Example Video Example ) Find the area enclosed by one loop of the four-leafed rose r = 11 cos(20). Solution 0 = - The curve r = 11 cos(20) is sketched in the figure above. Notice from the figure that the region enclosed by the right loop is swept out by a ray that rotates from 0 = - * to e = *. Therefore A = 1,2 do gives us the following. A = de do + cos(40)) de

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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Example
Video Example )
Find the area enclosed by one loop of the four-leafed rose r = 11 cos(20).
Solution
9.
2 de gives us the following.
2
The curve r = 11 cos(20) is sketched in the figure above. Notice from the figure that the region enclosed by the right loop is swept out by a ray that rotates from 0 = -
* to 0 =
Therefore A =
4
4
▪ n / 4
1 2
do
A =
-n/4
(1 / 4
1
de
-1/4
▪n / 4
= 121
de
n/ 4
= 121
(1 + cos(40)) d0
J7/ 4
121
2
||
Transcribed Image Text:Example Video Example ) Find the area enclosed by one loop of the four-leafed rose r = 11 cos(20). Solution 9. 2 de gives us the following. 2 The curve r = 11 cos(20) is sketched in the figure above. Notice from the figure that the region enclosed by the right loop is swept out by a ray that rotates from 0 = - * to 0 = Therefore A = 4 4 ▪ n / 4 1 2 do A = -n/4 (1 / 4 1 de -1/4 ▪n / 4 = 121 de n/ 4 = 121 (1 + cos(40)) d0 J7/ 4 121 2 ||
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