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Consider a circular lawn with a radius of 10 ft. Assumethat a sprinkler distributes water in a radial fashion according to the formula

$$f(r) = \frac{r}{16} - \frac{r^2}{160}$$

(measured in cubic ft of water per hour per square ft of lawn), where $r$ is the distance in ft from the sprinkler. Find the amount of water that is distributed in 1 hour in the following two annular regions.

$$A = \{(r,\theta): 4 \leq r \leq 5, 0 \leq θ \leq 2\pi\}$$

$$B = \{(r,\theta): 9 \leq r \leq 10, 0 \leq θ\leq 2\pi\}$$

Is the distribution of water uniform? Determine the amount ofwater the entire lawn receives in 1 hour.

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I think I just have to use a double integral for f(r) over those two regions, but I don't that's right. – thebottle394 Apr 24 '13 at 4:23

For any region we want to integrate $f$ over that region to give us the amount of water, in $ft^{3}$, distributed in one hour.

So, in general we have the amount of water is $$ \int_{D}f(r)dA $$ where D is the region being watered.

For region A:

Amount of water is $$ \int_{0}^{2\pi}\int_{4}^{5}(\frac{r}{16} - \frac{r^{2}}{160})rdrd\theta $$

For region B:

Amount of water is $$ \int_{0}^{2\pi}\int_{9}^{10}(\frac{r}{16} - \frac{r^{2}}{160})rdrd\theta $$

and for the whole lawn:

Amount of water is $$ \int_{0}^{2\pi}\int_{0}^{10}(\frac{r}{16} - \frac{r^{2}}{160})rdrd\theta $$

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...and the "angle integral" is separable, so you can just "pull out" $\int_{0}^{2\pi} d\theta = 2 \pi$ for all three. You get to do this because the "sprinkler function" has no dependence on direction of spraying. – RecklessReckoner Apr 24 '13 at 4:30

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