The graph of r =1+sin 0 is shown here. The first quadrant region of the graph is shaded. We could find the area of the shaded region by integrating . (1+ sin 0)² d0. Positions labeled a and b on the figure correspond to values of 0 that are lower and upper limits of integration. [ Select ] v pi OR 1 OR O a = • b= [Select ] pi OR 2 OR pi/2

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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The graph of r =1+sin 0 is shown here. The first quadrant region of the graph is shaded.
We could find the area of the shaded region by integrating . (1+ sin 0)² d0.
Positions labeled a and b on the figure correspond to values of 0 that are lower and upper limits of integration.
[ Select ]
v pi OR 1 OR O
a =
• b= [Select ]
pi OR 2 OR pi/2
Transcribed Image Text:The graph of r =1+sin 0 is shown here. The first quadrant region of the graph is shaded. We could find the area of the shaded region by integrating . (1+ sin 0)² d0. Positions labeled a and b on the figure correspond to values of 0 that are lower and upper limits of integration. [ Select ] v pi OR 1 OR O a = • b= [Select ] pi OR 2 OR pi/2
The graph of r =1+sin 0 is shown here.
The right half region of the graph is shaded.
We could find the area of the shaded region by integrating (1+ sin 0)² d0.
Positions labeled a and b on the figure correspond to values of 0 that are lower and upper limits of integration.
[ Select ]
|-pi OR O OR -pi/2
• b= [Select ]
pi OR 2 OR pi/2
Transcribed Image Text:The graph of r =1+sin 0 is shown here. The right half region of the graph is shaded. We could find the area of the shaded region by integrating (1+ sin 0)² d0. Positions labeled a and b on the figure correspond to values of 0 that are lower and upper limits of integration. [ Select ] |-pi OR O OR -pi/2 • b= [Select ] pi OR 2 OR pi/2
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