# MATH Area Between Curves

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1 MATH - Area Between Curves Philippe Laval September, 8 Abstract This handout discusses techniques used to nd the area of regions which lie between two curves. Area Between Curves. Theor Given two functions f () and g () we are interested in nding the area of the region bounded b these two curves, assuming such a region eists. In some instances, we ma need to specif additional parameters such as vertical lines to more clearl de ne the region. Let us assume we want to nd the area of the region shown in gure. The region is delimited b the curves f (), g (), and the vertical lines a and b. We use a technique similar to the technique used when we de ned the de nite integral. First, we divide the interval [a; b] into n subintervals of equal length. The length of each subinterval will be b a n. This introduces n + points we label a,,, :::, n b. Each interval is of the form [ i ; i+ ] and has width i. In each interval [ i ; i+ ] we pick a point we call i and we form the rectangle of width and height f ( i ) g ( i ). This is shown on gure. We can then approimate the area of the region b adding the area of each rectangle. This gives us AREA nx f ( i ) i g ( i )

2 Figure : Finding the area of the region Figure : Approimating the area of the region using rectangles

3 As for the de nite integral, the eact area is obtained b taking the limit as n!. In other words, we have AREA lim n! nx f ( i ) i g ( i ) This is precisel the de nition of the de nite integral of the function f () g () between a and b. In other words, we have AREA Z b a (f () g ()) d For this to work, it is important that f and g be such that the graph of f is above the graph of g. For this reason, we prefer to use the notation top and bottom. In the case of gure, we would have top f () and bottom g (). Therefore, we have the formula: AREA Z b a ( top bottom ) d where a and b correspond to the range of values the variable must have to cover the region. Both top and bottom are functions of. The region should be graphed so that we know which function is the top function ( top ) and which function is the bottom function ( bottom ). If the graphs of f () and g () intersect, the top and bottom functions ma switch. In this case, we would epress the area of the region as several integrals. One wa to understand what the integral represent is to picture covering the entire area with those vertical rectangles of width d and height top bottom. The product ( top bottom ) d gives the area of each rectangle. We add the area of these rectangles for between a and b. In some instances, this wa of covering a region ma not be ver practical. Consider the region shown in gure. In this case, one can see that nding top and bottom is going to be di cult. The two functions will change several times as we tr to cover the whole region. Instead, it would be easier to cover the region with thin horizontal rectangles as shown in gure. The technique used is similar to the previous case. What we did for, we now do for. We have to epress the curves bounding the region in terms of. We see that the region is bounded b the curves F () and G (),

4 Figure : Finding the area of the region Figure : Approimating the area of the region

5 and the horizontal lines c and d. We divide the interval [c; d] into n subintervals of equal length. The length of each subinterval will be. We approimate the area of the region b adding the area of each rectangle. Thus we have nx AREA F (i ) G (i ) i To get the eact area, we then take the limit as n!. This gives us AREA lim n! nx F (i ) i G ( i ) This is precisel the de nition of the de nite integral of the function F () G () between c and d. In other words, we have AREA Z d c (F () G ()) d For this to work, it is important that F and G be such that the graph of F is to the right of the graph of G. For this reason, we prefer to use the notation right and left. In the case of gure, we would have right F () and left G (). Therefore, we have the formula: AREA Z d c ( right left ) d where c and d correspond to the range of values must have to cover the region. This time, right and left will be functions of. Remark In some instances, either integral can be used. In general, what determines which integral to use is whether it is easier to cover the region with vertical narrow strips or horizontal narrow strips. The eamples below should help clarif this. Remark In all the eamples below, we need to gure out two things:. If we use a narrow vertical strip ( d integral) (a) What is the height of a tpical rectangle in other words what are top and bottom? The will be epressed as functions of. 5

6 Figure 5: Graph of sin and cos (b) What is the range of values for?. If we use a narrow horizontal strip ( d integral) (a) What is the width of a tpical rectangle in other words what are right and left? The will be epressed as functions of. (b) What is the range of values for? Eample Epress the area of the region bounded b sin, cos,, and using one or more d integrals, then evaluate the integral(s). The curves are shown in Figure 5. The point where sin and cos meet has coordinates d integral, we have AREA Z ; p!. If we use a ( top bottom ) d. Since top and bottom 6

7 change whether we are to the left or to the right of the point of intersection, we need to use two integrals to epress the area. AREA Z Z ( top bottom ) d (cos sin ) d + Z (sin cos ) d (sin + cos )j + ( cos sin )j p p!! + ( + ) + ( ) p p p!! Eample Epress the area of the region bounded b the line and the parabola + 6, using one or more d integrals, then evaluate the integral(s). The graph of the two curves is shown below: First, we nd the points of intersection b solving + 6 From the second equation, we get +. If we substitute in the rst equation, we obtain + 8 or 8 which we can factor as ( ) ( + ). Therefore, the solutions are or. Hence, the points of intersection are ( ; ) and (5; ). Using a d integral, we see that: AREA Z ( right left ) d. Looking at the picture, we see that right corresponds to the line. Since on 7

8 the line ; we have that right +. Also, left corresponds to the parabola. On the parabola + 6, therefore left We replace in the integral to nd the area. Z AREA + + d Z + + d Eample Epress the area of the region bounded b f () and g () + using one or more integrals, then evaluate the integrals. First, we draw a picture of the region.. A tpical rectangle of width d and height g () f () can be used to cover the region. We still need to nd the limits of integration. For this, we need 8

9 to nd the points at which the two curves intersect. This is done b solving + Therefore, the area of the region is A Z Z Z Z + + d + d + d + d 8 Eample Epress the area of the region bounded b, e,, and using one or more integrals, then evaluate the integrals. First, we draw a picture of the region

10 A tpical rectangle of width d and height e can be used to cover the region. This time, we are given the values of. The area is A e e Z (e ) d e Eample 5 Epress the area of the region bounded b and using one or more integrals, then evaluate the integrals. First, we draw a picture of the region. A tpical rectangle of width d and height can be used to cover the region. This time, we are not given the limit values for, we need to nd them. The correspond to the -coordinates of the points of intersection between and. To nd these, we solve

11 Thus, the area is: A Z Z d d Eample 6 Show that the area of a circle of radius r is r. First, we draw a picture. Since ever circle of radius r has the same area, we consider the circle of radius r centered at the origin The upper half of the circle is the function p r. The lower half is the function p r. The picture shows a circle of radius. A tpical rectangle of width d and height top bottom p p r r p r

12 To cover the circle, ranges from r to r. Thus, the area is A Z r r Z r p r d p r d p r + r sin r r r sin r r Eample 7 Epress the area of the region bounded b + 5,,, and using one or more integrals, then evaluate the integrals. First, we draw the region If we tr to proceed as before, we see that we will have to use several integrals as top and bottom change throughout the region. Instead, we use a thin horizontal rectangle of height d to cover the region. The area of such a rectangle will be ( right left ) d. We need to epress right and left as a function of (since we now have a d integral). This is eas, on the right, the

13 rectangle ends on the curve, so right. On the left, the rectangle ends on the line + 5, therefore 5, thus left 5. Finall, since we have a d integral, we need thew range of values for (not ). The are given to us, ranges from to. It follows that the area is: A Z Z 8 ( right left ) d + 5 d Things to know Be able to graph the region delimited b a set of given curves and epress its area using d or d integrals. Related problems assigned #,, 5, 7,,, 5 on pages 6, 7

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