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calculate \iint_{S} f(x, y, z) d S for the given surface and hinetion 1 . \mathrm{m}(1 / 3)21) Port of the plone x+y+z=1, where x, y, z \geqslant 0;f(x, y, z)=zS:\begin{array}{l}x+y+z=1 \\\Rightarrow z=1-x-y \\\Rightarrow f(x, y)=1-x-y \text { fre } f(x, y)=-1 \text { and }(x, y)=-1\end{array}\begin{array}{l}\partial s=\sqrt{1+\left(\frac{\partial z}{\partial x}\right)^{2}+\left(\frac{\partial z}{\partial y}\right)^{2}} \partial x \partial y \\\Rightarrow \partial s=\sqrt{1+(-1)^{2}+(-1)^{2}} \partial x \partial y \\\Rightarrow \partial s=\sqrt{1+1+1} \partial x \partial y \\\Rightarrow \partial s=\sqrt{3} \partial x \partial y\end{array}Let 2=0^{\circ}x+y+z=1z x+y=1\Rightarrow y=1-xi.e \quad 0 \leqslant y \leqslant 1-xIf y=0,(2 / 3)\begin{aligned}& y=1-x \\\Rightarrow & 0=1-x \\\Rightarrow & x=1\end{aligned}i.e 0 \leq x \leq 1\begin{aligned}\iint_{S} f(x, y, z) d s & =\iint_{S} z d S \\& =\int_{x=0}^{x=1} \int_{y=0}^{y=1-x}[(1-x-y) \sqrt{3} d y \partial x \\& =\sqrt{3} \int_{x=0}^{x=1} \int_{y=0}^{y=1-x}(1-x-y) d y \partial x \\& =\sqrt{3} \int_{x=0}^{x=1}\left([y]_{0}^{1-x}-x[y]_{0}^{1-x}-\frac{1}{2}\left[y^{2}\right]_{0}^{1-x}\right) \partial x \\& =\sqrt{3} \int_{x=0}^{x=1}\left[(1-x)-x(1-x)-\left[\frac{1-x)^{2}}{2}\right] \partial x\right.\end{aligned}\begin{array}{l}=\sqrt{3} \int_{x=0}^{x=1}\left(1-x-x+x^{2}-\frac{\left(1+x^{2}-2 x\right)}{2}\right) d x \quad(3 / 3) \\ =\sqrt{3} \int_{i=0}^{x=1}\left(1-2 x+x^{2}-\frac{1}{2}-\frac{x^{2}}{2}+x\right) d x \\ =\sqrt{3} \int_{x=0}^{x=1}\left(\frac{1}{2}-x+\frac{x^{2}}{2}\right) d x \quad\left[\because \int x^{n} y_{x}=\frac{x^{n+1}}{n+1}\right] \\ =\sqrt{3}\left[\frac{1}{2}[x]_{0}^{1}-\frac{1}{2}\left[x^{2}\right]_{0}^{1}+\frac{1}{6}\left[x^{3}\right]_{0}^{1}\right] \\ =\sqrt{3}\left[\frac{1}{2}-\frac{1}{2}+\frac{1}{6}\right] \\ =\frac{\sqrt{3}}{6} \text { Ans } \\ \text { ie } \iint_{S} f(x, y, z) d s=\frac{\sqrt{3}}{6} \\\end{array} ...