Thomas' Calculus 13th Edition

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

Chapter 15: Multiple Integrals - Section 15.2 - Double Integrals over General Regions - Exercises 15.2 - Page 884: 79


$x^2+2y^2 \leq 4$ This is the region inside the ellipse $x^2+2y^2=4$

Work Step by Step

We have $\iint_{R} (4-x^2-2y^2) dA$ The integral must be a maximum on the region when $4-x^2-2y^2\gt 0$ We will have to multiply with $-1$, a negative number, so that the direction of the inequality reverses. So, $4-x^2-2y^2\leq 0$ or, $x^2+2y^2 \leq 4$ This is the region inside the ellipse $x^2+2y^2=4$
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