Calculus (3rd Edition)

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

Chapter 17 - Line and Surface Integrals - Chapter Review Exercises - Page 971: 32

Answer

The total mass is $\mathop \smallint \limits_C^{} {x^2}y{\rm{d}}s = \frac{{40}}{3}$

Work Step by Step

We have the mass density $\rho \left( {x,y} \right) = {x^2}y$ g/cm. In this case, we have two segments: ${{\bf{r}}_1}\left( t \right) = \left( {2t,2} \right)$ and ${{\bf{r}}_2}\left( t \right) = \left( {2,2 - 2t} \right)$ for $0 \le t \le 1$ So, the total mass is the sum of the mass for each segment: $\mathop \smallint \limits_C^{} {x^2}y{\rm{d}}s = \mathop \smallint \limits_{{C_1}}^{} {x^2}y{\rm{d}}s + \mathop \smallint \limits_{{C_2}}^{} {x^2}y{\rm{d}}s$ 1. The first segment: ${{\bf{r}}_1}\left( t \right) = \left( {2t,2} \right)$ for $0 \le t \le 1$. We obtain $\rho \left( {{{\bf{r}}_1}\left( t \right)} \right) = 8{t^2}$ $ds = ||{{\bf{r}}_1}'\left( t \right)||dt = \sqrt {\left( {2,0} \right)\cdot\left( {2,0} \right)} dt = 2dt$ By Eq. (4) in Section 17.2: $\mathop \smallint \limits_C^{} \rho \left( {x,y} \right){\rm{d}}s = \mathop \smallint \limits_a^b \rho \left( {{\bf{r}}\left( t \right)} \right)||{\bf{r}}'\left( t \right)||{\rm{d}}t$ $\mathop \smallint \limits_{{C_1}}^{} {x^2}y{\rm{d}}s = 16\mathop \smallint \limits_0^1 {t^2}{\rm{d}}t = \frac{{16}}{3}{t^3}|_0^1 = \frac{{16}}{3}$ 2. The second segment: ${{\bf{r}}_2}\left( t \right) = \left( {2,2 - 2t} \right)$ for $0 \le t \le 1$. We obtain $\rho \left( {{{\bf{r}}_2}\left( t \right)} \right) = 4\left( {2 - 2t} \right) = 8 - 8t = 8\left( {1 - t} \right)$ $ds = ||{{\bf{r}}_2}'\left( t \right)||dt = \sqrt {\left( {0, - 2} \right)\cdot\left( {0, - 2} \right)} dt = 2dt$ By Eq. (4) in Section 17.2: $\mathop \smallint \limits_C^{} \rho \left( {x,y} \right){\rm{d}}s = \mathop \smallint \limits_a^b \rho \left( {{\bf{r}}\left( t \right)} \right)||{\bf{r}}'\left( t \right)||{\rm{d}}t$ $\mathop \smallint \limits_{{C_2}}^{} {x^2}y{\rm{d}}s = 16\mathop \smallint \limits_0^1 \left( {1 - t} \right){\rm{d}}t = 16\left( {t - \frac{1}{2}{t^2}} \right)|_0^1 = 8$ So, the total mass is $\mathop \smallint \limits_C^{} {x^2}y{\rm{d}}s = \frac{{16}}{3} + 8 = \frac{{40}}{3}$
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