A pond is approximately circular with a diameter of 400 feet

A pond is approximately circular, with a diameter of 400 feet. Starting at the center, the depth of the water is measured every 25 feet and recorded in the table (see figure) 25 50 75 100 125 150 175 200 0 25 5075100125 150 175 200 Depth 20 19 19 17151410 6 0 20 1% -12 10 50 100 150 200 Distance from center (a) Use Simpson\'s Rule to approximate the volume of water in the pond. (Round your answer to the nearest whole number.) 1366593 ft3 (b) Use the regression capabilities of a graphing utility to find a quadratic model for the depths recorded in the table. (Round your coefficients to six decimal places.) d=

Solution

Solution :

( a )

Diameter of pond = 400 feet, radius of pond = 200 feet.

For each 25 feet from centre, depth of pond is given. I gather then that we should assume that depth is the same at x feet from centre in all directions.

If we draw a circle x feet from centre of pond, that circle has circumference = 2x
Depth of pond at that distance from centre is given by the points given in table.
So we have a cylinder with surface area = 2x * depth

Call this function f(x). Then we get:
x ............ f(x) = 2x * depth(from table)
0 ............ 2*0*20 ........ = 0
25 .......... 2*25*19 ...... = 950
50 .......... 2*50*19 ...... = 1900
75 .......... 2*75*17 ...... = 2550
100......... 2*100*15 .... = 3000
125 ........ 2*125*14 .... = 3500
150 ........ 2*150*10 .... = 3000
175 ........ 2*175*6 ...... = 2100
200 ........ 2*200*0 ...... = 0

Now if we take surface area of each cylinder and multiply it by width of subinterval, we will get an approximation to the volume of pond.

For each subinterval [a,b], volume can be approximate using Simpson\'s Rule as
= (b-a)/6 [f(a) + 4f((a+b)/2) + f(b)]

Since each subinterval requires a value for function f(x) at midpoint of [a, b], we need to take subintervals that are 50 units wide (since we have measures for each 25 feet)

For each subinterval [a,b], we measure volume as follows:

a ....... b .... (a+b)/2 ..... (b-a)/6 [f(a) + 4f((a+b)/2) + f(b)]
0 ....... 50 ...... 25 .......... 50/6 [f(0) + 4f(25) + f(50)]
50 ... 100 ...... 75 .......... 50/6 [f(50) + 4f(75) + f(100)]
100 . 150 .... 125 .......... 50/6 [f(100) + 4f(125) + f(150)]
150 . 200 .... 175 .......... 50/6 [f(150) + 4f(175) + f(200)]

So we get :

V = 50/6 [f(0) + 4f(25) + f(50)] + 50/6 [f(50) + 4f(75) + f(100)]
..... + 50/6 [f(100) + 4f(125) + f(150)] + 50/6 [f(150) + 4f(175) + f(200)]

V = 50/6 [f(0) + 4f(25) + 2f(50) + 4f(75) + 2f(100) + 4f(125) + 2f(150) + 4f(175) + f(200)]

V = 50/6 [0 + 4*950 + 2*1900 + 4*2550 + 2*3000 + 4*3500 + 2*3000 + 4*2100 + 0]

V = 50/6 [52200]

V = 435,000

V 1,366,593 ft³

( b )