# Is this correct?

## 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.

## 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:

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SammyS
Staff Emeritus
Homework Helper
Gold Member

## 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.

## 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 .

Mark44
Mentor

## 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.

## 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

LCKurtz
Homework Helper
Gold Member
@qq545282501: Have you figured out the correct limit yet? What is your final answer?

@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?

SammyS
Staff Emeritus
Homework Helper
Gold Member
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.

qq545282501
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

Mark44
Mentor
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

qq545282501