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 1001: 20



Work Step by Step

Consider $\vec{r} (x,y) =x i+x^2 j+z k$ So, $\vec{r_x} \times \vec{r_y}=2x i-j$ Now, $F \cdot n d \theta =F \cdot \dfrac{\vec{r_x} \times \vec{r_y}}{|\vec{r_x} \times \vec{r_y}|} \ dz \ dx =-x^2 \ dz \ dx$ $|\vec{r_x} \times \vec{r_y}| =\sqrt {(2x)^2+(-1)^2}=\sqrt {4x^2+1}$ Now, $\iint_{S} F (x,y,z) \ d \theta=\int_{-1}^{1} \int_{0}^{2} -x^2 \ dz \ dx$ or, $=\int_{-1}^{1}[(-x^2 ) z ]_{0}^{2} \ dz \ dx$ or, $=\int_{-1}^{1}-2x^2 \ dx$ or, $=-2 [\dfrac{x^3}{3}]_{-1}^1$ Thus, we evaluate the integral as: $\int_{-1}^{1} \int_{0}^{2} -x^2 \ dz \ dx=-\dfrac{4}{3}$
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