Area of a Curved FigureDate: 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/ |
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