Calculus (3rd Edition)

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

Chapter 16 - Multiple Integration - 16.5 Applications of Multiple Integrals - Exercises - Page 890: 9

Answer

The total mass of the atmosphere contained: $2.593 \times {10^{10}}$ kg.

Work Step by Step

Let the $z$-coordinate represents the altitude $h$. So, we have the mass density of the atmosphere given by $\delta \left( z \right) = a{{\rm{e}}^{ - bz}}$ $kg/k{m^3}$, where $a = 1.225 \times {10^9}$ and $b = 0.13$. We have ${\cal W}$, the cone-shaped region defined by $\sqrt {{x^2} + {y^2}} \le z \le 3$. From the figure attached, we see that ${\cal W}$ can be considered as a $z$-simple region bounded below by the cone ${z^2} = {x^2} + {y^2}$ and bounded above by the plane $z=3$. The projection of ${\cal W}$ onto the $xy$-plane is ${\cal D}$, a disk of radius $3$, ${x^2} + {y^2} \le 9$. We choose to evaluate the integral in cylindrical coordinates. So, the cone-shaped region is represented by $r \le z \le 3$. Thus, the description of ${\cal W}$ in cylindrical coordinates: ${\cal W} = \left\{ {\left( {r,\theta ,z} \right)|0 \le r \le 3,0 \le \theta \le 2\pi ,r \le z \le 3} \right\}$ Using Eq. (1), the total mass is given by ${\rm{total{\ }mass}} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} \delta \left( {x,y,z} \right){\rm{d}}V$ $ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^3 \mathop \smallint \limits_{z = r}^3 \delta \left( {r\cos \theta ,r\sin \theta ,z} \right)r{\rm{d}}z{\rm{d}}r{\rm{d}}\theta $ $ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^3 \mathop \smallint \limits_{z = r}^3 a{{\rm{e}}^{ - bz}}r{\rm{d}}z{\rm{d}}r{\rm{d}}\theta $ $ = - \frac{a}{b}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^3 \left( {{{\rm{e}}^{ - bz}}|_r^3} \right)r{\rm{d}}r{\rm{d}}\theta $ $ = - \frac{a}{b}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^3 \left( {{{\rm{e}}^{ - 3b}} - {{\rm{e}}^{ - br}}} \right)r{\rm{d}}r{\rm{d}}\theta $ $ = - \frac{a}{b}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^3 \left( {{{\rm{e}}^{ - 3b}}r - r{{\rm{e}}^{ - br}}} \right){\rm{d}}r{\rm{d}}\theta $ $ = - \frac{a}{b}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\frac{1}{2}{{\rm{e}}^{ - 3b}}{r^2}|_0^3} \right){\rm{d}}\theta + \frac{a}{b}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^3 r{{\rm{e}}^{ - br}}{\rm{d}}r{\rm{d}}\theta $ $ = - \frac{{9a}}{{2b}}\mathop \smallint \limits_{\theta = 0}^{2\pi } {{\rm{e}}^{ - 3b}}{\rm{d}}\theta + \frac{a}{b}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^3 r{{\rm{e}}^{ - br}}{\rm{d}}r{\rm{d}}\theta $ Consider the inner integral of the second integral on the right-hand side: $\mathop \smallint \limits_{r = 0}^3 r{{\rm{e}}^{ - br}}{\rm{d}}r$ Write $u=r$ and $dv = {{\rm{e}}^{ - br}}dr$. So, $du = dr$ and $v = - \frac{1}{b}{{\rm{e}}^{ - br}}$. Using Integration by Parts Formula (Section 8.1), $\smallint u{\rm{d}}v = uv - \smallint v{\rm{d}}u$ we get $\mathop \smallint \limits_{r = 0}^3 r{{\rm{e}}^{ - br}}{\rm{d}}r = - \left( {\frac{r}{b}{{\rm{e}}^{ - br}}|_0^3} \right) + \frac{1}{b}\mathop \smallint \limits_{r = 0}^3 {{\rm{e}}^{ - br}}{\rm{d}}r$ $\mathop \smallint \limits_{r = 0}^3 r{{\rm{e}}^{ - br}}{\rm{d}}r = - \frac{3}{b}{{\rm{e}}^{ - 3b}} - \frac{1}{{{b^2}}}\left( {{{\rm{e}}^{ - br}}|_0^3} \right)$ $\mathop \smallint \limits_{r = 0}^3 r{{\rm{e}}^{ - br}}{\rm{d}}r = - \frac{3}{b}{{\rm{e}}^{ - 3b}} - \frac{1}{{{b^2}}}\left( {{{\rm{e}}^{ - 3b}} - 1} \right)$ $\mathop \smallint \limits_{r = 0}^3 r{{\rm{e}}^{ - br}}{\rm{d}}r = - \left( {\frac{3}{b} + \frac{1}{{{b^2}}}} \right){{\rm{e}}^{ - 3b}} + \frac{1}{{{b^2}}}$ $\mathop \smallint \limits_{r = 0}^3 r{{\rm{e}}^{ - br}}{\rm{d}}r = \frac{1}{{{b^2}}}\left( {1 - \left( {3b + 1} \right){{\rm{e}}^{ - 3b}}} \right)$ Substituting back this result in the original integral gives ${\rm{total{\ }mass}} = - \frac{{9a}}{{2b}}\mathop \smallint \limits_{\theta = 0}^{2\pi } {{\rm{e}}^{ - 3b}}{\rm{d}}\theta + \frac{a}{b}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^3 r{{\rm{e}}^{ - br}}{\rm{d}}r{\rm{d}}\theta $ $ = - \frac{{9a}}{{2b}}\mathop \smallint \limits_{\theta = 0}^{2\pi } {{\rm{e}}^{ - 3b}}{\rm{d}}\theta + \frac{a}{b}\mathop \smallint \limits_{\theta = 0}^{2\pi } \frac{1}{{{b^2}}}\left( {1 - \left( {3b + 1} \right){{\rm{e}}^{ - 3b}}} \right){\rm{d}}\theta $ $ = - \frac{{9a}}{{2b}}{{\rm{e}}^{ - 3b}}\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta + \frac{a}{{{b^3}}}\left( {1 - \left( {3b + 1} \right){{\rm{e}}^{ - 3b}}} \right)\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta $ $ = - \frac{{9a\pi }}{b}{{\rm{e}}^{ - 3b}} + \frac{{2a\pi }}{{{b^3}}}\left( {1 - \left( {3b + 1} \right){{\rm{e}}^{ - 3b}}} \right)$ Substituting $a = 1.225 \times {10^9}$ and $b = 0.13$ in the last equation above we get ${\rm{total{\ }mass}} = 2.593 \times {10^{10}}$ Thus, the total mass of the atmosphere contained: $2.593 \times {10^{10}}$ kg.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.