Presentation on theme: "APPLICATIONS OF INTEGRATION"— Presentation transcript:
1 APPLICATIONS OF INTEGRATION 6APPLICATIONS OF INTEGRATION
2 6.2 Volumes In this section, we will learn about: APPLICATIONS OF INTEGRATION6.2VolumesIn this section, we will learn about:Using integration to find outthe volume of a solid.
3 In trying to find the volume of a solid, VOLUMESIn trying to find the volume of a solid,we face the same type of problem asin finding areas.
4 We have an intuitive idea of what volume means. VOLUMESWe have an intuitive idea of what volumemeans.However, we must make this idea preciseby using calculus to give an exact definitionof volume.
5 We start with a simple type of solid VOLUMESWe start with a simple type of solidcalled a cylinder or, more precisely,a right cylinder.
6 As illustrated, a cylinder is bounded by CYLINDERSAs illustrated, a cylinder is bounded bya plane region B1, called the base, anda congruent region B2 in a parallel plane.The cylinder consists of all points on line segments perpendicular to the base and join B1 to B2.
7 If the area of the base is A and the height of CYLINDERSIf the area of the base is A and the height ofthe cylinder (the distance from B1 to B2) is h,then the volume V of the cylinder is definedas:V = Ah
8 In particular, if the base is a circle with CYLINDERSIn particular, if the base is a circle withradius r, then the cylinder is a circularcylinder with volume V = πr2h.
9 RECTANGULAR PARALLELEPIPEDS If the base is a rectangle with length l andwidth w, then the cylinder is a rectangular box(also called a rectangular parallelepiped) withvolume V = lwh.
10 For a solid S that isn’t a cylinder, we first IRREGULAR SOLIDSFor a solid S that isn’t a cylinder, we first‘cut’ S into pieces and approximate eachpiece by a cylinder.We estimate the volume of S by adding the volumes of the cylinders.We arrive at the exact volume of S through a limiting process in which the number of pieces becomes large.
11 We start by intersecting S with a plane IRREGULAR SOLIDSWe start by intersecting S with a planeand obtaining a plane region that is calleda cross-section of S.
12 Let A(x) be the area of the cross-section of S IRREGULAR SOLIDSLet A(x) be the area of the cross-section of Sin a plane Px perpendicular to the x-axis andpassing through the point x, where a ≤ x ≤ b.Think of slicing S with a knife through x and computing the area of this slice.
13 The cross-sectional area A(x) will vary as x increases from a to b. IRREGULAR SOLIDSThe cross-sectional area A(x) will varyas x increases from a to b.
14 We divide S into n ‘slabs’ of equal width ∆x IRREGULAR SOLIDSWe divide S into n ‘slabs’ of equal width ∆xusing the planes Px1, Px2, to slice the solid.Think of slicing a loaf of bread.
15 If we choose sample points xi* in [xi - 1, xi], we IRREGULAR SOLIDSIf we choose sample points xi* in [xi - 1, xi], wecan approximate the i th slab Si (the part of Sthat lies between the planes and ) by acylinder with base area A(xi*) and ‘height’ ∆x.
16 The volume of this cylinder is A(xi*). IRREGULAR SOLIDSThe volume of this cylinder is A(xi*).So, an approximation to our intuitiveconception of the volume of the i th slab Siis:
17 Adding the volumes of these slabs, we get an IRREGULAR SOLIDSAdding the volumes of these slabs, we get anapproximation to the total volume (that is,what we think of intuitively as the volume):This approximation appears to become better and better as n → ∞.Think of the slices as becoming thinner and thinner.
18 Therefore, we define the volume as the limit of these sums as n → ∞). IRREGULAR SOLIDSTherefore, we define the volume as the limitof these sums as n → ∞).However, we recognize the limit of Riemannsums as a definite integral and so we havethe following definition.
19 Let S be a solid that lies between x = a and x = b. DEFINITION OF VOLUMELet S be a solid that lies between x = aand x = b.If the cross-sectional area of S in the plane Px,through x and perpendicular to the x-axis,is A(x), where A is a continuous function, thenthe volume of S is:
20 When we use the volume formula , it is important to remember VOLUMESWhen we use the volume formula, it is important to rememberthat A(x) is the area of a movingcross-section obtained by slicing throughx perpendicular to the x-axis.
21 Notice that, for a cylinder, the cross-sectional VOLUMESNotice that, for a cylinder, the cross-sectionalarea is constant: A(x) = A for all x.So, our definition of volume gives:This agrees with the formula V = Ah.
22 Show that the volume of a sphere of radius r is SPHERESExample 1Show that the volume of a sphereof radius r is
23 If we place the sphere so that its center is Example 1If we place the sphere so that its center isat the origin, then the plane Px intersectsthe sphere in a circle whose radius, from thePythagorean Theorem,is:
24 So, the cross-sectional area is: SPHERESExample 1So, the cross-sectional area is:
25 Using the definition of volume with a = -r and b = r, we have: SPHERESExample 1Using the definition of volume with a = -r andb = r, we have:(The integrand is even.)
26 The figure illustrates the definition of volume SPHERESThe figure illustrates the definition of volumewhen the solid is a sphere with radius r = 1.From the example, we know that the volume of the sphere isThe slabs are circular cylinders, or disks.
27 The three parts show the geometric interpretations of the Riemann sums SPHERESThe three parts show the geometricinterpretations of the Riemann sumswhen n = 5, 10,and 20 if we choose the sample points xi*to be the midpoints .
28 Notice that as we increase the number SPHERESNotice that as we increase the numberof approximating cylinders, the correspondingRiemann sums become closer to the truevolume.
29 Find the volume of the solid obtained by VOLUMESExample 2Find the volume of the solid obtained byrotating about the x-axis the region underthe curve from 0 to 1.Illustrate the definition of volume by sketchinga typical approximating cylinder.
30 The region is shown in the first figure. VOLUMESExample 2The region is shown in the first figure.If we rotate about the x-axis, we get the solidshown in the next figure.When we slice through the point x, we get a disk with radius
31 The area of the cross-section is: VOLUMESExample 2The area of the cross-section is:The volume of the approximating cylinder(a disk with thickness ∆x) is:
32 The solid lies between x = 0 and x = 1. So, its volume is: VOLUMESExample 2The solid lies between x = 0 and x = 1.So, its volume is:
33 Find the volume of the solid obtained VOLUMESExample 3Find the volume of the solid obtainedby rotating the region bounded by y = x3,y = 8, and x = 0 about the y-axis.
34 As the region is rotated about the y-axis, it VOLUMESExample 3As the region is rotated about the y-axis, itmakes sense to slice the solid perpendicularto the y-axis and thus to integrate withrespect to y.Slicing at height y, we get a circular disk with radius x, where
35 So, the area of a cross-section through y is: VOLUMESExample 3So, the area of a cross-section through y is:The volume of the approximatingcylinder is:
36 Since the solid lies between y = 0 and y = 8, its volume is: VOLUMESExample 3Since the solid lies between y = 0 andy = 8, its volume is:
37 The region R enclosed by the curves y = x VOLUMESExample 4The region R enclosed by the curves y = xand y = x2 is rotated about the x-axis.Find the volume of the resulting solid.
38 The curves y = x and y = x2 intersect at the points (0, 0) and (1, 1). VOLUMESExample 4The curves y = x and y = x2 intersect atthe points (0, 0) and (1, 1).The region between them, the solid of rotation, and cross-section perpendicular to the x-axis are shown.
39 A cross-section in the plane Px has the shape VOLUMESExample 4A cross-section in the plane Px has the shapeof a washer (an annular ring) with innerradius x2 and outer radius x.
40 Thus, we find the cross-sectional area by VOLUMESExample 4Thus, we find the cross-sectional area bysubtracting the area of the inner circle fromthe area of the outer circle:
46 The solids in Examples 1–5 are all called solids of revolution because they are obtained by revolving a regionabout a line.
47 In general, we calculate the volume of SOLIDS OF REVOLUTIONIn general, we calculate the volume ofa solid of revolution by using the basicdefining formula
48 We find the cross-sectional area A(x) or A(y) in one of the following SOLIDS OF REVOLUTIONWe find the cross-sectional areaA(x) or A(y) in one of the followingtwo ways.
49 If the cross-section is a disk, we find WAY 1If the cross-section is a disk, we findthe radius of the disk (in terms of x or y)and use:A = π(radius)2
50 If the cross-section is a washer, we first find WAY 2If the cross-section is a washer, we first findthe inner radius rin and outer radius rout froma sketch.Then, we subtract the area of the inner disk from the area of the outer disk to obtain: A = π(outer radius)2 – π(outer radius)2
51 Find the volume of the solid obtained SOLIDS OF REVOLUTIONExample 6Find the volume of the solid obtainedby rotating the region in Example 4about the line x = -1.
52 The figure shows the horizontal cross-section. SOLIDS OF REVOLUTIONExample 6The figure shows the horizontal cross-section.It is a washer with inner radius 1 + y andouter radius
53 So, the cross-sectional area is: SOLIDS OF REVOLUTIONExample 6So, the cross-sectional area is:
55 In the following examples, we find VOLUMESIn the following examples, we findthe volumes of three solids that arenot solids of revolution.
56 The figure shows a solid with a circular base VOLUMESExample 7The figure shows a solid with a circular baseof radius 1. Parallel cross-sectionsperpendicular to the base are equilateraltriangles.Find the volume of the solid.
57 Let’s take the circle to be x2 + y2 = 1. VOLUMESExample 7Let’s take the circle to be x2 + y2 = 1.The solid, its base, and a typical cross-sectionat a distance x from the origin are shown.
58 As B lies on the circle, we have So, the base of the triangle ABC is VOLUMESExample 7As B lies on the circle, we haveSo, the base of the triangle ABC is|AB| =
59 Since the triangle is equilateral, we see that its height is VOLUMESExample 7Since the triangle is equilateral, we seethat its height is
60 Thus, the cross-sectional area is : VOLUMESExample 7Thus, the cross-sectional area is :
61 The volume of the solid is: VOLUMESExample 7The volume of the solid is:
62 Find the volume of a pyramid whose base is a square with side L VOLUMESExample 8Find the volume of a pyramidwhose base is a square with side Land whose height is h.
63 We place the origin O at the vertex VOLUMESExample 8We place the origin O at the vertexof the pyramid and the x-axis along itscentral axis.Any plane Px that passes through x and is perpendicular to the x-axis intersects the pyramid in a square with side of length s.
64 We can express s in terms of x by observing VOLUMESExample 8We can express s in terms of x by observingfrom the similar triangles thatTherefore, s = Lx/hAnother method is to observe that the line OP has slope L/(2h)So, its equation is y = Lx/(2h)
65 Thus, the cross-sectional area is: VOLUMESExample 8Thus, the cross-sectional area is:
66 The pyramid lies between x = 0 and x = h. So, its volume is: VOLUMESExample 8The pyramid lies between x = 0 and x = h.So, its volume is:
67 In the example, we didn’t need to place NOTEIn the example, we didn’t need to placethe vertex of the pyramid at the origin.We did so merely to make the equations simple.
68 Instead, if we had placed the center of NOTEInstead, if we had placed the center ofthe base at the origin and the vertex onthe positive y-axis, as in the figure, you canverify that we would haveobtained the integral:
69 A wedge is cut out of a circular cylinder of VOLUMESExample 9A wedge is cut out of a circular cylinder ofradius 4 by two planes. One plane isperpendicular to the axis of the cylinder.The other intersects the first at an angle of 30°along a diameter of the cylinder.Find the volume of the wedge.
70 If we place the x-axis along the diameter VOLUMESExample 9If we place the x-axis along the diameterwhere the planes meet, then the base ofthe solid is a semicirclewith equation-4 ≤ x ≤ 4
71 A cross-section perpendicular to the x-axis at VOLUMESExample 9A cross-section perpendicular to the x-axis ata distance x from the origin is a triangle ABC,whose base is and whose heightis |BC| = y tan 30° =
72 Thus, the cross-sectional area is: VOLUMESExample 9Thus, the cross-sectional area is:
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