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)#.
