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### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# Sketch the solid whose volume is given by the integrated integral $$\int\limits_0^1 {\int\limits_0^1 {\left( {4 - x - 2y} \right)} } dxdy$$

Given:- Double integrated integral

$$\int\limits_0^1 {\int\limits_0^1 {\left( {4 - x - 2y} \right)} } dxdy$$

To draw:- Solid volume is given by integrated integral

See the step by step solution

## Step (1):- Defining volume

The volume before the surface z=g(x,y) and below the region D is $$\int {\int\limits_D {\left( {4 - x - 2y} \right)dxdy} }$$represents the volume below the plane z=4-x-2y and a square $$x \ge 0,y \le 1$$

## Step (2):- Calculating volume by solving integral

$$\begin{array}{l}\int\limits_0^1 {\left( {\int\limits_0^1 {4 - x - 2y} } \right)dxdy} \\\int\limits_0^1 {\left( {4x - \frac{{{x^2}}}{2} - 2yx} \right)} _0^1dy\\\int\limits_0^1 {\left( {4\left( 1 \right) - \frac{{{{\left( 1 \right)}^2}}}{2} - 2\left( 0 \right)\left( 1 \right) - 0} \right)} _0^1dy\\\int\limits_0^1 {\left( {\frac{7}{2} - 2y} \right)} dy\end{array}$$

## Step (3):- Integrating w.r.t y

$$\begin{array}{l}\int\limits_0^1 {\left( {\frac{7}{2} - 2y} \right)} dy = \left( {\frac{7}{2}y - 2.\frac{{{y^2}}}{2}} \right)_0^1\\\frac{7}{2} - 2.\frac{{{1^2}}}{2} - 0\\\frac{7}{2} - 1 = \frac{5}{2}\end{array}$$

Therefore, the volume of the solid is $$\frac{5}{2}$$cubic units.