Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 16 - Multiple Integration - 16.3 Triple Integrals - Exercises - Page 871: 21

Answer

The volume of the solid region: $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} 1{\rm{d}}V = \frac{1}{{12}}$.

Work Step by Step

We have a solid region ${\cal W}$ in the first octant $x \ge 0$, $y \ge 0$, $z \ge 0$ bounded by the planes $x+y+z=1$ and $x+y+2z=1$. Referring to the figure attached, we can consider ${\cal W}$ as a $z$-simple region bounded below by the plane $z = \frac{1}{2}\left( {1 - x - y} \right)$ and bounded above by the plane $z=1-x-y$. So, the volume is equal to the iterated integral: $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} 1{\rm{d}}V = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left( {\mathop \smallint \limits_{z = \left( {1 - x - y} \right)/2}^{1 - x - y} {\rm{d}}z} \right){\rm{d}}A$, where ${\cal D}$ is the projection of ${\cal W}$ onto the $xy$-plane. To evaluate the integral over ${\cal D}$, we must find the boundary of ${\cal D}$. Step 1. Find the boundary of ${\cal D}$ We see that both $x+y+z=1$ and $x+y+2z=1$ intersect the $xy$-plane ($z=0$) at the line $x+y=1$. Thus, the projection of ${\cal W}$ onto the $xy$-plane is a triangle ${\cal D}$ in the first quadrant bounded by the line $y=1-x$. Step 2. Express ${\cal D}$ as a simple domain The domain ${\cal D}$ is both vertically and horizontally simple region. However, we choose to describe ${\cal D}$ as a vertically simple region. Thus, ${\cal D}$ is defined by ${\cal D} = \left\{ {\left( {x,y} \right)|0 \le x \le 1,0 \le y \le 1 - x} \right\}$ Step 3. Evaluate the triple integral as an iterated integral $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} 1{\rm{d}}V = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left( {\mathop \smallint \limits_{z = \left( {1 - x - y} \right)/2}^{1 - x - y} {\rm{d}}z} \right){\rm{d}}A$ $ = \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 0}^{1 - x} \left( {\mathop \smallint \limits_{z = \left( {1 - x - y} \right)/2}^{1 - x - y} {\rm{d}}z} \right){\rm{d}}y{\rm{d}}x$ $ = \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 0}^{1 - x} \left( {z|_{\left( {1 - x - y} \right)/2}^{1 - x - y}} \right){\rm{d}}y{\rm{d}}x$ $ = \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 0}^{1 - x} \left( {\frac{1}{2} - \frac{1}{2}x - \frac{1}{2}y} \right){\rm{d}}y{\rm{d}}x$ $ = \mathop \smallint \limits_{x = 0}^1 \left( {\left( {\frac{1}{2}y - \frac{1}{2}xy - \frac{1}{4}{y^2}} \right)|_0^{1 - x}} \right){\rm{d}}x$ $ = \mathop \smallint \limits_{x = 0}^1 \left( {\frac{1}{2}\left( {1 - x} \right) - \frac{1}{2}x\left( {1 - x} \right) - \frac{1}{4}{{\left( {1 - x} \right)}^2}} \right){\rm{d}}x$ $ = \mathop \smallint \limits_{x = 0}^1 \left( {\frac{1}{2} - \frac{1}{2}x - \frac{1}{2}x + \frac{1}{2}{x^2} - \frac{1}{4} + \frac{1}{2}x - \frac{1}{4}{x^2}} \right){\rm{d}}x$ $ = \mathop \smallint \limits_{x = 0}^1 \left( {\frac{1}{4} - \frac{1}{2}x + \frac{1}{4}{x^2}} \right){\rm{d}}x$ $ = \left( {\frac{1}{4}x - \frac{1}{4}{x^2} + \frac{1}{{12}}{x^3}} \right)|_0^1$ $ = \frac{1}{4} - \frac{1}{4} + \frac{1}{{12}} = \frac{1}{{12}}$ Thus, the volume of the solid region: $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} 1{\rm{d}}V = \frac{1}{{12}}$.
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