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 - Section 16.1 - Line Integrals - Exercises 16.1 - Page 944: 31


$\dfrac{1}{6} (17^{3/2}-1)$

Work Step by Step

As we know that the surface area is: $ds=\sqrt{(dx)^2+(dy)^2} dt$ or, $ds=\sqrt{1+4x^2} dx$ Now, $S=\int_C (x+\sqrt y) ds=\int_{0}^{2} x+\sqrt {x^2}(\sqrt{1+4x^2} dx)$ or, $S= 2 \int_{0}^{2} (x\sqrt{1+4x^2}) dx$ Let $1+4x^2 =p \implies x dx =\dfrac{1}{8} dp$ or, $S=\dfrac{1}{6} [(p)^{3/2}]_{1}^{17}$ Thus, surface area $S=\dfrac{1}{6} (17^{3/2}-1)$
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