Question

A sprinkler distributes water in a circular pattern, supplying water to a depth of e^-r feet per hour at a distance of r feet from the sprinkler.
A) what is the total amount of water supplied per hour inside of a circle of radius 11?
B) what is the total amount of water that goes throughout the sprinkler per hour?

Answer 1

Answer:

A) 12,5651 feet/hour

B) 12,5663 feet/hour

Step-by-step explanation:

A) Being

f$f(x)=e^{-r}$

the amount of water distributed at a distance r. But, the total amount of water distributed inside a circle of radius r , is the sum of all the water distributed from 0 until r. That is

$W(R=11)=\int\limits^R_{-R}\int\limits^R_{-R} {f(x,y)} \, dxdy = \int\limits^{2\pi} _{0}\int\limits^R_{0} {f(x,y)} \, r drd\alpha = \int\limits^{2\pi} _{0}\int\limits^R_{0} e^{-r} \, r drd\alpha = 2\pi [2-(R+1)e^{-R} ] = 2\pi [2-(11+1)e^{-11} ] = 12,5651$

B) the total ammount of water that goes out of the sprinkler will be distributed to different distances according to f(r) , therefore it will be the sum of all the ammount of water at all the distances.

$W(R=\infty)= \lim_{R \to \infty} 2\pi [2-(R+1)e^{-R} ] =2\pi [2- \lim_{R \to \infty}(R+1)e^{-R} ] = 2\pi [2- \lim_{R \to \infty}(R+1)/e^{R} ] = 2\pi [2- \lim_{R \to \infty} 1/e^{R} ] = 2\pi [2- 0] = 4\pi = 12,5663$