Double integrals over general regions.

In summary: A, use polar coordinates to find the double integrals In summary, Deveno is trying to solve a homework problem involving equations in polar coordinates. He has drawn the circles and set the two equations to equal each other. He is unsure how to find the intervals of θ and is not sure about the intervals of r. He asks for any help. He integrates the polar form of the equations from θ=0
  • #1
juronimo
5
0

Homework Statement



The domain D is the intersection of two disks x^2 +y^2 = 1 and x^2 + (y-1)^2 =1

use polar coordinates to find the double integral ∫∫(x)dA

Homework Equations



x = rcosθ y = rsinθ r^2 = x^2 + y^2

The Attempt at a Solution



I have drawn the circles. also i set the two circle equations to equal each other.

x^2 +y^2 = x^2 + (y-1)^2

which yielded points of intersection.. [-√(3/4), (1/2)] and [√(3/4),(1/2)]

however i am unsure how to find my intervals of θ and I'm not sure of my intervals of r. I think it may be 0≤r≤1.


any help would be appreciated, even just to put me on the right track.. I have been looking over this problem for 45 minutes now
 
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  • #2
im looking at it now. could i use simple geometry to find the intervals of theta?

Pythagorean theorem to get the angle on each side with respect to the x axis?
 
  • #3
it should be obvious that from symmetry, we can just let θ run from 0 to π/2, and double the result.

also, after θ = π/6, r is just a constant function r(θ) = 1.

so the real challenge is capturing that thin sliver under the line y = (1/√3)x. the lower limit of r will be 0, of course, but the upper limit will be the circle centered at (0,1).

use x = rcos(θ) and y = rsin(θ) to derive a formula for r in terms of θ for the upper circle (it should be a fairly nice function of θ), which will give you your upper limit for r.

then just integrate the polar form (times r dr dθ, as the polar form of dA) of your function from θ = 0 to θ = π/6, and r = 0 to (upper limit from above). after that, you just have a pie slice (nifty pun, eh?) of π/3 radians, to integrate your function over (and then double everything).
 
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  • #4
I agree with Deveno. Do this integral in three parts. For [itex]\theta[/itex] from 0 to [itex]\pi/6[/itex], your boundary is the upper circle, [itex]x^2+ (y-1)^2= 1[/itex] or [itex]x^2+ y^2- 2y+ 1= r^2- 2r sin(\theta)+ 1= 1[/itex] which reduces to [itex]r^2= 2rsin(\theta)[/itex]. Since r is not 0 anywhere except (0, 0), we must have [itex]r= 2sin(\theta)[/itex]. That is, for [itex]\theta[/itex] from 0 to [itex]\pi/2[/itex], integrate with r from 0 to [itex]2sin(\theta)[/itex].

For [itex]\theta[/itex] from 0 to [itex]\pi- \pi/6= 5\pi/6[/itex], the boundary is the circle [itex]x^2+ y^2= r^2= 1[/itex] so the "r" integral is from 0 to 1. Finally, for [itex]\theta[/itex] from [itex]5\pi/6[/itex] to [itex]\pi[/itex], use r from 0 to [itex]2sin(\theta)[/itex] again.
 
  • #5
I'm adding all 3 integrals up to compute my final area?
 
  • #6
i see it more of as, computing 0 to pi/2 and having my upperbound be my bottom circle.
and then computing 0 to pi/6 and having my upperbound be my top circle.

subtracting the second from the first and then doubling it.
 
  • #7
juronimo said:
i see it more of as, computing 0 to pi/2 and having my upper-bound be my bottom circle.
and then computing 0 to pi/6 and having my upper-bound be my top circle.

subtracting the second from the first and then doubling it.
What do you mean by the "upper circle"? ... lower circle?
 
  • #8
by "upper circle" i believe he means the one centered at (0,1).

however, if he wants to compute an integral over an area to be subtracted from the integral over the (lower) semi-circle, the "lower bound" for r, needs to be
r = 2 sin(θ), because the area to be subtracted is the part "outside" the upper circle.

see here (and ignore the funky scaling these are circles, not ellipses): http://www.wolframalpha.com/input/?i=plot{(x^2+y^2=1),(x^2+(y-1)^2=1),(y=(1/√3)x)}
 
  • #9
correct, i want to compute the area of the oval type shape in the middle.

i took the integral with 0 ≤ r ≤ 2sin(θ) 0≤ θ ≤ ∏/2

∫∫ (r2cosθ) dr dθ

and got 2/3


then i took the integral
(5/6)∏ ≤θ ≤ ∏ 0 ≤ r ≤ 2sinθ

∫∫(r2cosθ) dr dθ

and got -1/24



i took the third integral.
0 ≤ θ ≤ (5/6)∏ 0≤r≤1
∫∫ (r2cosθ) dr dθ

and got -1/6.



im stumped after that, i feel as though i would add them all up.

doing so i got my total area to be 11/24
 
  • #10
i think you are taking the integral over the wrong regions.

above the line y = x/√3, the region in question is entirely within the unit circle. this line corresponds to θ= π/6 in the upper two quadrants.

so we first split our integral:

[tex]\int \int_D x dA = \int \int_D r^2cos(\theta) dr d\theta[/tex]

[tex]=2\left(\int_0^{\pi/6}\int_0^{2sin(\theta)} r^2cos(\theta) drd\theta + \int_{\pi/6}^{\pi/2}\int_0^1 r^2cos(\theta) drd\theta\right)[/tex]

that is, we are splitting half of D into:

[tex]D_1 = \{(x,y) : x^2+y^2 \leq 1,\ y > \frac{x}{\sqrt{3}},\ x \geq 0\}[/tex] and

[tex]D_2 = \{(x,y) : x^2 + (y-1)^2 \leq 1,\ y \leq \frac{x}{\sqrt{3}}\}[/tex]

these are disjoint, so the integral over half of D is the integral over D1 + the integral over D2.

i don't believe your answer is correct.
 

1. What is a double integral over a general region?

A double integral over a general region is a mathematical tool used to calculate the area under a surface or between two surfaces in three-dimensional space. It involves finding the volume between two surfaces by integrating over the region of interest.

2. How is a double integral over a general region different from a regular double integral?

A double integral over a general region is different from a regular double integral because the region of integration is not limited to a rectangular or square shape. Instead, it can be any shape in two-dimensional space, making it a more versatile tool for calculating volume.

3. What are some common applications of double integrals over general regions?

Double integrals over general regions are commonly used in physics, engineering, and other scientific fields to calculate volumes of irregularly shaped objects or to find the mass, center of mass, or moment of inertia of an object. They are also used in economics and statistics to determine probabilities and expected values.

4. How do you set up a double integral over a general region?

To set up a double integral over a general region, you first need to determine the limits of integration for both the inner and outer integrals. This involves finding the equations of the curves that form the boundaries of the region. Then, you can use the appropriate integration method (such as rectangular, polar, or cylindrical coordinates) to calculate the integral.

5. Are there any limitations or challenges to using double integrals over general regions?

One limitation of using double integrals over general regions is that the boundaries of the region must be known and described by equations. This may be difficult or impossible for very complex or irregularly shaped regions. Additionally, setting up the integral and calculating the limits of integration can be challenging and time-consuming for certain regions.

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