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 16.6 - Surface Integrals - Exercises 16.6 - Page 1000: 1


$\dfrac{17 \sqrt {17}-1}{4}$

Work Step by Step

Consider $\vec{r} (x,z) =x i+x^2 j+zk$ So, $\vec{r_x} \times \vec{r_z}=2x i-j +0 k$ and $|\vec{r_x} \times \vec{r_z}| =\sqrt {4x^2+1}$ Now, $\int_{0}^2 \int_{0}^3 6 (x,y,z) \ d \theta =\int_{0}^2 \int_{0}^3 x \sqrt {4x^2+1} \ dz \ dx=\int_0^2 (3x \sqrt {4x^2+1} ) \ dx$ Consider $a =4x^2+1$ and $da=8x dx$ or, $=\dfrac{3}{8} \int_{1}^{17} \sqrt a da$ or, $=\dfrac{3}{8} [\dfrac{2}{3} a^{3/2}]_{1}^{17}$ or, $=\dfrac{17 \sqrt {17}-1}{4}$
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