Chapter 5 Applications of Integration
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1 MA111 Application of Integration Asst.Prof.Dr.Supranee Lisawadi 1 Chapter 5 Applications of Integration Section 5.1 Area Between Two Curves In this section we use integrals to find areas of regions that lie between the graphs of two functions. Consider the region that lies between two curves y = f and y = g and between the vertical lines = a and = b, where f and g are continuous functions and f g for all in [ a, b ]. The area A of this region is b A = [ f g ]d a
2 MA111 Application of Integration Asst.Prof.Dr.Supranee Lisawadi Note: [ ]d g f d g d f g f A b a b a b a = = = under area under area If we are asked to find the area where f g for some values of but g f for other values of, we have the following epression of A. Area Between Curves: The area between the curves y = f and y = g and between = a and = b is = b a d g f A Note: = f g if f g g f if g f g f
3 MA111 Application of Integration Asst.Prof.Dr.Supranee Lisawadi 3 Eample 1: Find the area bounded by y =, y = 3, = 0, and =. Eample : Find the area enclosed by y = -cos, y = sin, = π, and = π
4 MA111 Application of Integration Asst.Prof.Dr.Supranee Lisawadi 4 Eample 3: Find the area enclosed by y =, y =.
5 MA111 Application of Integration Asst.Prof.Dr.Supranee Lisawadi 5 Eample 4: Find the area enclosed by y =, + y = 0, + y = 3.
6 MA111 Application of Integration Asst.Prof.Dr.Supranee Lisawadi 6 Some regions are best treated by regarding as a function of y. If a region is bounded by curves with equations = f y, = gy, y = c, and y = d, where f and g are continuous and f y gy for c y d, then the area is d [ f y g y ] A = dy Eample 5: Find the area enclosed by 4 + y = 1, = y. c
7 MA111 Application of Integration Asst.Prof.Dr.Supranee Lisawadi 7 Section 5. Volumes by Slicing; Disks and Washers Definition of Volume: Let S be a solid that lies between = a and = b. If the cross-sectional area of S in the plane P, through and perpendicular to the -ais, is A, where A is continuous function, then the volume of S is n i = 1 n * i V = lim A Δ = A d b b a When we use the volume formula V = A d it is important to remember that A is the a area of a moving cross-section obtained by slicing through perpendicular to the -ais. Notes: 1. The solids in Eamples 1-5 below are called solids of revolution because they are obtained by revolving about a line.. If the cross-section is a disk see Eample 1-3, we find the radius of the disk in terms of or y and use A = π radius 3. If the cross-section is a washer see Eample 4 and 5, we find the inner radius r in and outer radius r out from a sketch and compute the area of the washer by subtracting the area of the inner disk from the area of the outer disk: A = πouter radius πinner radius
8 MA111 Application of Integration Asst.Prof.Dr.Supranee Lisawadi 8 Eample 1: Find the volume of the solid obtained by rotating about the -ais the region under the curve y = from 0 to 4.
9 MA111 Application of Integration Asst.Prof.Dr.Supranee Lisawadi 9 Eample : Find the volume of the solid obtained by rotating the region bounded by y= 3, y= 7, and = 0 about the y-ais.
10 MA111 Application of Integration Asst.Prof.Dr.Supranee Lisawadi 10 Eample 3: Find the volume of the solid obtained by rotating the region bounded by y= 1/, = 1, = and y= 0about the -ais.
11 MA111 Application of Integration Asst.Prof.Dr.Supranee Lisawadi 11 Eample 4: The region R enclosed by the curves y =, y = 3 and 0 is rotating about the -ais. Find the volume of the resulting solid.
12 MA111 Application of Integration Asst.Prof.Dr.Supranee Lisawadi 1 Eample 5: Find the volume of the solid obtained by rotating the region in Eample 4 about the line y =. Eample 6: Find the volume of the solid obtained by rotating the region in Eample 4 about the line = 1.
13 MA111 Application of Integration Asst.Prof.Dr.Supranee Lisawadi 13 Section 5.3 Volumes by Cylindrical Shells Method of cylindrical shells: The volume V of cylindrical shell with inner radius r 1, outer radius r, and height h is calculated by subtracting the volume V 1 of the inner cylinder from the volume V of the outer cylinder: V = V V 1 = πrh πrh 1 1 = π r r h = π r + r r r h 1 1 r + r1 h r r 1 = π 1 If we let Δ r = r r the thickness of the shell and 1 r = r + r1 the average radius of the shell, then this formula for the volume of a cylindrical shell becomes V = π rhδ r and it can be remembered as or V = [circumference] [height] [thickness] V = π [radius] [height] [thickness]
14 MA111 Application of Integration Asst.Prof.Dr.Supranee Lisawadi 14 Now let S be the solid obtained by rotating about the y-ais the region bounded by y= f [where 0 f ], y= 0, = a, and = b, where b> a 0. The volume of the solid obtained by rotating about the y-ais the region under the curve y= f from a to b, is b { V = π f d { a radius height thickness Similarly, the volume of the solid obtained by rotating about the -ais the region bounded by curve g y =, the y-ais, the line y = c, and the line y = d, is d { V = π y g y dy { c radius height thickness
15 MA111 Application of Integration Asst.Prof.Dr.Supranee Lisawadi 15 Eample 1: Find the volume of the solid obtained by rotating about the y-ais the region bounded by y = 3 and y = 0.
16 MA111 Application of Integration Asst.Prof.Dr.Supranee Lisawadi 16 Eample : Find the volume of the solid obtained by rotating about the y-ais the region bounded by y = and y =. Use both slicing and cylindrical shells methods.
17 MA111 Application of Integration Asst.Prof.Dr.Supranee Lisawadi 17 Eample 3: Find the volume of the solid obtained by rotating the region bounded by and = y i about the -ais and ii about y = 1. y =
18 MA111 Application of Integration Asst.Prof.Dr.Supranee Lisawadi 18 Eample 4: Find the volume of the solid obtained by rotating the region bounded by y= 4 and y = 3 about = 1.
19 MA111 Application of Integration Asst.Prof.Dr.Supranee Lisawadi 19 Section 5.4 Arc Length What do we mean by the length of a curve? We might think of fitting a piece of string to the curve in Figure 1 and then measuring the string against a ruler. But that might be difficult to do with much accuracy if we have a complicated curve. If the curve is a polygon, we can easily find its length; we just add the lengths of the line segments that form the polygon. We are going to define the length of a general curve by first approimating it by a polygon and then taking a limit as the number of segments of the polygon is inscribed See Figure. Now suppose that a curve C is defined by the equation y= f, where f is continuous and a b. We obtain a polygonal approimation to C by dividing the interval [, ] ab in to n subintervals with endpoints 1,,..., n and equal width Δ. If y i = f, then the point, i P y lies on C and the polygon with vertices P0, P1,..., P n, i i i illustrated in Figure 3, is an approimation to C.
20 MA111 Application of Integration Asst.Prof.Dr.Supranee Lisawadi 0 The length L of C is approimately the length of this polygon and the approimation gets better as we let n increases See Figure 4. Therefore, we define the length L of C with equation y= f, a b, as the limit of the lengths of these inscribed polygons if the limit eists: L= lim Pi 1Pi n n i = 1 The definition of arc length given above is not very convenient for computational purposes, but we can derive an integral formula for L in the case where f has a continuous derivative. [Such a function f is called smooth because a small change in produces a.] small change in f If we let Δ yi = yi yi, then 1 i 1 i i i 1 i i 1 P P = + y y = Δ + Δ y By applying the Mean Value Theorem to f on the interval [ ] number that is Thus we have Therefore, * i between i 1 and i such that, i 1 * = f f f i i 1 i i i 1 i * i Δ y = f Δ. * Pi 1Pi = Δ + Δ y = Δ + f i Δ * i * i = 1+ f Δ = 1 + f Δ since Δ > 0 i 1 i i n n i= 1 i= 1 * n n L= lim P P = lim 1+ f Δ i, we find that there is a
21 MA111 Application of Integration Asst.Prof.Dr.Supranee Lisawadi 1 We recognize this epression as being equal to b a 1+ f d by the definition of a definite integral. This integral eists because the function 1 g = + f is continuous. Thus we have proved the following theorem: The Arc Length Formula: If f is continuous on [ ab,, ] then the length of the curve y= f, a b, is b L= 1+ f d a Eample 1: Find the length of the arc of the semicubical parabola 1,1 and 4,8. y = between the point 3
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