An agricultural sprinkler distributes water in a circular pattern of radius 120 ft. It supplies water to a depth of e−r feet per hour at a distance of r feet from the sprinkler. (Do not substitute numerical values; use variables only.) (a) If 0 < R ≤ 120, what is the total amount of water supplied per hour to the region inside the circle of radius R centered at the sprinkler? ft3 (b) Determine an expression for the average amount of water per hour per square foot supplied to the region inside the circle of radius R. ft3 (per hour per square foot)

Answer:a) [tex]\bf 2\pi(1-e^{-120})\;(ft)^3[/tex] cubic feet per hourb) [tex]\bf \frac{2(1-e^{-R})}{R^2}[/tex] cubic feet per hour per squared footStep-by-step explanation:The region inside the circle R in polar coordinates can be written as [tex]\bf R=\left \{ (r,\theta)|0\leq r\leq R,0\leq \theta\leq 2\pi  \right \}[/tex] (a) If 0 < R ≤ 120, what is the total amount of water supplied per hour to the region inside the circle of radius R centered at the sprinkler? [tex]\bf \int_{0}^{120}\int_{0}^{2\pi}e^{-r}drd\theta[/tex] Since the exponential is a continuous function we can split the integral [tex]\bf \int_{0}^{120}\int_{0}^{2\pi}e^{-r}drd\theta=\int_{0}^{120}e^{-r}dr\int_{0}^{2\pi}d\theta=\\(1-e^{-120})2\pi[/tex] and the total amount of water supplied per hour to the region inside the circle of radius R centered at the sprinkler equals [tex]\bf \boxed{2\pi(1-e^{-120})\;(ft)^3}[/tex] (b) Determine an expression for the average amount of water per hour per square foot supplied to the region inside the circle of radius R. This would the total amount of water received by the region inside R divided by the area of the region [tex]\bf \frac{1}{\pi R^2}\left (\int_{0}^{R}\int_{0}^{2\pi}e^{-r}drd\theta\right )=\frac{2\pi(1-e^{-R})}{\pi R^2}=\frac{2(1-e^{-R})}{R^2}[/tex] and the average amount of water per hour per square foot supplied to the region inside the circle of radius R is [tex]\bf \boxed{\frac{2(1-e^{-R})}{R^2}}[/tex] cubic feet per hour per square foot.