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Thanvir Ahmed Mr. Scot AcreAP Calculus 22 February 2016Solids of Revolution Calculating the Area Under and Between Curves:Integrals and derivatives serve as the building blocks for calculus. Arguably, integrals and derivatives are two of the most fundamental objects to all of calculus. Integrals are also referred to as antiderivatives or primitives (Integral). An integral is essentially the area under any given curve. The concepts of calculus, were discovered independently by Sir Isaac Newton and Gottfried Leibniz (Simmons). Sir Isaac Newton, the man who is most often referred to as the Father of Physics, demonstrated how these concepts can be applied to physics. For example, integrating an acceleration vs time graph will calculate the velocity. Likewise, integrating a velocity vs time graph will calculate the displacement or distance. Solving physics problems is not the only function of integrals. Integrals, definite integrals in particular, can be used to compute the surface area of different shapes and volumes of solids. The two types of integrals are definite and indefinite. A definite integral is an integral that has a starting and an end point, meaning that it is constrained and its numerical value can be calculated. However, an indefinite integral is an integral that does not have a starting and an end point, meaning that its numerical value cannot be calculated. The definite integral computes the area under a curve by cutting the bounded region into infinitely many rectangular sections with extremely small widths (the width approaches, but is not equal to zero). The area of each individual rectangular section is calculated using the rectangle area formula, which is simply the length multiplied by the width. These individual rectangles are then added together in order to compute the area of the definite integral. The definite integral is able to give an exact value, even though the graph may be curved or oddly shaped, by having an infinite amount of rectangles. The more rectangles that are used in the integral, the closer the left165100000value is to the exact value. Figure 1. Indefinite and Definite IntegralsFigure 1 shows the general notation of both an indefinite integral (left) and a definite integral (right). In the notation for the definite integral, the variables a and b represent the lower and upper bounds of the integral. This means to integrate from a to b. The indefinite integral lacks this constraint. The dx refers to the type of cut that is used to calculate the widths of the rectangles that are used in both the definite and indefinite integral. Likewise, the f(x) refers to the lengths of the rectangles. The definite integral above is read as “the integral of f(x) from a to b with respect to x”. A dx cut is perpendicular to the x-axis. Similarly, the integral can be taken with respect to the y-axis using a dy cut, which is perpendicular to the y-axis. To use a dy cut, the function must be manipulated so that it is in terms of y. The area between two curves is calculated by subtracting the smaller area from the larger one. Subtracting the integrals of the curves from one another eliminates the overlapping areas, leaving only the bounded area between the two curves. The following figures demonstrate how to calculate the area under individual curves, as well as the area between two curves. The first problem asks to calculate the area between two curves, R. The two functions that are being used for this problem are y= x and y= x3. 27336751762125bbleft1714500bb47625001695450dxdx33718501685925aa49434751276350666751695450aa3295650-952500-95250008382001714500dxdx10191751295400Figure 2. Integrals of Both Functions Figure 2 shows the first step in order to calculate the area between the two functions. It is not necessary to graph the two functions against each other, but it serves as a visual aid. Both functions are shown with the dx cut, and one of the infinitely many rectangles that are subsequently used in order to calculate the integral. The a and b are just placed at 0 and 10 for both of these, these a and b values are used in these graphs to visually represent an integral. For the area between these functions the a and b values become the x values for the intersections points of the two functions. Since these are both definite integrals, they are both in the abfxdx standard notation, with the corresponding functions for f(x) and the a and b values. left12700001066800299085RR95250342265aa2624455365760bbFigure 3. Area Between Two Curves Figure 3 shows the two different functions graphed together. By looking at the graph alone, the y= x function has more area than the y= x3 function for the portion in which they overlap. The intersection points of these two functions are (0,0) and (9,3). This means that 0 is the lower bound for the integral, or the a value, and 9 is the upper bound, or the b value. Since the integral of the y= x3 funtion is smaller for the domain of 0 to 9, it is subtracted from the integral of y= x from 0 to 9 in order to calculate the area between the two curves. This area is visually represented as the region between the two curves, labeled as R. The calculation for R is shown in the following figure. Area Between Curves (R)=abf1xdx- abf2xdx R=09x dx- 09x3dxR=4.5 Units2 Figure 4. Integral Setup and Calculation for AreaFigure 4 shows the setup for the integral used to compute the area of R. The area between the two curves is calculated by subtracting the smaller integral from the larger one for the given domain. This domain is determined by the upper and lower bounds, which are essentially just the intersection points of the two curves, as shown in in Figure 3. After the setup is complete, the area can be computed by using the Fundamental Theorem of Calculus, or a calculator. The area between the two curves, or R, is calculated to equal 4.5 Units2. Calculating Volumes of Solids:Besides calculating area, integrals can also be used to calculate the volumes of different solids. They are specifically used to calculate the volumes of three-dimensionally revolved solids. The three different methods used to do this are the disk method, the ring method, and the shell method. The different methods are used in different scenarios, and have requirements that must be met before calculating the volume of the specified solid. The different methods are further explained below. The Disk and Ring Method: 1905096520 Figure 5. Disk MethodFigure 5 is a visual representation of the disk method. By name, the disk method indicates that it uses disks to calculate the volume. This method forms a solid by using infinitely many disks with small heights. The heights approach zero, but are not equal to zero. It is important to note that the disk method can only be used if the cuts are perpendicular to the axis of rotation. A disk is essentially a cylinder, and consequently uses the same formula to calculate volume, which is V= π × radius2 × height. In this representation, the integral was calculated using a dx cut. This means that dx acts as the height of the disk. The radius in this case is the y-value. This formula only computes the volume for one disk. In order to calculate the volume of the entire solid, an integral must be used. The integral to be used for the disk method is shown below. abπ(r2)dx Figure 6. Volume Formula Setup for Disk Methodleft2138680OuterInnerOuterInnerFigure 6 shows the setup that is used to calculate the volume using the disk method. The volume formula being used is V= π × radius2 × height, which is just the volume for a cylinder, as disks are essentially cylinders with very small heights. The height is dx, and the radius, or r, is just y value or the function in terms of x. So in order to calculate the volume of the solid shown in Figure 5, r would be substituted for 3x. If there is a dy cut instead, then the radius would have to be in terms of y as well. Figure 7. Ring MethodFigure 7 is a visual representation of the disk method, also known as the washer method. The picture on the left shows that the solid consists of an outer and an inner solid. The inner solid is subtracted from the outer solid in order to calculate the volume. This is essentially using the disk method twice, and subtracting. There are infinitely many rings. These rings are the result of subtracting the inner solid from the outer one. The same formula is used for the disk method is once again used to for the ring method, the only difference being that it is used twice for the ring method, and the smaller volume is subtracted from the larger volume. 9810753209290dxdx31638393146425(9,3)(9,3)-2000253265805The second problem asks to calculate the volume of a revolved solid. This particular solid is formed by revolving the area between the two functions y= x and y= x3 around the line y=-2. This area, R, was graphed in Figure 3, and calculated to equal 4.5 Units2 in Figure 4. The first step to calculate the volume of this solid is to first graph the different functions. This step has already been completed in Figure 3. Similar to the method that was used to calculate area, the y= x function has the larger volume for the domain in which the functions overlap. This results in y= x being the outer solid, and y= x3 being the inner solid. In order to calculate the volume of the revolved area between the two functions, the volumes of the inner solid must be subtracted from that of the outer solid. -20002566675(0,0)(0,0)4067175208915y = -2y = -216192475565Figure 8. Solid Model Figure 8 shows the solid that is created using the disk method. The area between the two curves, R, is rotated around the line y=-2. The intersection points of the two curves are (0,0) and (9,3). Consequently, the lower bound or a value is 0, and the upper bound or b value is 9. Volume of Solid (V)=abπf1x2dx- abπf2x2dx V=09πx +22 dx- 09πx3+22dxV=98.96 Units3 Figure 9. Integral Setup and Calculation for VolumeFigure 9 shows the setup of the integral that was used to calculate the volume of the revolved solid in Problem 2. Since there is a dx cut being used, the radii must be expressed in terms of x. Both radii must be adjusted, as the axis of rotation is below the x-axis, at y=-2. Adding 2 to each radius changes this to the x-axis, and allows the volume to be calculated. After the setup is complete, the volume can be computed by using the Fundamental Theorem of Calculus, or a calculator. The volume of the solid formed by revolving the area between the two functions is calculated to equal 98.96 Units3. The Shell Method: When the cut made is parallel to the desired axis of rotation, there are two options. The first option is to rearrange the function so the cut can be made perpendicular to the axis of rotation, and then use the disk method to calculate the volume. The second option is to calculate the volume by using cylindrical shells, this is known as the shell method. For the shell method, the cuts must be parallel to the axis of rotation. Figure 10. Volume by Cylindrical ShellsFigure 10 shows an example of a cylindrical shell. The cuts are made parallel to the axis of rotation. Meaning that there are there are dx cuts being rotated around the y-axis. Each shell is hollow, so the volume of the shell is calculated similar to a rectangle. When the shell is rolled out, the length is the radius r multiplied by 2π. The radius is the distance from any given x to the axis of rotation. The height is the height of the function, in this case just y or f(x). The shell also has a thickness of dx. To calculate the volume of the solid, all of the shells must be summed together. The formula is shown in the following figure. Volume of Solid (V)=ab(2×π×r×h) dxFigure 11. Volume Formula for Shell MethodFigure 11 shows the formula used to calculate the volume of a solid using the shell method. In the case of the solid shown in Figure 10, the radius, r, can be substituted for x. The height is changed to y or f(x) for the solid in Figure 10. Everything must be in terms of x, as there is a dx cut. For the shell method, the cut must be parallel to the axis of rotation.The Cross Section Method:Another method that can be used to calculate the volume of a solid is the cross section method. The cross section method does not find the volume of a solid of revolution. It is used when a solid is composed of cross sections of similar shape. The cross section can be of any shape such as squares, rectangles, triangles, etc. In order to use the cross section method, the shapes must be perpendicular to the x-axis or y-axis. The definite integral is also used to calculate these volumes. The cross section method thinly slices the object into an infinite number of cross sections or slabs. The volume of each cross section is found and added up to find the total volume.29718007747000Figure 12. Volume by Cross Sectional AreaFigure 12 shows two diagrams of a solids with cross sections of squares and cross sections of semicircles. Many other cross sections could be used, such as triangles or trapezoids. The area of each cross section is found with a thickness of dx or dy, depending on the scenario. This works because the cross sections are proportional to one another. No matter the cross section, each is solved the same way. The steps necessary to calculate the volume of solids using the cross section method is shown in the following figure. Volume of Solid (V)= abAreaxdxFigure 13. Volume Formula for Cross Section MethodFigure 13 shows the formula used to calculate the volume of a solid by using the cross section method. In this equation, Area(x) stands for the area of one of the cross sections. If the cross sections are squares, then Area(x) would equal x2, like the case in Figure 12 with the square cross section. If the cross section is a semicircle, then Area(x) would equal 12πr2, like the case in Figure 12 with the semicircle cross section. With semicircles, r would be substituted for x or y, depending on the scenario. -1809752531110x-x3x-x3dx00x-x3x-x3dxThe third problem asks to calculate the volume of a solid formed by isosceles right triangles. The region R, the area between the two functions y= x and y= x3 graphed in Figure 3, is the base of this solid. For this solid, the cross sections perpendicular to the x-axis are isosceles right triangles with one leg on the base. The first step to calculate the volume of this solid is to first graph the different functions. This step has already been completed in Figure 3. The area of one triangle must be found in order to calculate the total volume. This calculation is shown in the following figures, along with an illustration of the solid itself. Figure 14. Triangular Solid Formed by Cross Sections Figure 14 shows a model of the solid made, as well as the dimensions of each cross section. The area of one cross section is 12(x-x3)2. The a and b values, or lower and upper bounds, are still the same, at 0 and 9. The calculation for the volume of this solid is shown in the following figure.Volume of Solid (V)= abAreaxdxV=0912x-x32dxV=1.35 Units3 Figure 15. Integral Setup and Calculation for Volume Figure 15 shows the setup of the integral that was used to calculate the volume of the solid in Problem 3, shown in Figure 14. Since this is a dx cut being used, the area is in terms of x. After the setup is complete, the volume can be computed by using the Fundamental Theorem of Calculus, or a calculator. The volume of the solid formed by using isosceles right triangles as cross sections perpendicular to the x-axis on the region R is equal to 1.35 Units2.In conclusion, the area between functions can be calculated as well as the volumes of many different three dimensional solids by utilizing integrals. Integrals are one of the most fundamental objects in all of calculus. These integrals work by cutting a figure into an infinite number of thin sections, and then summing them together to get the area, and this area can be rotated in order to form a solid. In the case of finding the area within bounded region, the area of each piece is summed up. Integrals and solids of revolution are not just found in text books, but can be applied to the real world in careers such as engineering and architectural designing, and serve as a staircase to higher levels of calculus and learning. Works CitedDawkins, Paul. "Calculus I - Volumes of Solids of Revolution / Method of Rings." Calculus I - Volumes of Solids of Revolution / Method of Rings. Paul Dawkins, 15 May 2013. Web. 21 Feb. 2016. <, Rafael. Volume by Shells. Digital image.?Calculus I. Bakersfield College, 2 Sept. 2003. Web. 19 Feb. 2016. <;."Integral." Wolfram MathWorld. Wolfram Research, Inc., 5 Feb. 2016. Web. 18 Feb. 2016. <;."MyWorld." Volumes. Atom, 5 Aug. 2012. Web. 21 Feb. 2016. <, G. "History of Calculus." History of Calculus. McGraw-Hill, 10 July 2010. Web. 18 Feb. 2016. <;.“Solids of Cross Sections.” Mesa Community College. Web 6 February. <;. ................
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