Question

(a) I wanted to determine how much water my sprinkler was using, so I set out a bunch of empty cat food cans at various distances from the sprinkler and noted how high the water was in each can after one hour. My sprinkler reaches from 0 feet to 16 feet away from the sprinkler. The data are given in Table 1. The sprinkler distributes water in a circular pattern, so I assumed that points the same distance from the sprinkler received the same amounts of water. One way to use the data in Table 1 to estimate how many cubic feet of water my lawn got from my sprinkler in one hour is to write down an integral for the volume of water using the method of shells, and then use a right-endpoint Riemann sum to approximate this integral. Give your answer in decimal form rounded to the nearest cubic foot.

(b) My neighbor decided to collect the same data for her watering. Her sprinkler reaches from 0 feet to 21 feet away from the sprinkler, but she did not set out the cans at evenly spaced distances and so the calculation is a bit more complicated. The data are given in Table 2.Use the data in Table 2 to estimate how many cubic feet of water her lawn got from her sprinkler in one hour. Again, write down an integral for the volume of water using the method of shells, and then use a right-endpoint Riemann sum with subintervals of different lengths to approximate this integral. Give your answer in decimal form rounded to the nearest cubic foot.

Answer 1

a) 108 ft

Explanation:

2pi (x*f(xi)........)*deltax

x is the point in the table

so for the first three points

2pi(2*f(2)+4*f(4)+6*f(6).....)*2

and so on for the rest

make sure you convert f(x) into feet, for example (2, 2.1) do 2.1/12

first point in equation

2pi(2*(2.1/12)+....)*2