Answer
$\sqrt 6$
Work Step by Step
Here, $r_u=1+j; r_v=1-j+k$
Then
$|r_u \times r_v|=\sqrt 6$
The surface ares is given by: $S=\iint_S |r_u \times r_v| du dv$
Thus, $S=\int_0^1 \int_0^1 \sqrt 6 du dv =\sqrt 6(1-0)=\sqrt 6$
University Calculus: Early Transcendentals (3rd Edition)
by Hass, Joel R.; Weir, Maurice D.; Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0321999584
ISBN 13:
978-0-32199-958-0
Chapter 15 - Practice Exercises - Page 909: 25
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