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Find the surface area of The part of the plane $$5 x+3 y-z+6=0$$
that lies above the rectangle $$[1,4] \times[2,6]$$

$$5 x+3 y-z+6=0$$

$$z=5 x+3 y+6=f(x, y)$$

$$A(s)=\int_{R}\int \sqrt{{f_{x}^{2}\left(x, y\right)+} {f_{y}^{2}\left(x, y\right)+1}} d A$$

$$f_{x}\left(x, {y}\right)=\frac{d z}{d x}=5$$

$$f_{y}(x, y)=3$$

$$A(s)=\int_{y_{1}=2}^{y_{2}=6} \int_{x_{1}=1}^{x_{2}=4}(\sqrt{(5)^{2}+(3)^{2}+1} $$

$$A(s)=\int_{2}^{6} \int_{1}^{1} \sqrt{35} d x d y=\sqrt{35} *(4-1)(6-2)$$

$$A(s)=12 \sqrt{35} $$

Find the surface area of The part of the surface $$z=x y$$ that lies within $$x^{2}+y^{2}=1$$

$$z=x y \quad , \quad x^{2}+y^{2}=1$$

$$x^{2}+y^{2}=r^{2}, 0 \rightarrow 2 \pi$$

$$A(s)=\int_{R} \int \sqrt{f_{x}^{2}(x, y)+f_{y}^{2}(x, y)+1}$$

$$=\int_{0}^{2 \pi} \int_{0}^{1} \sqrt{{y^{2}+x^{2}}+1} \:dxdy$$

$$=\int_{0}^{2 \pi} \int_{0}^{1} \sqrt{r^{2}+1} \: rdrd \theta$$

$$\int_{0}^{2 \pi} |_{0}^{1} \frac{\left(r^{2}+1\right)^{\frac{3}{2}}}{3 / 2 * 2 r} r d \theta$$

$$=\int_{0}^{2 \pi} {(2)^{\frac{3}{2}}-(1)^{\frac{3}{2}}} d \theta\left(\frac{1}{3}\right)$$

$$A(s)=\frac{1}{3}(2 \pi-0)(2 \sqrt{2}-1)=3,83$$

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