Double Integral of function in region bounded by two circles

  • #1
songoku
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Homework Statement
Let R be the region which lies inside the circle and outside the circle
a) Sketch the two circles and shade the region R in your diagram
b) Evaluate the integral
Relevant Equations


The polar form of is and polar form of is

1732704918952.png


My idea is to divide the working into two parts:
1) find the integral in 1st quadrant and multiply by 2 to include the region in 4th quadrant
2) find the integral in 2nd quadrant and multiply by 2 to include the region in 3rd quadrant

Integral in 1st quadrant:



Integral in 2nd quadrant:


Am I correct? Thanks
 
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  • #2
My strategy is integral in red - integral in blue.
 
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  • #3
anuttarasammyak said:
My strategy is integral in red - integral in blue.
1) When doing integral for blue part, I got zero. What is my mistake?

2) Why can't I use to get polar form of ? The center is (0, 0) so and are both zero and when I plug it in I get

Thanks
 
  • #4
The blue part seems to be

where r and are from center of blue circle whose radius is a.
 
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  • #5
anuttarasammyak said:
For blue I got

where r is radius of blue circle, is radius angle measured at its center.
My attempt for blue circle:






Where is mistake in my working?

I also don't understand why you multiply by 2. Is integral from 0 to already represents the whole circle?

Thanks
 
  • #6
My bad. I had reposted #4.

In your approach denominator in the integrand be

Which is same with mine.
 
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  • #7
songoku said:
1) When doing integral for blue part, I got zero. What is my mistake?

must be positive. is negative for .

The parametrization of is therefore

2) Why can't I use to get polar form of ? The center is (0, 0) so and are both zero and when I plug it in I get

Thanks
The general equation for a conic section of eccentricity and semi-latus rectum with a focus at the orign is not but If you expand the denominator in binomial series for you will see that for it has far more terms in its fourier series than just the first three given above. For it is constant, as one would expect.
 
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  • #8
anuttarasammyak said:
The blue part seems to be

where r and are from center of blue circle whose radius is a.
1) Why do we need to integrate it from o to ? From , shouldn't we integrate from 0 to and multiply by 2?

2) Is it possible to integrate the expression I got in the last line of post #5?

3) Why is my method in OP not working?

Thanks
 
  • #9
songoku said:
1) Why do we need to integrate it from o to 2π? From r=10cos⁡θ, shouldn't we integrate from 0 to π2 and multiply by 2?

2) Is it possible to integrate the expression I got in the last line of post #5?

3) Why is my method in OP not working?
1) Please find attached the figure to show the integration
1732858015038.png

2) Sure, but it is not a solution to the problem.
3) Let us check your method and mine. What is the result value ?
 
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  • #10
anuttarasammyak said:
1) Please find attached the figure to show the integration
View attachment 353962
The is not between and horizontal?

1732872120929.png
 
  • #11
songoku said:
The θ is not between r and horizontal?
The angle in my integral is as figured whatever name it could have.
 
  • #12
anuttarasammyak said:
The angle in my integral is as figured whatever name it could have.
If I use , the angle is always measured from the center of circle?
 
  • #13
Yes, that seems to be same as my figure.
 
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  • #14
If you do parametrize the blue circle as then you end up with which is a much less tractable integral than using the parametrization
 
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  • #15
songoku said:
Homework Statement: Let R be the region which lies inside the circle and outside the circle
a) Sketch the two circles and shade the region R in your diagram
b) Evaluate the integral
Relevant Equations:



The polar form of is and polar form of is

View attachment 353914

My idea is to divide the working into two parts:
1) find the integral in 1st quadrant and multiply by 2 to include the region in 4th quadrant
2) find the integral in 2nd quadrant and multiply by 2 to include the region in 3rd quadrant

Integral in 1st quadrant:



Integral in 2nd quadrant:


Am I correct? Thanks
What you have in the OP should work just fine.

Was there some problem with this?
 
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  • #16
SammyS said:
What you have in the OP should work just fine.

Was there some problem with this?
I am just not sure whether my working is correct.

Thank you for all the help and explanation anuttarasammyak, pasmith, SammyS
 
  • #17
@songoku your way works. What value have you got ? I would like to comapre your way and mine to get some good insights.
 
  • #18
anuttarasammyak said:
@songoku your way works. What value have you got ? I would like to comapre your way and mine to get some good insights.
I got
 
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  • #19
@songoku Thanks.
where a=5. Contribution of the blue part is 4a. That should be equal to my integral so


We get troublesome definite integral value by your smart way.
 
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  • #20
anuttarasammyak said:
@songoku Thanks.
where a=5. Contribution of the blue part is 4a. That should be equal to my integral so


We get troublesome definite integral value by your smart way.
In the same manner we may easily expand it to

for p>0
 
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