Calculus (3rd Edition)

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

Chapter 17 - Line and Surface Integrals - 17.4 Parametrized Surfaces and Surface Integrals - Exercises - Page 957: 2

Answer

We show that $G\left( {r,\theta } \right) = \left( {r\cos \theta ,r\sin \theta ,1 - {r^2}} \right)$ parametrizes the paraboloid $z = 1 - {x^2} - {y^2}$. The grid curves of this parametrization (see the figure attached): 1. $\theta $-grid curve: $G\left( {{r_0},\theta } \right) = \left( {{r_0}\cos \theta ,{r_0}\sin \theta ,1 - {r_0}^2} \right)$ This is a circle of radius ${r_0}$ at the height $z = 1 - {r_0}^2$. 2. $r$-grid curve: $G\left( {r,{\theta _0}} \right) = \left( {r\cos {\theta _0},r\sin {\theta _0},1 - {r^2}} \right)$ This is a parabola with maximum point at $\left( {0,0,1} \right)$.

Work Step by Step

We have $x = r\cos \theta $, $y = r\sin \theta $, and $z = 1 - {r^2}$. Since $z = 1 - {x^2} - {y^2} = 1 - {r^2}{\cos ^2}\theta - {r^2}{\sin ^2}\theta = 1 - {r^2}$ So, $G\left( {r,\theta } \right) = \left( {r\cos \theta ,r\sin \theta ,1 - {r^2}} \right)$ parametrizes the paraboloid $z = 1 - {x^2} - {y^2}$. The grid curves of this parametrization through $P = \left( {{r_0},{\theta _0}} \right)$ are (see the figure attached) 1. $\theta $-grid curve: $G\left( {{r_0},\theta } \right) = \left( {{r_0}\cos \theta ,{r_0}\sin \theta ,1 - {r_0}^2} \right)$ This is a circle of radius ${r_0}$ at the height $z = 1 - {r_0}^2$. 2. $r$-grid curve: $G\left( {r,{\theta _0}} \right) = \left( {r\cos {\theta _0},r\sin {\theta _0},1 - {r^2}} \right)$ This is a parabola with maximum point at $\left( {0,0,1} \right)$.
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