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 ##x^2+y^2=100## and outside the circle ##x^2-10x+y^2=0##
a) Sketch the two circles and shade the region R in your diagram
b) Evaluate the integral ##\iint_R \frac{1}{\sqrt{x^2+y^2}}~dA##
Relevant Equations
##\iint_R f(x,y)~dA=\int_{\alpha}^{\beta} \int^{b}_{a} f(r,\theta) r~dr~d\theta##

##r=2a \cos\theta+2b\sin\theta##
The polar form of ##x^2+y^2=100## is ##r=10## and polar form of ##x^2-10x+y^2=0## is ##r=10 \cos\theta##

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:
$$\int_{0}^{\frac{\pi}{2}} \int^{10}_{10\cos\theta} \frac{1}{r}r~dr~d\theta$$
$$=\int_{0}^{\frac{\pi}{2}} \int^{10}_{10\cos\theta} dr~d\theta$$

Integral in 2nd quadrant:
$$\int_{\frac{\pi}{2}}^{\pi} \int^{10}_{0} dr~d\theta$$

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 ##\int_{0}^{\pi} \int_{0}^{10\cos\theta} dr~d\theta## for blue part, I got zero. What is my mistake?

2) Why can't I use ##r=2a\cos\theta +2b\sin\theta## to get polar form of ##x^2+y^2=100##? The center is (0, 0) so ##a## and ##b## are both zero and when I plug it in I get ##r=0##

Thanks
 
  • #4
The blue part seems to be
[tex]\int_0^{2\pi} d\theta \int_0^a dr \frac{r}{\sqrt{a^2+r^2+2ar\cos\theta}} [/tex]
where r and ##\theta## are from center of blue circle whose radius is a.
 
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  • #5
anuttarasammyak said:
For blue I got
[tex]2\int_0^\pi \frac{1}{2r\cos \frac{\theta}{2}}\frac{1}{2}r^2 d\theta[/tex]
where r is radius of blue circle, ##\theta## is radius angle measured at its center.
My attempt for blue circle:
$$\iint_R \frac{1}{\sqrt{x^2+y^2}} dA$$
$$=\int_{0}^{\pi} \int_{0}^{10\cos\theta} \frac{1}{\sqrt{(5+5\cos\theta)^2 + (5\sin\theta)^2}}r~dr~d\theta$$
$$=\int_{0}^{\pi} \int_{0}^{10\cos\theta} \frac{1}{\sqrt{50+50\cos\theta}}r~dr~d\theta$$
$$=\int_{0}^{\pi} \frac{1}{\sqrt{50}} \frac{1}{\sqrt{1+\cos\theta}} \left.\frac{1}{2}r^2 \right|_{0}^{10\cos\theta}d~\theta$$
$$=\int_{0}^{\pi} \frac{5\cos^2 \theta}{\cos \frac{\theta}{2}}d~\theta$$

Where is mistake in my working?

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

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

In your approach denominator in the integrand be
[tex]\sqrt{(a+r cos \theta)^2+(r sin \theta)^2}[/tex]
Which is same with mine.
[tex]\int_0^a dr [/tex]
 
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  • #7
songoku said:
1) When doing integral ##\int_{0}^{\pi} \int_{0}^{10\cos\theta} dr~d\theta## for blue part, I got zero. What is my mistake?

[itex]r[/itex] must be positive. [itex]\cos \theta[/itex] is negative for [itex]\frac\pi2 < \theta \leq \pi[/itex].

The parametrization of [itex]x^2 - 10xy + y^2 = 0[/itex] is therefore [tex]r = R(\theta) = \begin{cases}
0 & -\pi \leq \theta < -\tfrac12 \pi \\
10\cos \theta & -\tfrac12 \pi \leq \theta \leq \tfrac 12 \pi \\
0 & \tfrac12 \pi < \theta \leq \pi \end{cases}[/tex]

2) Why can't I use ##r=2a\cos\theta +2b\sin\theta## to get polar form of ##x^2+y^2=100##? The center is (0, 0) so ##a## and ##b## are both zero and when I plug it in I get ##r=0##

Thanks
The general equation for a conic section of eccentricity [itex]e[/itex] and semi-latus rectum [itex]\ell[/itex] with a focus at the orign is not [itex]r(\theta) = a\cos \theta + b \sin \theta + c[/itex] but [tex]
r(\theta) = \frac{\ell}{1 + e\cos(\theta - \theta_0)}.[/tex] If you expand the denominator in binomial series for [itex]e < 1[/itex] you will see that for [itex]e > 0[/itex] it has far more terms in its fourier series than just the first three given above. For [itex]e = 0[/itex] it is constant, as one would expect.
 
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  • #8
anuttarasammyak said:
The blue part seems to be
[tex]\int_0^{2\pi} d\theta \int_0^a dr \frac{r}{\sqrt{a^2+r^2+2ar\cos\theta}} [/tex]
where r and ##\theta## are from center of blue circle whose radius is a.
1) Why do we need to integrate it from o to ##2\pi##? From ##r=10\cos\theta##, shouldn't we integrate from 0 to ##\frac{\pi}{2}## 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 ##\theta## is not between ##r## 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 ##x=a+r\cos\theta##, 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 [itex]\{(5 + r\cos \theta, r \sin \theta) : r \in [0,5]. \theta \in [0, 2\pi)\}[/itex] then you end up with [tex]\iint_{\mbox{blue circle}} \frac{1}{\sqrt{x^2 + y^2}}\,dx\,dy =
\int_0^{2\pi} \int_0^5 \frac{r}{\sqrt{(r + 5 \cos \theta)^2 + 25\sin^2 \theta}}\,dr\,d\theta[/tex] which is a much less tractable integral than [itex]\int_{-\pi/2}^{\pi/2} \int_0^{10 \cos \theta}\,dr\,d\theta[/itex] using the parametrization [itex]\{ (r \cos \theta, r \sin \theta) : 0 \leq r \leq 10 \cos \theta, -\frac \pi 2 \leq \theta \leq \frac \pi 2 \}.[/itex]
 
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  • #15
songoku said:
Homework Statement: Let R be the region which lies inside the circle ##x^2+y^2=100## and outside the circle ##x^2-10x+y^2=0##
a) Sketch the two circles and shade the region R in your diagram
b) Evaluate the integral ##\iint_R \frac{1}{\sqrt{x^2+y^2}}~dA##
Relevant Equations: ##\iint_R f(x,y)~dA=\int_{\alpha}^{\beta} \int^{b}_{a} f(r,\theta) r~dr~d\theta##

##r=2a \cos\theta+2b\sin\theta##

The polar form of ##x^2+y^2=100## is ##r=10## and polar form of ##x^2-10x+y^2=0## is ##r=10 \cos\theta##

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:
$$\int_{0}^{\frac{\pi}{2}} \int^{10}_{10\cos\theta} \frac{1}{r}r~dr~d\theta$$
$$=\int_{0}^{\frac{\pi}{2}} \int^{10}_{10\cos\theta} dr~d\theta$$

Integral in 2nd quadrant:
$$\int_{\frac{\pi}{2}}^{\pi} \int^{10}_{0} dr~d\theta$$

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 ##20\pi-20##
 
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  • #19
@songoku Thanks.
[tex]20\pi-20=4a\pi-4a[/tex] where a=5. Contribution of the blue part is 4a. That should be equal to my integral so
[tex]a \int_0^{2\pi} d\theta \int_0^1 dx \frac{x}{(1+x^2-2x \cos\theta)^{\frac{1}{2}}}=4a[/tex]
[tex] \int_0^{2\pi} d\theta \int_0^1 dx \frac{x}{(1+x^2-2x \cos\theta)^{\frac{1}{2}}}=4[/tex]
We get troublesome definite integral value by your smart way.
 
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  • #20
anuttarasammyak said:
@songoku Thanks.
[tex]20\pi-20=4a\pi-4a[/tex] where a=5. Contribution of the blue part is 4a. That should be equal to my integral so
[tex]a \int_0^{2\pi} d\theta \int_0^1 dx \frac{x}{(1+x^2-2x \cos\theta)^{\frac{1}{2}}}=4a[/tex]
[tex] \int_0^{2\pi} d\theta \int_0^1 dx \frac{x}{(1+x^2-2x \cos\theta)^{\frac{1}{2}}}=4[/tex]
We get troublesome definite integral value by your smart way.
In the same manner we may easily expand it to
[tex] \int_0^{2\pi} d\theta \int_0^1 dx \frac{x}{(1+x^2-2x \cos\theta)^{\frac{1-p}{2}}}=\frac{2^{2+p}}{1+p}\int_0^\frac{\pi}{2}\cos^{1+p}\theta \ d\ \theta[/tex]
for p>0
 
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