notation for the definite integral of the sample equation is as follows: 4 โซ โ๐ฅ ๐๐ฅ 0 Definite integrals calculate not only the area under one curve, but also the area between two curves. The method used to solve for this area is very similar to the first approach. To begin, there needs to be a pair of graphed functions. It is essential that the two equations are graphed because the upper and lower boundaries for the definite integral are defaulted to where the two graphs intersect each other. For the sample Jordan 4 ๐ฅ equations, ๐ฆ = โ๐ฅ and ๐ฆ = 3, the intersection points are located at (0, 0) and (9, 3). See Figure 3 for the graph of these two functions. Figure 3. The area between two curves. To find the area between these specific functions, the lower and upper boundaries for the x values will be set to 0 and 9, respectively, because the shaded area in Figure 3 lies within the closed interval between those points. To solve for the area between any two curves, the equation of the lower lying function must be subtracted from that of the upper function. This is the equivalent of subtracting the region under the lower curve from the total region to solve for the area of the space that lies between the two. In other words, the equation ๐ฆ = ๐ฅ 3 must be subtracted from the equation ๐ฆ = โ๐ฅ to create one function to insert into the f(x) portion of the definite integral. Furthermore, the notation and calculation of the definite integral is as follows: Jordan 5 9 โซ โ๐ฅ โ 0 ๐ฅ ๐๐ฅ 3 = 9 = 4.5 2 When the placeholders of the integral are replaced with the appropriate values and functions, the area of the region between the two curves is computed as 4.5 units 2. The power of definite integrals transcends the simple ability to calculate areas under and between graphs. With integrals it is possible to rotate the regions under curves and not only make transform into 3 dimensional shapes, but also find the volumes of said shapes. Three methods โ disk, ring, and shell โ are commonly applied when performing such a task. The first method โ the disk method โ is typically a mode of volume calculation for solids with dx or dy cuts that are perpendicular to their axes of rotation. In other words, the disk method is customarily used when the solid that the equation creates either revolves around the x-axis and uses the dx symbol to signify a change in x values or revolves around the y-axis and uses the dy symbol mark changes in y values. See Figure 4 for an example. Jordan 6 Figure 4. Graph of perpendicular dx cuts and axis Figure 4 shows that the lines designating the dx cuts on the graph of the sample equation, ๐ฆ = โ๐ฅ, are perpendicular to the x-axis, the axis of rotation. This is a situation in which the disk method would be used to find the volume of solid of revolution. In the disk method, each sectioned off slab created by the dx or dy cut โ in this case, dx โ forms a circular disk. By finding the volume of each of the disks and adding them, the definite integral calculates the volume of the entire solid. Figure 5 models a singular disk in the solid of the rotated sample equation. Jordan 7 Figure 5. A disk in the solid of revolution. Because the disk is circular, the equation to find its volume is equivalent to that of a cylinder, ๐๐ 2 โ, where r is the radius of the disk and h is the width. To calculate the volume of a single disk, the r value must be replaced with the equation of the function that creates the solid and the designated cut for the solid must take the place of the h symbol. For the sample, the equation is ๐(โ๐ฅ)2 ๐๐ฅ. Because the integral is the sum of all of these disk volumes on an interval, its equation โ and the integral of the sample- is as follows: ๐ โซ ๐๐ 2 โ ๐ ๐ ๐๐ โซ ๐(โ๐ฅ)2 ๐๐ฅ ๐ Jordan 8 With definite integration through the disk method, the volume of the solid of revolution may be computed within a designated interval โ hence the upper and lower boundaries indicated by the a and b. The second method โ the ring method โ is much like the disk, however it is typically applied in situations where the area between two curves is rotated around an axis to form a solid. In the ring method, there are two equations which causes the disk, once sliced, to have a hole in the center, causing the shape to resemble a ring. See Figure 6 for a depiction of the ring that is created. Figure 6. A ring in the solid of revolution. Jordan 9 Because the ring is also circular, the equation to find its volume is much like that of the cylinder in the disk method, however the volume of the inner circle that is created must be subtracted to compute an accurate value. Therefore the equation for the volume of a singular ring is ๐๐ 2 โ โ ๐๐ 2 โ, where R represents the radius of the outer circle and r stands for the radius of the inner circle. When applying this equation to a definite integral, the equation of the upper lying equation must be substituted for the symbol R while that of the lower equation must be substituted for the r. The upper and lower boundaries for the interval should be where the two graphs intersect as well. Much like in the disk method, the h ought to be replaced by the dx or dy cut that is being used to slice the solid. The equation for the integral that computes the volume is as follows: ๐ โซ ๐๐ 2 โ โ ๐๐ 2 โ ๐ However, it is also feasible to rotate such a region about an axis that is not the x(or y-) axis. To complete this task, a minor adjustment must be made to the equation of the definite integral. When the axis is below the rotating region, the difference between that axis and the x- (or y-) axis must be added onto the radius of the circles. For ๐ฅ example, if the region bounded by the sample equations, ๐ฆ = โ๐ฅ and ๐ฆ = 3, is rotated about the horizontal line ๐ฆ = โ2 instead of the x-axis, 2 units must be added onto the radii of the inner and outer circles of the ring. The equation for the definite integral would then be as follows: 9 2 ๐ฅ โซ ๐(โ๐ฅ + 2)2 ๐๐ฅ โ ๐ ( + 2) ๐๐ฅ 3 0 = 31.5๐ โ 98.96 Jordan 10 Again, the upper and lower boundaries are set to 0 and 9 because these are the x values that correlate with the points in which the two graphs intersect. The volume of the solid of revolution created by the rotated region between the two curves is computed as approximately 98.96 units3. The third method โ the shell method โ does not quite follow the patterns of disk and ring. It is typically a mode of volume calculation for solids with dx or dy cuts that are parallel to their axes of rotation. In other words, the shell method is customarily used when the solid that the equation creates either revolves around the x-axis and uses the dy symbol to signify a change in x values or revolves around the y-axis and uses the dx symbol mark changes in y values. The sections in the shell method are best understood if thought of as a tube or hollow cylinder that is cut along the side to lay flat. See Figure 7 for an illustration of a shell section. Figure 7. A shell in the solid of revolution. Jordan 11 The equation to find the volume of a singular shell like the one shown in Figure 7 is 2๐(๐)(โ๐๐๐โ๐ก)(๐ค๐๐๐กโ), where the r represents the radius of the shell. To compute the volume, the equations of the functions should be inserted into the equation for the height of the shell and the corresponding dx or dy should replace the width. Like the previous two methods, the equation of the lower function should be subtracted from that of the upper as it is incorporated into the volume equation as the height of the shell. Also, the variable r must be replaced with whatever variable or function that dictates the width of the shell. In the case of the two sample equations, this variable is simply x. The equation of the definite integral that evaluates the total volume of this solid of revolution created by the two sample functions is as follows: 9 ๐ฅ โซ 2๐(๐ฅ) (โ๐ฅ โ ) ๐๐ฅ 3 0 The disk, ring, and shell methods are but three processes in which the volume of a solid of revolution may be computed. However, there are times when one is required to find the volume of a solid that is not formed by the rotation of a function or two about a line. Such figures take the shapes of the most common of geometrical shapes, such as cubes, wedges, spheres, etc. There is one last technique that has the ability and versatility to calculate the volume of any designated solid. Said technique is called the cross-section method. In this method, the shape of the cross section is dictated by the overall shape of the solid. Moreover, the computed volume may vary because the cross-sectional shape is determined by the mathematician. Jordan 12 Similar to the previous processes, the region under or between two curves may be incorporated into the volume of the solid. In many cases, said region represents one face of the solid. For instance, the area between the two sample equations, ๐ฆ = โ๐ฅ and ๐ฅ ๐ฆ = 3, will serve as the base of a solid made with isosceles right triangles. See Figure 8 for a diagram. Figure 8. Slabs in the cross-section method. In Figure 8, the base legs of the isosceles triangles span the region bounded by the sample functions. The complete solid is composed of multiple isosceles triangles such as those depicted in the figure above. Each triangle has varying lengths for their legs because the distance between the two functions vary along the interval. The integral, like in the previous methods of volume calculation, is the sum of all of these crosssectional triangles. Therefore, to find the volume of the complete solid formed by the equations, the volumes of each and every triangle along this interval must be compounded into one value. Jordan 13 To calculate this value for any singular cross-section, the formula for the volume of the shape of that cross section should be considered. For example, the volume equation for a triangle would be best to use in the sample problem. This equation is as follows: 1 ๐๐๐๐ข๐๐ = (๐๐๐ ๐)(โ๐๐โ๐ก)(๐ค๐๐๐กโ) 2 Because the slabs used for the cross-sections are isosceles triangles, the lengths for both legs are congruent and, therefore, simply squaring the value of the base will be a more straightforward approach. The input value for the base of the triangle correlates to the region of the graph ๐ฅ that lies between the two curves, ๐ฆ = โ๐ฅ and ๐ฆ = 3. However, because the length of this area varies the equations of the functions will be substituted into the volume function instead. The dx or dy cut should fill the placeholder for width as well. Much like the integrals of the past methods of volume calculation, the lower and upper boundaries for the interval are set to 0 and 9 respectively. The equation and solution for this integral is as follows: 9 โซ 0 1 ๐ฅ (โ๐ฅ โ )2 ๐๐ฅ 2 3 = 1.35 The value correlating to the volume of the entire solid formed by the region between the two sample curves is computed as approximately 1.35 units3. To conclude, there are multiple methods and approaches to use when calculating the area or volume of areas under and between curves. From disk to ring to shell to Jordan 14 cross-section, the only necessities for accurate computations of these values are the equations themselves and the knowledge behind the application. None of the aforementioned methods, however, would have been useful without definite integration, and yet these are still but a few applications of the function. Jordan 15 Works Cited "Definite Integral." Wolfram Alpha: Computational Knowledge Engine. WolframAlpha, n.d. Web. 20 Feb. 2015. <http://www.wolframalpha.com/input/?i=%28sqrt%28x%29%2B2%29%2B%2B%28x%2F3%2B%2B2%29>. Foerster, Paul A. "Volume of a Solid by Plane Slicing." Calculus: Concepts and Applications. Berkeley, CA: Key Curriculum, 1998. 242-51. Print. Y=sqrt(x). Digital image. WolframAplha: Computational Knowledge Engine. WolframAlpha, n.d. Web. 20 Feb. 2015. <http://www.wolframalpha.com/input/?i=y=sqrt(x)&lk=4&num=1>.