Double Integral of function in region bounded by two circles

[songoku]

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 $r=10 \cos\theta$

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

 

[anuttarasammyak]

My strategy is integral in red - integral in blue.

 

[songoku]

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

 

[anuttarasammyak]

The blue part seems to be
$$\int_0^{2\pi} d\theta \int_0^a dr \frac{r}{\sqrt{a^2+r^2+2ar\cos\theta}}$$
where r and $\theta$ are from center of blue circle whose radius is a.

 

[songoku]

For blue I got
$$2\int_0^\pi \frac{1}{2r\cos \frac{\theta}{2}}\frac{1}{2}r^2 d\theta$$
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

 

[anuttarasammyak]

My bad. I had reposted #4.

In your approach denominator in the integrand be
$$\sqrt{(a+r cos \theta)^2+(r sin \theta)^2}$$
Which is same with mine.
$$\int_0^a dr$$

 

[pasmith]

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?

$r$ must be positive. $\cos \theta$ is negative for $\frac\pi2 < \theta \leq \pi$.

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

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 $e$ and semi-latus rectum $\ell$ with a focus at the orign is not $r(\theta) = a\cos \theta + b \sin \theta + c$ but $$r(\theta) = \frac{\ell}{1 + e\cos(\theta - \theta_0)}.$$ If you expand the denominator in binomial series for $e < 1$ you will see that for $e > 0$ it has far more terms in its fourier series than just the first three given above. For $e = 0$ it is constant, as one would expect.

 

[songoku]

The blue part seems to be
$$\int_0^{2\pi} d\theta \int_0^a dr \frac{r}{\sqrt{a^2+r^2+2ar\cos\theta}}$$
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

 

[anuttarasammyak]

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

2) Sure, but it is not a solution to the problem.
3) Let us check your method and mine. What is the result value ?

 

[songoku]

1) Please find attached the figure to show the integration
View attachment 353962

The $\theta$ is not between $r$ and horizontal?

 

[anuttarasammyak]

The θ is not between r and horizontal?

The angle in my integral is as figured whatever name it could have.

 

[songoku]

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?

 

[anuttarasammyak]

Yes, that seems to be same as my figure.

 

[pasmith]

If you do parametrize the blue circle as $\{(5 + r\cos \theta, r \sin \theta) : r \in [0,5]. \theta \in [0, 2\pi)\}$ then you end up with $$\iint_{\mbox{blue circle}} \frac{1}{\sqrt{x^2 + y^2}}\,dx\,dy = \int_0^{2\pi} \int_0^5 \frac{r}{\sqrt{(r + 5 \cos \theta)^2 + 25\sin^2 \theta}}\,dr\,d\theta$$ which is a much less tractable integral than $\int_{-\pi/2}^{\pi/2} \int_0^{10 \cos \theta}\,dr\,d\theta$ using the parametrization $\{ (r \cos \theta, r \sin \theta) : 0 \leq r \leq 10 \cos \theta, -\frac \pi 2 \leq \theta \leq \frac \pi 2 \}.$

 

[SammyS]

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?

 

[songoku]

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

 

[anuttarasammyak]

@songoku your way works. What value have you got ? I would like to comapre your way and mine to get some good insights.

 

[songoku]

@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 $20\pi-20$

 

[anuttarasammyak]

@songoku Thanks.
$$20\pi-20=4a\pi-4a$$ where a=5. Contribution of the blue part is 4a. That should be equal to my integral so
$$a \int_0^{2\pi} d\theta \int_0^1 dx \frac{x}{(1+x^2-2x \cos\theta)^{\frac{1}{2}}}=4a$$
$$\int_0^{2\pi} d\theta \int_0^1 dx \frac{x}{(1+x^2-2x \cos\theta)^{\frac{1}{2}}}=4$$
We get troublesome definite integral value by your smart way.

 

[anuttarasammyak]

@songoku Thanks.
$$20\pi-20=4a\pi-4a$$ where a=5. Contribution of the blue part is 4a. That should be equal to my integral so
$$a \int_0^{2\pi} d\theta \int_0^1 dx \frac{x}{(1+x^2-2x \cos\theta)^{\frac{1}{2}}}=4a$$
$$\int_0^{2\pi} d\theta \int_0^1 dx \frac{x}{(1+x^2-2x \cos\theta)^{\frac{1}{2}}}=4$$
We get troublesome definite integral value by your smart way.

In the same manner we may easily expand it to
$$\int_0^{2\pi} d\theta \int_0^1 dx \frac{x}{(1+x^2-2x \cos\theta)^{\frac{1-p}{2}}}=\frac{2^{2+p}}{1+p}\int_0^\frac{\pi}{2}\cos^{1+p}\theta \ d\ \theta$$
for p>0