Suppose R is the shaded region in the figure. AS an iterated integral in polar coordinates, B /| f(x, y) dA = f(r cos(0), r sin(0)) r dr de R with limits of integration 1 A = -pi/4 -1 B = -(3pi)/4 -2 C = -3 D = sqrt(18) -4 (Click on graph to enlarge)
Suppose R is the shaded region in the figure. AS an iterated integral in polar coordinates, B /| f(x, y) dA = f(r cos(0), r sin(0)) r dr de R with limits of integration 1 A = -pi/4 -1 B = -(3pi)/4 -2 C = -3 D = sqrt(18) -4 (Click on graph to enlarge)
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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
Transcribed Image Text:Suppose R is the shaded region in the figure. As an iterated integral in polar coordinates,
B
D
/| f(x, y) dA =
f(r cos(0), r sin(0)) r dr d0
R
with limits of integration
A = -pi/4
-1
B = -(3pi)/4
-2
C =
-3
D = sqrt(18)
-4
-3
(Click on graph to enlarge)
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