Double Integral of two concentric circles

In summary: The Attempt at a SolutionSo I understand that this is two concentric circles(an elementary region) which I can break down into two halves. So what I attempted was to break it into two times the first integral of x from 2^(1/2) to 0 and the y from (1-x^2)^(1/2) to (2-x^2)^(1/2) so for my final solution I got (pi + 2)/4 the correct solution was pi/2 so I looked up the solution and they used the limit of x from 1 to 2^(1/2). That didn't make a whole lot of sense to me
  • #1
robbondo
90
0

Homework Statement


Let D be the region given as the set of (x,y) where 1 <! x^2+y^2 <! 2 and y !<0. Is D an elementary region? Evaluate [tex]\int\int_{D} f(x,y) dA[/tex] where f(x,y) = 1+xy.


Homework Equations




The Attempt at a Solution



So I understand that this is two concentric circles(an elementary region) which I can break down into two halves. So what I attempted was to break it into two times the first integral
of x from 2^(1/2) to 0 and the y from (1-x^2)^(1/2) to (2-x^2)^(1/2) so for my final solution I got (pi + 2)/4 the correct solution was pi/2 so I looked up the solution and they used the limit of x from 1 to 2^(1/2). That didn't make a whole lot of sense to me because x actually goes from 0 to 2^(1/2) doesn't it? Can someone attempt to explain why they chose their limits of integration the way they did? Is my method correct and maybe I just screwed up during the stupid Trig. Substitutions which is also very likely. Thanks~
 
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  • #2
I would do this in polar coordinates, where dA = rdrdtheta, this should make it easier to do the limits of integration and the integration itself. I'm not sure what y!<0 means..., but you r limit on r will be from 1 to sqrt(2)
 
  • #3
I think it can be because they might have different limits from y.

y from (1-x^2)^(1/2) to (2-x^2)^(1/2), and from 0 to (2-x^2)^(1/2)

Why don't you use cylindrical co-od!?
 
  • #4
Sorry I meant 1 <= x^2+y^2 <= 2 and y <= 0. So even if I switch to different coordinate system the main conceptual issue I'm having is why you take the integral of x from square root of two to 1 instead of square root of two to 0.
 
  • #5
if you went from square root of two to 0 you would get the entire part of the circle whose radius spans from the origin to sqrt(2). to change this, you integrate from 1 to sqrt(2). this way you will have a circle with a hole in it that has radius 1. that is what you want, righ?
 
  • #6
I thought that because the limits of integration for y were from (1-x^2)^1/2 to (2-x^2)^1/2 would prevent that from happening...
 
  • #7
Ahh, i see what you're saying, I didn't notice that and was thinking in polar.

Edit, but still x will be from 1 to sqrt(2). You will see this if you draw a picture.
 
  • #8
I guess I'm still a little dissatisfied with the reasoning as to why the limits of integration of x go from 1 to root 2. I asked my TA this afternoon and she assured me that I should use the integral from -root two to root two for x. Now though If I don't use symmetry I'm going to have to break the integral up into two parts. I was told that I cannot use symmetry for this problem because the function f(x) may not be symmetric. I feel like this is something I should have in my book, but it's nowhere to be found. All the example programs deal with solid shapes, not anything like this problem. The weird thing is that the solution I have does use symmetry and does have x from 1 to root 2 and they get the correct answer, so I'm sure that's right but WHY... So far all I've heard is to use polar coordinates which doesn't really anwer my basic question...
 
  • #9
what are their y-limits?

this may help
http://www.ekmpowershop1.com/ekmps/shops/tyrrell123/images/doughnut.gif
I see the way you are thinking.
I think x will go from 1 to sqrt 2
only if you take y as radius of many co-centric circles
and find volume for each tiny area ..
and those co-centric circles go from 1 to sqrt 2

In that dough nut, try to think of many circle with very small intervals, and radius y
 

1. What is a double integral of two concentric circles?

A double integral of two concentric circles is a mathematical concept that involves calculating the volume between two circles that share the same center point. It is represented by a double integral symbol (∫∫) and is a type of multiple integral used in calculus.

2. How is a double integral of two concentric circles calculated?

To calculate the double integral of two concentric circles, the outer and inner radii of the circles must be known. The integral is then solved using the formula: ∫∫ f(x,y) dA = ∫ a b ∫ c d f(x,y) dy dx, where a and b are the limits of integration for the outer circle and c and d are the limits of integration for the inner circle.

3. What is the significance of a double integral of two concentric circles?

The double integral of two concentric circles is often used in physics and engineering to calculate the mass, volume, or surface area of an object with a circular cross-section. It is also used in higher-level calculus to solve more complex problems involving multiple variables.

4. Can the double integral of two concentric circles be extended to three or more circles?

Yes, the concept of a double integral can be extended to three or more concentric circles. This is known as a triple or multiple integral and is used to calculate the volume between multiple concentric shapes.

5. What are some real-life applications of the double integral of two concentric circles?

The double integral of two concentric circles has many real-life applications, such as calculating the volume of a cylindrical tank, determining the mass of a rotating disk, and finding the area of a circular lens. It is also used in fields such as engineering, physics, and economics to solve various mathematical problems.

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