Find Area of Region Inside Cardioid & Outside Circle

[qq545282501]

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:

 

[SammyS]

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 .

 

[Mark44]

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 $-5\pi/6$.

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]

@qq545282501: Have you figured out the correct limit yet? What is your final answer?

 

[qq545282501]

@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 don't 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]

I think @SammyS is correct, -7pi/6 should work, I think it as it goes from -7pi/6 to pi/6, counterclockwise. but i don't 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]

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