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Area Between Curves

Date: 01/11/97 at 00:40:11
From: Anonymous
Subject: The Application of integrals

I have no trouble finding the area between curves when there are only 
two curves, but this problem involves three. I get an answer, but it 
is off by .02. The book states it as a fraction, so I assume that my 
number should be more accurate.  The problem is:

Find the area bounded on the right by x+y = 2, on the left by y = x^2, 
and below by the x-axis. 

Thanks for the help. 


Date: 01/11/97 at 06:17:19
From: Doctor Pete
Subject: Re: The Application of integrals


To find the area of the region described above, you need to do three 

1) Sketch the graphs of all three boundaries, and the region enclosed.
2) Determine the intersection points of the boundaries.
3) Find the most effective way of dividing up the region, if
   necessary, in order to make the resulting integrals simple to

In the above case, if you draw the graph, you will notice that the 
region is a bit like a triangle, except that one of its sides is 
curved inwards (this is the part which is bounded by the parabola).  
There are a few ways to approach this, but in either case, you need to 
know where the parabola intersects the line x+y = 2.  (The other two 
vertices of the region are easy; they are (0,0) and (2,0).)  To find 
this third intersection point, we solve the simultaneous system of 

     x+y = 2
       y = x^2

This gives x^2 + x - 2 = 0, which factors into (x-1)(x+2) = 0.  Hence 
x = 1 or -2; but the point we wish to find lies in the first quadrant, 
hence x = 1, y = 1 is the third point.

Now, with this, we can integrate the region in two ways:

Method 1:  Divide up the region into two parts, along the vertical 
line x = 1.  This is seen by noting that the region is equivalent to 
the area under two separate curves, i.e., y = x^2 from x = 0 to x = 1, 
and y = -x + 2 from x = 1 to x = 2.

Method 2:  We look at the figure "sideways," along the y-direction, 
and see that we can express the area as the integral of the difference 
of the function x = -y + 2 and x = y^(1/2).  The limits of integration 
are seen to be from y = 0 to y = 1.

If we use Method 1, we obtain:

     Area = Int[x^2 dx, 0, 1] + Int[-x+2 dx, 1, 2]
          = (1^3)/3 + (-(2^2)/2 + 2(2)) - (-(1^2)/2 + 2(1))
          = 1/3 + (-2 + 4) - (-1/2 + 2)
          = 1/3 + 2 - 3/2
          = 5/6

If we use Method 2, we obtain:

     Area = Int[((-y+2) - y^(1/2)) dy, 0, 1]
          = (-(1^2)/2 + 2(1) - 2/3 (1^(3/2))) - 0
          = -1/2 + 2 - 2/3
          = 3/2 - 2/3
          = 5/6

In both cases, the same result is obtained.  Note that I did not write 
it as 0.83333....

You will note that in Method 2, we did not need to consider 3 curves, 
because we took advantage of the fact that the x-axis is a straight 
line which is perpendicular to the axis we integrated over.  Thus we 
were able to reconsider the region as one obtained by the difference 
between two curves, simply by "rotating" the picture mentally.  
However, Method 1 works equally well, and in fact, is more desirable 
in cases where the bounded curves are not always so nice.  For 
example, find the area of the region bounded by the curves:

     y = x/2
     y = x
     xy = 1
     xy = 4

You will note that in order to find the area, we must divide the 
region up into at least 3 separate pieces, and in each case, view 
these pieces as differences of two curves; one above, and one below.  
I will leave it as an exercise for you to calculate, and I strongly 
suggest you at least attempt to solve it.

-Doctor Pete,  The Math Forum
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Associated Topics:
High School Calculus

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