If 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, what is the total amount of water supplied per hour inside of a circle of radius 11?

It's `2 pi - 24 pi e^(-11) (ft^3)/h approx 6.2819 (ft^3)/h`

We have the information that the height of water supplied per hour at a given distance `r` is `e^(-r)`. The volume of water supplied per hour at a given region is simply the integral of this quantity over this region. Calling `z(r)=e^(-r)`, we can see that this problem can be reinterpreted as the problem of determining the volume of the region under a surface defined by `z(r)` (wich represents the height supplied per hour at a given distance), using polar coordinates.

This "volume" can be calculated by the integral:

`int_Omega e^(-r) dA = int_0^(R)int_0^(2 pi)e^(-r) r d theta dr,`

where `Omega` is the region over wich you are integrating, in this case, the circle of radius `R`.

The term `r` appears due to the use of polar coordinates.

The integral in `theta` is very easy to solve:

`int_0^R int_0^(2 pi)e^(-r) r d theta dr = int_0^R r e^(-r) [theta]_0^(2 pi) dr = 2pi int_0^R r e^(-r) dr`

The integral in `r` is not too difficult either, we just need integration by parts.

`2pi int_0^R r e^(-r) dr = 2 pi ([-r e^(-r)]_0^R - int_0^R e^(-r) dr) = 2 pi (- R e^(-R) - e^(-r)|_0^R) = 2 pi [1 - e^(-R) (1 + R)]`

In your question, the circle has radius `11`. Plugging this value in the result we get the answer:

`2 pi [1 - e^(-11) (12)] = 2 pi - 24 pi e^(-11) approx 6.2819`

which is, as stated in your question, in `(ft^3)/(h)`.