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### Area of a Curved Figure

```
Date: 07/26/97 at 10:55:49
From: Thomas Hamby
Subject: Finding the area of a curved figure without the use of
Calculus

I have a shape like this:

a) y = x^2+2  in the first quadrant when y (less then or = to) 2
b) x = 0      when y (greater than or = to) 0 and
y (less than or = to) 2
c) y = 0      when x (greater than or = to) 0 and
x (less than or = to) 5
d) x = 5      in the first quadrant and y (less than or = to) 27

This produces a figure with points A (0,0)  B(0,2)  C(5,27)  D(5,0)

I need to know how to find the area of this shape WITHOUT calculus.  I
know the answer won't be exact but the area is somewhere between 50-55
units.  I need a good method of solving the area that will get me as
close as possible to a calculus method.

Thanks for the help!
```

```
Date: 07/29/97 at 09:21:14
From: Doctor Rob
Subject: Re: Finding the area of a curved figure without the use of
Calculus

Choose a positive integer k and set a = 5/k, which will be the width
of some rectangles. Draw vertical lines at x = 0, a, 2*a, ..., k*a.

Find the intersections of these lines with the curve:  B = P0(0, 2),
P1(a, a^2+2), P2(2*a, 4*a^2 + 2), ..., Pk(k*a, k^2*a^2 + 2) = C(5, 27)

Now from P0, P1, ..., Pk-1 draw horizontal lines to the right until
they intersect the next vertical line. This will give you a series of
rectangles all of width a and of varying heights. They form a kind of
staircase-type figure. Compute the areas of these rectangles and add
them up.  You get:

L(k) = a*2 + a*(a^2 + 2) + a*(4*a^2 + 2) + ... + a*([k-1]^2*a^2 + 2),
= a*[2*k + a^2*(0^2 + 1^2 + 2^2 + ... + (k-1)^2)],
= 2*a*k + a^3*(k - 1)*k*(2*k - 1)/6,
= 10 + (5 - 5/k)*5*(10 - 5/k)/6,
= 10 + (250 - 375/k + 125/k^2)/6,
= (125/k^2 - 375/k + 310)/6.

Here we had to know the value of the sum 0^2 + 1^2 + ... + (k-1)^2.
This holds for any positive integer k.  The area L(k) is smaller than
the area X you are looking for, since the union of these rectangles is
contained in it.

Now from P1, P2, ..., Pk draw horizontal lines to the left until they
intersect the neighboring vertical line. This will give you another
series of rectangles all of width a and of varying heights. Compute
the areas of these rectangles and add them up.  You get:

U(k) = a*(a^2 + 2) + a*(4*a^2 + 2) + ... + a*(k^2*a^2 + 2),
= 2*a*k + a^3*(1^2 + 2^2 + ... + k^2),
= 10 + a^3*k*(k + 1)*(2*k + 1)/6,
= 10 + 5*(5 + 5/k)*(10 + 5/k)/6,
= (125/k^2 + 375/k + 310)/6.

Here we had to know the value of the sum 1^2 + 2^2 + ... + k^2.
This also holds for any positive integer k. The area U(k) is larger
than the area X you are looking for, since the union of these
rectangles contains it.

Another possiblity is to use instead of U(k), the sum of areas of
trapezoids formed by drawing segments P0P1, P1P2, ..., Pk-1Pk, which
will also be greater than X.

Thus your area X satisfies the inequality L(k) < X < U(k). This will
allow you to estimate it to any degree of accuracy you wish. You can
start with k = 5 and a = 1, if you like, and progress to k = 50,
k = 500, k = 5000, and so on.

In calculus, this is exactly the approach that is used to compute the
area using a definite integral. The next step is to let k->infinity,
and note that L(k) and U(k) both approach 310/6 = 155/3 = 51.666666...
This forces X = 155/3. Limits are, however, out of bounds here, since
you are not allowed the use of calculus.

-Doctor Rob,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Calculus
High School Conic Sections/Circles
High School Geometry

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