Group Work ~ Appendix H2
1. Find the area of the shaded region.
b. r  sin  4 
r  4  3sin 
a.

1
2

2
14
2
 sin 4  d

20
  4  3sin   d

2
2


1

2
2
 16  24sin   9sin

2

 d

2


1
2
2
 1  cos 
2

 16  24sin   9 


 d

2
14
  1  cos8 d
40



1
2
2

9
9

 16  24sin   2  2 cos  d


2
1
9
9

16  24 cos     sin  

2
2
2

1
  25  9
2

1 4 1  cos8
d
2 0
2
 2
 2
1
sin 8 


4
8 

16

0
4
Sketch the graph of the limacon r 
2.
1
 cos  , then find the are inside the larger loop and
2
outside the smaller loop.
2

0
2 3

4 3
2
11

  cos  

22

2 3
2
2 3
1
1

2

  cos   cos  d

2 2 3  4

2 3

2
11
11


  cos   d     cos   d
22
22


2 3
1
1
1

  cos    cos 2  1 d

2 2 3  4
2


1 3
1

  sin   sin 2 

2 4
4


1  3  4

2  4  3



2

3  3

2
8
4  3 3
8
2 3
Consider the 2n leafed rose r  cos n , where n is even.
a.
Find the area of one leaf.
b.
Find the area of all 2n leaves. What fraction of the circle r  1 does the entire
rose fill up?
a. The leaf begins and ends when the radius is
 3 2 n
1   sin  2n 
 
 
zero. Thus we have.

4
2n
2n

0  cos n
1  3  

3
   
n 
n 
4  2n 2n 
2
2

 3

 ,
4n
2n 2n

The area of one petal is
, where n is even.
3
4n
2n
1
cos 2  n d
2 
3.
2n
3

2 3
2 1
4 

 sin
  2sin
3 2
3 

1
1
3


  cos   cos 2 d

2 2 3  4
2

1

4
2 / 3
2n
  cos  2n   1d
2n
b. Since the rose r  cos n (where n is even) has 2n petals and the area of one petal is

, the
4n
  
area of the entire rose is (2n)    . This means that any rose with an even number of petals
 4n  2
is half the area of a unit circle.
4.
Repeat question 3 where n is odd.
The same logic applies as in problem number three. Thus the area of one petal is

but n is
4n
  
odd. Furthermore, the area of the entire rose is (n)    since r  cos n has only n petals
 4n  4
when n is odd.
5.
Use polar coordinates to find the area of the region inside the circle r  1 and to the right
1
of x  .
2
Recall: x  r cos
1
sec 
Therefore, x  is r 
in polar coordinates.
2
2
sec 
1
The circle r  1 intersects the line r 
when sec  2 or cos   . Thus at
2
2
 
   , . To find the area between the circle and to the right of the line, we can take
3 3
 
the area of the circle from    , and subtract it from the area of the line from
3 3
 
  , .
3 3
 3
 3
1
1
 sec  
1d   
 d

2  3
2  3  2 
1
 
2
 3
1
 
2
 3


3

2
 3
1
  sec 2  d
 3
8  3
1
 tan 
 3
8
3
4
 3
 3