10. calculate (iint_{r} y e^{-x y} d a), where (r=0,5 \times0,6). answer: submit answer next item

Explanation:

Step1: Set up the double - integral

Since (R = [0,5]\times[0,6]), the double - integral (\iint_{R}ye^{-xy}dA=\int_{0}^{6}\int_{0}^{5}ye^{-xy}dxdy).

Step2: Integrate with respect to (x)

Let (u=-xy), then (du=-y dx). When (x = 0), (u = 0); when (x = 5), (u=-5y). (\int_{0}^{5}ye^{-xy}dx=-\int_{0}^{-5y}e^{u}du=\left[-e^{u}\right]_{0}^{-5y}=1 - e^{-5y}).

Step3: Integrate the result with respect to (y)

(\int_{0}^{6}(1 - e^{-5y})dy=\int_{0}^{6}1dy-\int_{0}^{6}e^{-5y}dy). (\int_{0}^{6}1dy=\left[y\right]{0}^{6}=6). For (\int{0}^{6}e^{-5y}dy), let (t=-5y), (dt=-5dy). When (y = 0), (t = 0); when (y = 6), (t=-30). Then (\int_{0}^{6}e^{-5y}dy=-\frac{1}{5}\int_{0}^{-30}e^{t}dt=\frac{1}{5}\left[1 - e^{-30}\right]).

So (\int_{0}^{6}(1 - e^{-5y})dy=6-\frac{1}{5}(1 - e^{-30})=6-\frac{1}{5}+\frac{1}{5}e^{-30}=\frac{29}{5}+\frac{1}{5}e^{-30}\approx\frac{29}{5}=5.8).

Answer:

(6-\frac{1}{5}(1 - e^{-30})) (or approximately (5.8))