Calculus (3rd Edition)

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

Chapter 16 - Multiple Integration - 16.6 Change of Variables - Exercises - Page 905: 16

Answer

$-\frac{1}{2}.$

Work Step by Step

We have $$ \operatorname{Jac}(G)=\frac{\partial(x, y)}{\partial(u, v)}=\left|\begin{array}{ll} {\frac{\partial x}{\partial u}} & {\frac{\partial x}{\partial v}} \\ {\frac{\partial y}{\partial u}} & {\frac{\partial y}{\partial v}} \end{array}\right|=\left|\begin{array}{ll} { v/u} & {\ln u} \\ {2u/v} & {-u^2/v^2} \end{array}\right| =-(u/v)-(2u(\ln u)/v). $$ Now at the point $(u,v)= (1,2)$, we have $ \operatorname{Jac}(G)=-(1/2)-0=-\frac{1}{2}.$
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