Thomas' Calculus 13th Edition

Published by Pearson

Chapter 16: Integrals and Vector Fields - Section 16.6 - Surface Integrals - Exercises 16.6 - Page 1000: 13

Answer

$2$

Work Step by Step

The surface integral can be found as: $dS= \sqrt {(f_x)^2 +(f_y)^2 +1} \ dy \ dx=\sqrt {(-2)^2 +(-2)^2 +1} \ dy \ dx =3$ We set up the integral and solve the surface integral as follows: $\iint_{S} G (x,y,z) \ dS=\int_0^1 \int_0^{1-x} (2-x-y) (3) \ dy \ dx$ or, $=3 \int_0^1 [2y-xy-\dfrac{y^2}{2}]_0^{1-x} \ dx$ or, $=3 \int_0^1 [2-2x-x+x^2-\dfrac{(1-2x+x^2)}{2}] \ dx$ or, $=3 \int_0^1 (2-2x-x+x^2-\dfrac{1}{2}+x-\dfrac{x^2}{2}] \ dx$ or, $=3 \int_0^1 (\dfrac{x^2}{2}-2x+\dfrac{3}{2}] \ dx$ or, $=3 \times \dfrac{2}{3}$ or, $=2$

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