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Expert-verified Found in: Page 1119 ### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861 # Find the area of S is the portion of the plane with equation x = y + z that lies above the region in the xy-plane that is bounded by y = x, y = 5, y = 10, and the y-axis.

$\mathrm{Hence}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{triangle}\mathrm{is}=\frac{75\sqrt{3}}{2}$

See the step by step solution

## Step 1. Given

Equation of plane x=y+ z

## Step 2.  Formula for finding the area of surface

$\phantom{\rule{0ex}{0ex}}\mathrm{If}\mathrm{a}\mathrm{surface}\mathrm{S}\mathrm{is}\mathrm{given}\mathrm{by}\mathrm{z}=\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)\mathrm{for}\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)\in \mathrm{D}\subset {\mathrm{R}}^{2}\phantom{\rule{0ex}{0ex}}\mathrm{Then}\mathrm{the}\mathrm{surface}\mathrm{area}\mathrm{of}\mathrm{smooth}\mathrm{surface}\mathrm{is}\phantom{\rule{0ex}{0ex}}\begin{array}{r}{\int }_{S} dS={\int }_{S} \sqrt{{\left(\frac{\mathrm{\partial }z}{\mathrm{\partial }x}\right)}^{2}+{\left(\frac{\mathrm{\partial }z}{\mathrm{\partial }y}\right)}^{2}+1dA}\\ ={\iint }_{D} \sqrt{{\left(\frac{\mathrm{\partial }z}{\mathrm{\partial }x}\right)}^{2}+{\left(\frac{\mathrm{\partial }z}{\mathrm{\partial }y}\right)}^{2}+1dA\cdots \cdots \left(1}\end{array}$

## Step 3.  Finding the partial derivative

$x=y+z\phantom{\rule{0ex}{0ex}}then,\phantom{\rule{0ex}{0ex}}z=x-y\phantom{\rule{0ex}{0ex}}now,firstfind\phantom{\rule{0ex}{0ex}}\begin{array}{r}\frac{\mathrm{\partial }z}{\mathrm{\partial }x}=\frac{\mathrm{\partial }}{\mathrm{\partial }x}\left(x-y\right)\\ =1\end{array}\phantom{\rule{0ex}{0ex}}and\phantom{\rule{0ex}{0ex}}\begin{array}{r}\frac{\mathrm{\partial }z}{\mathrm{\partial }\mathrm{y}}=\frac{\mathrm{\partial }}{\mathrm{\partial }\mathrm{y}}\left(x-y\right)\\ =-1\end{array}$

## Step 4. Graph viewed it as a x-simple, then the region of integration will be

$D=\left\{\left(x,y\right)|0\le x\le y,5\le y\le 10\right\}$

## Step 5. Finding the area

$Theareaofsurfaceis\phantom{\rule{0ex}{0ex}}\begin{array}{r}{\int }_{S} dS={\iint }_{D} \sqrt{{\left(\frac{\mathrm{\partial }z}{\mathrm{\partial }x}\right)}^{2}+{\left(\frac{\mathrm{\partial }z}{\mathrm{\partial }y}\right)}^{2}+1}dA\\ ={\int }_{5}^{10} {\int }_{0}^{y} \sqrt{{1}^{2}+-{1}^{2}+1}dxdy\\ ={\int }_{5}^{10} {\int }_{0}^{y}\sqrt{3}dxdy\\ =\sqrt{3}{\int }_{5}^{10} {\int }_{0}^{y} dxdy\\ =\sqrt{3}{\int }_{5}^{10} {\int }_{0}^{y} dxdy\\ =\sqrt{3}{\int }_{5}^{10} \left[x{\right]}_{0}^{y}dy\\ =\sqrt{3}{\int }_{5}^{10} \left[y-0\right]dy\\ =\sqrt{3}{\int }_{5}^{10} ydy\\ =\sqrt{3}{\int }_{5}^{10}y dy\\ =\sqrt{2}\left[\frac{{y}^{2}}{2}{\right]}_{5}^{10}\\ =\sqrt{2}\left[\frac{{10}^{2}}{2}-\frac{{5}^{2}}{2}\right]\\ \\ =\frac{75\sqrt{3}}{2}\\ \mathrm{Hence}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{triangle}\mathrm{is}=\frac{75\sqrt{3}}{2}\end{array}$ ### Want to see more solutions like these? 