MHB How do I find the area inside one circle and outside another?

[Dethrone]

Two circles lie in a plane. The circle of radius 1 meter overlaps the circle of smaller radius $r$ in such a way that their points of intersection are separated by distance $2r$. Show that the area inside the small circle and outside the large circle is largest when $r=(1+(2/\pi)^2)^{-1/2}$. Note: you don't need to know the anti-derivative of $\sqrt{1-x^2}$

Any hints as to where to start?

 

[MarkFL]

Where would you choose to place the circles in the plane?

What are your boundary conditions regarding $r$ and what is the area at these boundaries?

Can you set up an integral equation representing the described area we wish to maximize? What should you do then?

 

[Dethrone]

Where would you choose to place the circles in the plane?

What are your boundary conditions regarding $r$ and what is the area at these boundaries?

Can you set up an integral equation representing the described area we wish to maximize? What should you do then?

I guess I could place the center of the larger circle at the origin to simplify its equation? By boundary conditions, do you mean $0<r<\frac{1}{2}$ and $0<A<\frac{\pi}{4}$?

The desired area is the portion outside the larger circle but inside the smaller circle...I'm not sure how to represent that.

 

[MarkFL]

I think I would orient the smaller circle at the origin like so:

View attachment 3070

Any ideas on how to find the center of the larger circle?

 

[Dethrone]

That was exactly what I had in mind while I was waiting for you to post, but I couldn't figure out the center of the larger circle so I didn't post anything.

With renewed effort, I'll try again. Okay!
The equation of the larger circle:
$$x^2+(y-b)^2=1$$

Isolating for $b$:
$$b=y-\sqrt{1-x^2}$$

Still not sure what to do.

 

[Opalg]

Draw a line from the centre of the large circle to one of the points of intersection. That will give you a right-angled triangle. Pythagoras??

 

[Dethrone]

I was thinking about that too...but I have one side $r$, and I don't know the length of the side from the intersection to the center, nor the center to the origin.

 

[MarkFL]

I was thinking about that too...but I have one side $r$, and I don't know the length of the side from the intersection to the center, nor the center to the origin.

What is the length of any ray from the circle's center to a point on the circle?

 

[Dethrone]

Assuming you're referring to the large circle, any ray from the center to its side is length $r$. Ah, I see what that means...I'll set up the integral now!

 

[MarkFL]

Assuming you're referring to the large circle, any ray from the center to its side is length $r$. Ah, I see what that means...I'll set up the integral now!

Yes, and what is the radius of the larger circle? :D

 

[Dethrone]

Yes, and what is the radius of the larger circle? :D

I think you're referring to the ray from the origin to the center of the larger circle, since we've been given the radius already...:D

$$s=\sqrt{r^2+1}$$

Therefore, the equation of the larger circle:

$$x^2+(y+\sqrt{r^2+1})^2=1$$

Am I okay to proceed?

 

[MarkFL]

No, we have:

$$s^2+r^2=1\implies s=\sqrt{1-r^2}$$

It makes sense doesn't it that $s<1$ rather than $s>1$. :D

 

[Dethrone]

The equation of the smaller circle:

$$x^2+y^2=r^2$$

We want only the top half:

$$y=\sqrt{r^2-x^2}$$

The equation of the larger circle:

$$x^2+(y+\sqrt{1-r^2})$$

The top portion:

$$y=\sqrt{1-x^2}-\sqrt{1-r^2}$$

 

[MarkFL]

Okay, what are your limits of integration, and is there any symmetry you can employ?

 

[Dethrone]

So I have:

$$A=2 \int_{0}^{r} \sqrt{r^2-x^2}-\sqrt{1-x^2}+\sqrt{1-r^2}\,dx$$

Is that right?

 

[MarkFL]

So I have:

$$A=2 \int_{0}^{r} \sqrt{r^2-x^2}-\sqrt{1-x^2}+\sqrt{1-r^2}\,dx$$

Is that right?

Looks good to me. :D

 

[Dethrone]

But I still have to prove that the area inside the small circle and outside the large circle is largest when $r=(1+(2/\pi)^2)^{-1/2}$. I'm not sure how to do it. I've tried to differentiate the integral, but I'm not sure.

 

[MarkFL]

Okay, now to ease the computations, take the integral one term at a time. The first term simply represents the upper half of the area of the circle of radius $r$. For the other two terms, observe that one is not a function of $x$, and the other need not be integrated, since it will succumb to the derivative form of the FTOC.

Can you put all of this together to find $A'(r)$?

 

[Dethrone]

I'm confused, is it not:
$$A'(r)=2 \d{A}{r}\int_{0}^{r} \sqrt{r^2-x^2}-\sqrt{1-x^2}+\sqrt{1-r^2}\,dx$$
Applying FTOC:
$$=\sqrt{r^2-r^2}-\sqrt{1-r^2}+\sqrt{1-r^2}$$

 

[MarkFL]

I'm confused, is it not:
$$A'(r)=2 \d{A}{r}\int_{0}^{r} \sqrt{r^2-x^2}-\sqrt{1-x^2}+\sqrt{1-r^2}\,dx$$
Applying FTOC:
$$=\sqrt{r^2-r^2}-\sqrt{1-r^2}+\sqrt{1-r^2}$$

No, you cannot apply the derivative form of the FTOC to terms in the integrand that contain the variable with respect to which we are differentiating. Do you notice your result is simply zero?

 

[Dethrone]

That makes sense. So for the first integral, I will have to use a trig substitution following with $cos^2x=\frac{1}{2}(1+cos2x)$?

 

[Dethrone]

So I will have:

$$A'(r)=2\left[\frac{dA}{dr}\int_{0}^{r} \sqrt{r^2-x^2}\,dx-\frac{dA}{dr}\int_{0}^{r} \sqrt{1-x^2}\,dx+\d{A}{r}\int_{0}^{r} \sqrt{1-r^2} \,dx\right ]$$
$$=2\left[\d{A}{r}\frac{\pi r^2}{4}-\sqrt{1-r^2}+\d{A}{r}\left(r\sqrt{1-r^2}\right)\right]$$
$$=2\left[\frac{\pi r}{2}-\sqrt{1-r^2}+\sqrt{1+r^2}-\frac{r^2}{\sqrt{1-r^2}}\right]$$

(Whew)

My answer is off by a bit, where's my mistake? (Wondering) The question was very deceiving. "Note: you don't need to know the anti-derivative of $\sqrt{1-x^2}$". We needed to know the antiderivative of $\sqrt{r^2-x^2}$ which is the same thing, except with a variable...

EDIT: Found a mistake, but still another one.

 

[Deveno]

The first integral can be (cheaply) evaluated by noting it is the area under a semi-circle of radius $r$, that is:

$\displaystyle 2\int_0^r \sqrt{r^2 - x^2}\ dx = \dfrac{\pi r^2}{2}$.

As far as $x$ is concerned, the last integral is:

$\displaystyle 2\int_0^r f(r)\ dx = 2f(r)\int_0^r\ dx = 2f(r)r$, where $f(r)$ is a constant (depending on $r$).

Differentiating this with respect to $r$ gives:

$2f(r) + 2f'(r)r$

Now you can substitute in $f(r) = \sqrt{1-r^2}$. Make SURE you use the chain rule when differentiating to find $f'(r)$.

 

[MarkFL]

We didn't actually need to know the anti derivative of the first term as it represents area of the upper half of the smaller circle. :D

 

[Deveno]

So I will have:

$$A'(r)=2\left[\frac{dA}{dr}\int_{0}^{r} \sqrt{r^2-x^2}\,dx-\frac{dA}{dr}\int_{0}^{r} \sqrt{1-x^2}\,dx+\d{A}{r}\int_{0}^{r} \sqrt{1-r^2} \,dx\right ]$$
$$=2\left[\d{A}{r}\frac{\pi r^2}{4}-\sqrt{1-r^2}+\d{A}{r}\left(r\sqrt{1-r^2}\right)\right]$$
$$=2\left[\frac{\pi r}{2}-\sqrt{1-r^2}+\sqrt{1+r^2}-\frac{r^2}{\sqrt{1-r^2}}\right]$$

(Whew)

My answer is off by a bit, where's my mistake? (Wondering) The question was very deceiving. "Note: you don't need to know the anti-derivative of $\sqrt{1-x^2}$". We needed to know the antiderivative of $\sqrt{r^2-x^2}$ which is the same thing, except with a variable...

EDIT: Found a mistake, but still another one.

So far, what you have looks good, note the two middle terms cancel. Now set $A'(r) = 0$, to find the critical points.

 

[Dethrone]

The first integral can be (cheaply) evaluated by noting it is the area under a semi-circle of radius $r$, that is:

$\displaystyle 2\int_0^r \sqrt{r^2 - x^2}\ dx = \dfrac{\pi r^2}{2}$.

As far as $x$ is concerned, the last integral is:

$\displaystyle 2\int_0^r f(r)\ dx = 2f(r)\int_0^r\ dx = 2f(r)r$, where $f(r)$ is a constant (depending on $r$).

Differentiating this with respect to $r$ gives:

$2f(r) + 2f'(r)r$

Now you can substitute in $f(r) = \sqrt{1-r^2}$. Make SURE you use the chain rule when differentiating to find $f'(r)$.

$$\displaystyle 2\int_0^r \sqrt{r^2 - x^2}\ dx = \dfrac{\pi r^2}{2}$$
If $\sqrt{r^2-x^2}$ is the upper area of a circle, then it should be $\frac{\pi r^2}{2}$. But there are two of them, so shouldn't it just be just $\pi r^2$?

- - - Updated - - -

I get as my critical number:

$$r=\sqrt{\frac{1}{1+(\frac{2}{\pi})^2}}$$

which is slightly different than the one given in the question...
...wait a minute...
...it's the same...I misread the question and I thought I made a mistake...(Giggle)(Headbang)(Headbang)(Headbang)(Headbang)

Spoiler

I thought the question read:
$$r=\sqrt{\frac{1}{1-(\frac{2}{\pi})^2}}$$

EDIT: my brain is failing me. $$\displaystyle \int_0^r \sqrt{r^2 - x^2}\ dx = \dfrac{\pi r^2}{4}$$ represents $\frac{1}{4}$ of the circle, and so...
$$\displaystyle 2\int_0^r \sqrt{r^2 - x^2}\ dx = \dfrac{\pi r^2}{2}$$

Thanks everyone, I got it now :D (Yes)

 

[Deveno]

$$\displaystyle 2\int_0^r \sqrt{r^2 - x^2}\ dx = \dfrac{\pi r^2}{2}$$
If $\sqrt{r^2-x^2}$ is the upper area of a circle, then it should be $\frac{\pi r^2}{2}$. But there are two of them, so shouldn't it just be just $\pi r^2$?

The "2" comes from the fact that we are only integrating from 0 to $r$ and doubling this. This is fine, since:

$g(x) = \sqrt{r^2 - x^2}$ is an even function and the interval $[-r,r]$ is symmetric about the origin.

 

[MarkFL]

Here is a graph of the two circles, and you can adjust the radius of the smaller to see the area change. It is initialized at the optimum value:

[DESMOS=-2.1,2.1,-2.1,2.1]x^2+y^2=a^2;x^2+\left(y+\sqrt{1-a^2}\right)^2=1;a=0.843563608069[/DESMOS]

Click on the lower bound for the parameter $a$, and set the lower bound at 0 and the upper bound at 1, and the step size at 0.01, and then click the play button to animate it. :D

 

[Dethrone]

Thanks guys! :D

Mark! That was super cool...(Cool)