Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 16: Integrals and Vector Fields - Practice Exercises - Page 1029: 26



Work Step by Step

$$ Surface \space Area =\int_0^1 \int_0^1 [(u+v)(u-v)-v^2] \times \sqrt 6 \space du \space dv \\=\sqrt 6 \times \int_0^1 u^2-2v^2 \space du \space dv \\=\sqrt 6 \times \int_0^1 [\dfrac{u^3}{3}-2uv^2]_0^1 \space dv \\ = \sqrt 6[\dfrac{v}{3}-\dfrac{2v^3}{3}]_0^1 \\=-\sqrt{\dfrac{2}{3}}$$
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