Double Integral

Quote:

Originally Posted by Snooks02

A sprinkler distributes water in a circular pattern, supplying water to a depth of e^-r at a distance of r feet from the sprinkler.

Per hour?

[/quote]A. What is the total amount of water supplied per hour inside a circle of radius 18?
B. What is the total amount of water that goes through the sprinkler per hour[/QUOTE]

Imagine a small "piece" of that circle of area dA. The volume of that piece to a depth of $e^{-r}$ is [tex]e^{-r}dA[/itex] you want to integrate that.
In polar coordinates, dA is $r dr d\theta$ so for part (A) you need $\int_{r=0}^{16}\int_{\theta= 0}^{2\pi} e^{-r} r dr d\theta$ . For part (B) this is no restriction on r so it is $\int_{r=0}^\infty\int_{\theta= 0}^{2\pi} e^{-r} r dr d\theta$