# 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 $2 \pi - 24 \pi {e}^{- 11} \frac{f {t}^{3}}{h} \approx 6.2819 \frac{f {t}^{3}}{h}$

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 $z \left(r\right) = {e}^{- r}$, we can see that this problem can be reinterpreted as the problem of determining the volume of the region under a surface defined by $z \left(r\right)$ (wich represents the height supplied per hour at a given distance), using polar coordinates.

This "volume" can be calculated by the integral:

${\int}_{\Omega} {e}^{- r} \mathrm{dA} = {\int}_{0}^{R} {\int}_{0}^{2 \pi} {e}^{- r} r d \theta \mathrm{dr} ,$

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:

${\int}_{0}^{R} {\int}_{0}^{2 \pi} {e}^{- r} r d \theta \mathrm{dr} = {\int}_{0}^{R} r {e}^{- r} {\left[\theta\right]}_{0}^{2 \pi} \mathrm{dr} = 2 \pi {\int}_{0}^{R} r {e}^{- r} \mathrm{dr}$

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

$2 \pi {\int}_{0}^{R} r {e}^{- r} \mathrm{dr} = 2 \pi \left({\left[- r {e}^{- r}\right]}_{0}^{R} - {\int}_{0}^{R} {e}^{- r} \mathrm{dr}\right) = 2 \pi \left(- R {e}^{- R} - {e}^{- r} {|}_{0}^{R}\right) = 2 \pi \left[1 - {e}^{- R} \left(1 + R\right)\right]$

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

$2 \pi \left[1 - {e}^{- 11} \left(12\right)\right] = 2 \pi - 24 \pi {e}^{- 11} \approx 6.2819$

which is, as stated in your question, in $\frac{f {t}^{3}}{h}$.