Question

To find: The approximate value of\(\iint_{S}{{{e}^{-0.1(x+y+z)}}}dS \)

Short answer

The approximate value of \(\iint_{S}{{{e}^{-0.1(x+y+z)}}}dS \) is 49.09.

Step-by-step solution

Definition used:

"The surface integral of fover the surface Sas\(\iint{f}(x,y,z)dS=\underset{S}{\mathop{\lim }}\,\sum\limits_{\max }{\sum\limits_{{{\text{u}}{i}},{{\Delta }{{{v}{j}},\to 0}}}{\sum\limits{j=1}{f}}}(Pij)\Delta {{S}{ij}}$ where $\Delta {{S}{ij}}\approx \left| {{\mathbf{r}}{u}}\times {{\mathbf{r}}{v}} \right|\Delta {{u}{i}}\Delta {{v}{j}}\text{''}\)

From Figure, the parallel points are (0,2,3) and (2,2,3) in x-axis, (1,0,3) and (1,4,3) in y-axis and (1,2,0) and (1,2,6) in z-axis.

So, the surface areas are 24,12 and 8 for x,yand , zrespectively.

Find c\(\iint_{S} e^{-0.1(x+y+z)} d S \)..

..\(\begin{align}& \iint_{S}{{{e}^{-0.1(x+y+z)}}}dS\cong arra{{y}^{-0.7}}(24)+{{e}^{-0.4}}(12)+{{e}^{-0.8}}(12)+{{e}^{-0.3}}(8)+{{e}^{-0.9}}(8)\\&\cong 14.56+11.92+8.04+5.39+5.93+3.25\\&\cong49.09\end{align}\)

Thus, the approximate value of \(\iint_{S} e^{-0.1(x+y+z)} d S \) is $49.09$.