Is this correct?

  • #1

Homework Statement


set up a double integral to find the area of the region inside the cardioid r=2-2sin θ and outside the circle r=1.

Homework Equations




The Attempt at a Solution


setting the two equations equal to each other, I got theta = pi/6 , 5pi/6

Area= ∫-5pi/6 to pi/6 ∫1 to 2-2sinθ rdrdθ

I am not sure if my bounds for the lower angle is correct or not.

any help is appreciated

here is the graph:
Snapshot.jpg
 

Answers and Replies

  • #2
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,317
1,007

Homework Statement


set up a double integral to find the area of the region inside the cardioid r=2-2sin θ and outside the circle r=1.

Homework Equations




The Attempt at a Solution


setting the two equations equal to each other, I got theta = pi/6 , 5pi/6

Area= ∫-5pi/6 to pi/6 ∫1 to 2-2sinθ rdrdθ

I am not sure if my bounds for the lower angle is correct or not.

any help is appreciated

here is the graph:
View attachment 91714
The -5π/6 is not correct.

-5π/6 + 2π = 7π/6 is at the same location as -5π/6 .

You need a negative angle which terminates at the same place as 5π/6 .
 
  • #3
33,934
5,582

Homework Statement


set up a double integral to find the area of the region inside the cardioid r=2-2sin θ and outside the circle r=1.

Homework Equations




The Attempt at a Solution


setting the two equations equal to each other, I got theta = pi/6 , 5pi/6

Area= ∫-5pi/6 to pi/6 ∫1 to 2-2sinθ rdrdθ
Your graph shows intersections for ##\theta = \pi/6## and ##\theta = 5\pi/6##. Above you have ##-5\pi/6##.
qq545282501 said:
I am not sure if my bounds for the lower angle is correct or not.

any help is appreciated

here is the graph:
View attachment 91714
 
  • #4
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,554
765
@qq545282501: Have you figured out the correct limit yet? What is your final answer?
 
  • #5
@qq545282501: Have you figured out the correct limit yet? What is your final answer?
I think @SammyS is correct, -7pi/6 should work, I think it as it goes from -7pi/6 to pi/6, counterclockwise. but i dont understand the need to add 2pi, since 7pi/6 is the actual angle if you go clockwise from 0.
I didnt understand what @Mark44 said, or he is misunderstanding my question.

what do you think?
 
  • #6
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,317
1,007
I think @SammyS is correct, -7pi/6 should work, I think it as it goes from -7pi/6 to pi/6, counterclockwise. but i dont understand the need to add 2pi, since 7pi/6 is the actual angle if you go clockwise from 0.
I didnt understand what @Mark44 said, or he is misunderstanding my question.

what do you think?
Yes, "it goes from -7pi/6 to pi/6, counterclockwise".

If you are referring to my adding 2π, I was merely pointing out that -5π/6 is co-terminal with 7π/6 and that is incorrect for a limit of integration.
 
  • Like
Likes qq545282501
  • #7
Yes, "it goes from -7pi/6 to pi/6, counterclockwise".

If you are referring to my adding 2π, I was merely pointing out that -5π/6 is co-terminal with 7π/6 and that is incorrect for a limit of integration.
oh ! i see, anyway, thank you all ! big Love
 
  • #8
33,934
5,582
I didnt understand what @Mark44 said, or he is misunderstanding my question.
What I said was that in your drawing, you show one of the angles as being ##5\pi/6##, but in your integral (below), you are using ##-5\pi/6## as a limit of integration.
Area= ∫-5pi/6 to pi/6 ∫1 to 2-2sinθ rdrdθ
 
  • Like
Likes qq545282501

Related Threads on Is this correct?

  • Last Post
Replies
3
Views
878
  • Last Post
Replies
10
Views
1K
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
3
Views
995
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
2
Views
999
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
1K
Top