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Evaluate the surface integral.
$$\iint_{S}(x+y+z) d S$$
$$S$$ is the parallelogram with parametric equations $$x=u+v$$
$$y=u-v, z=1+2 u+v, 0 \leq u \leq 2,0 \leq v \leq 1$$

$$\int_{s} \int (x+y+z) d s$$

$$x=u+v \quad y=u-v \quad z=1+2 u+v$$

$$0 \leq v \leq 1 \quad 0 \leq u \leq 2$$

$$\int_{0}^{1} \int_{0}^{2}[u+v+u-v+1+2 u+v]\left|r_{u} \times r_{v}\right| d v d u$$

$$r_{u}=1\hat  i+1\hat j+2 \hat{k} \quad r_{v}=1 \hat{i}+(-1) \hat{j}+1 \hat{k} $$

$$r_{u} \times r_{v}=\left|\begin{array}{ccc}{i} & {j} & {k} \\ {1} & {1} & {2} \\ {1} & {-1} & {1}\end{array}\right|=3 \hat{i}+\hat{j}-2 \hat{k} $$

$$\left|r_{u} \times r_{v}\right|=\sqrt{(3)^{2}+(1)^{2}+(-2)^{2}}=\sqrt{14} $$

$$\int_{0}^{1} \int_{0}^{2}(4 u+v+1) * \sqrt{14} d u d v$$

$$\int_{0}^{1} |_{0}^{2}\left[2 u^{2}+u v+u\right] \sqrt{14} d v$$

$$=\sqrt{14} \int_{0}^{1} (10+2 v) d v=\sqrt{14} \: |_{0}^{1} 10 v+v^{2} $$

$$=11 \sqrt{14} $$

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