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• Notes

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 $x^{2}+y^{2}=1$

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

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

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

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

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

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

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

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