Area Between Curves. The idea: the area between curves y = f(x) and y = g(x) (if the graph of f(x) is above that of g(x)) for a x b is given by


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1 MATH 42, Fall 29 Examples from Section, Tue, 27 Oct 29 1 The First Hour Area Between Curves. The idea: the area between curves y = f(x) and y = g(x) (if the graph of f(x) is above that of g(x)) for a x b is given by b a f(x) g(x) dx = b a f(x) dx b a f(x) dx i.e. the difference between the area under the graph of f(x) and the area under the graph of g(x): But we have to be careful if the graphs of f and g cross each other. For example, let us say we want to find the area of the shaded region below: You could say that the area is between the curves y = sin (x) and y = for x 2π, which is correct. However, it would not be correct to simply apply 1
2 the formula above and say the area should be given by 2π sin x dx = ; this would be a mistake because the graph of y = sin x does not always lie above that of y = on the interval [, 2π]. Example. Find the area of the region enclosed by the curves y = cos x, y = 2 cos x for x 2π. Solution. First, we plot the functions: Since y = 2 cos x always lies above y = cos x over the interval [, 2π], the area between these curves is given by 2π (2 cos x) cos x dx = = 2[x] 2π 2[sin x] 2π = 2(2π ) 2(sin 2π sin ) = 4π 2π 2 2 cos x dx Example (parametrised curves). Find the area enclosed by the xaxis and the parametrised curve x = 1 + e t, y = t t 2. Solution. First, we want to check if these two curves cross each other. The xaxis is just another name for the curve y =. If these two curves intersect, then the y coordinates of the intersection points satisfy both y = and y = 2
3 t t 2. So we want to sole = t t 2, i.e. = t(1 t), which has t =, 1 as the solutions. So the two curves intersect at (x, y) = (, 1 + e ) = (, 2) and (x, y) = (1, 1 + e 1 ) = (1, 1 + e). Note that y = t t 2 is positive when < t < 1 (meaning the graph of the parametrised curve is above the xaxis on this interval) and negative when t < or t > 1 (the graph of the parametrised curve is below the xaxis on this these intervals). Also, for any t, e t is always >. We can use this information to plot the curve: The region bounded by these two curves is between x = 1 and x = 1 + e. So the area is = = x=1+e x=2 t=1 t= 1 y dx (t t 2 )e t dt (substitute x = 1 + e t and use y = t t 2 ) te t t 2 e t dt Integrating by parts with u = t 2, dv = e t dt (so du = 2t dt and v = e t ), we get t 2 e t dt = t 2 e t 2te t dt. 3
4 Integrating by parts with u = t, dv = e t dt (so du = dt and v = e t ), we get te t dt = te t e t dt. Hence 1 tet dt = [te t ] 1 1 et dt = e [e t ] 1 = 1. Also, 1 t2 e t dt = [ t 2 e t ] 1 1 2tet dt = ( e) 2( 1) = 2 e. Hence 1 tet dt 1 t2 e t dt = e = 3 e. 2 The Second Hour Example. Find the area enclosed by y = ln(x), xy = 4, x = 1 and x = 3. Solution. We can plot the curves given by y = ln(x) and xy = 4: The function y = ln(x) is always increasing. On the other hand, the part of the curve xy = 4 that is in the first quadrant, which can be thought of as the graph of y = 4/x, is always decreasing. It is probably not easy to work out that these two graphs intersect somewhere on 3 < x < 4 by hand, so you may assume that they do not intersect on 1 x 3. Given this, the area we are after is ln(x) dx x = [4 ln x (x ln x x)] 3 1 = ln Example. Find the area of the region between y = cos x and y = 1 cos x, on x π. (Before you start, try to work out how many times the curves cross each other.) 4
5 Solution. To figure out what the graph of y = 1 cos x, you can start with the graph of y = cos x. To get the graph of y = cos x, you would perform a reflection (of the graph of y = cos x) with respect to the xaxis. Then, to get the graph of y = 1 cos x, you would translate the graph of y = cos x upward by 1 unit. Then it is easy to see that the two curves y = cos x, y = 1 cos x intersect each other exactly once on the interval x π: Where exactly do they intersect? To answer this question, we need to solve cos x = 1 cos x, i.e. cos x = 1/2, i.e. x = π/3 (since x π). So the area between these two curves is π/3 3 cos x (1 cos x) dx + (1 cos x) cos x dx π/3 = [2 sin x x] π/3 + [x 2 sin x] π π/3 = π/3 Example. Let M denote the bounded region between y = and y = x 4 1. Let E 1, E 2 denote the regions inside the circles of radius 1/1 and centered at (1, 1), ( 1, 1), respectively. Let F be the region enclosed by the ellipse x2 9 + y2 9 = 1, which has parametrisation x = 3 cos t, y = sin t for t 2π. Find the area of the region outside M, E 1, E 2 but inside F. Solution. First, you can plot the four curves we are dealing with: (Yes, it does look like a face. :) ) 5
6 The area of M is given by (x4 1) dx = [ x5 5 x]1 1 = 2/5. The areas of E 1, E 2 are both equal to π(1/1) 2. How do we find the area of F? Consider the part of the ellipse above the xaxis. If we can find the area under this curve (or, in other words, the area between this curve and the xaxis), then the area of F would be twice this area. It would be easier to use the parametric equations of the ellipse to compute the area of F above the xaxis. Since this part of the ellipse is given by x = 3 cos t, y = 2 sin t but with t π, its area is t= t=pi (2 sin t)( 3 sin t) dt x=3 = 6 = 6 x= 3 π π ( = 6 y(t) dx π sin 2 t dt 1 cos 2 t dt ) cos 2t dt dt π 2 = 6( π 1 2t [sin 2 2 ] π π = 3π so the area of F is 2 6π. So the area of the region inside F but outside M, E 1, E 2 is 6π 2π(1/1) 2 2/5. 6
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