Integration of part of a radius gives a complex number....

[Adel Makram]

I am interested to find the length shown in red in the attached figure. I want this length as a function of d (shown in blue) and the angle θ. Then I will integrate this length to dθ from 0 to π/2.

Firstly, I used the law of the triangle to determine the length s which when subtracted from the radius r yields the desired length p.
s²+2sd cosθ+ d²-r²=0

Then I used a substitute a²=d²/ r²-d²
which gives, -sqrt(r²-d² ∫ sqrt {(a² cos²(θ)-1} dθ

Lastly, I used wolframe alfa to calculate the part under integration, here is what I got.( I used r=1 and d=1/2 r).

The result is a complex number ( attached). I don`t understand how the integration of a length which is a real number results into a complex number.

 

[BvU]

"law of the triangle" to determine the length s which when subtracted from the radius r yields the desired length p.
s2+2sd cosθ+ d2-r2=0

Could you explain this ? To me it looks as if $(s+d\cos\theta)^2 < r^2$ !

 

[Adel Makram]

Could you explain this ? To me it looks as if $(s+d\cos\theta)^2 < r^2$ !

Consider the triangle with sides r, d and s,

r²= d² + s² + 2 ds cosθ

This is a quadratic equation of s, given d and r.

 

[BvU]

Funny, I would say that $(s+d\cos\theta)^2 + z^2 = r^2$

 

[Adel Makram]

Funny, I would say that $(s+d\cos\theta)^2 + z^2 = r^2$

True, I am using this too later in the calculation of p.

Please see the word file I just attached.

 

[BvU]

Using it later is pointless if your first step is simply wrong
[edit] Oops -- looks like I'm off track. Let me review ...

 

[BvU]

About the question: you want to vary $\theta$ from $0$ to $\pi/2$. Is s a constant ? then doesn't d stay constant too ?

 

[Adel Makram]

About the question: you want to vary $\theta$ from $0$ to $\pi/2$. Is s a constant ? then doesn't d stay constant too ?

Yes $\theta$ varies from $0$ to $\pi/2$. d is a constant.

 

[BvU]

d was shown in blue. What happens when $\theta$ reaches ${\pi\over 2} + \arcsin ({2r\over d})$ ?

[edit] Sorry, ${\pi\over 2} + \arcsin ({d\over 2r})$ of course. This is when s = r and p goes through zero.

(to me that means d is approximately where z is in your drawing. meaning p = 0.

 

[Adel Makram]

d was shown in blue. What happens when $\theta$ reaches ${\pi\over 2} + \arcsin ({2r\over d})$ ?

(to me that means d is approximately where z is in your drawing. meaning p = 0.

I drew another circle with more clear lines. z is always perpendicular to r.

I didn`t get you at the second sentence. d is always ≤r

 

[BvU]

OK, so I see: $z = d\sin\theta, \quad s = \sqrt{r^2-z^2} - d\cos\theta, \quad p = r - s$ ( or $p = | r - s |$ since you want the length of p). The p = 0 occurs when $\theta$ reaches ${\pi\over 2} + \arcsin ({d\over 2r})$; that's when the bottom end of the blue line is on the circle as well.

Note only the + sign in your solution of the quadratic equation for s applies because s > 0. So $s = \sqrt{r^2-d^2\sin^2\theta} - d\cos\theta$

In your word document you want to integrate $$
\int_0^{\pi/2} r + d\cos\theta - \sqrt{d^2\cos^2\theta + r^2 - d^2 } \ d\theta$$ but you evaluate $$
\int_0^{\pi/2} - \sqrt{d^2\cos^2\theta - r^2 + d^2 } \ d\theta$$could that be the problem ?

By the way, $d^2\cos^2\theta + r^2 - d^2 = r^2 - d^2\sin^2\theta$ ; as long as $d \le r$ taking the square root should be OK.

 

[Adel Makram]

OK, so I see: $z = d\sin\theta, \quad s = \sqrt{r^2-z^2} - d\cos\theta, \quad p = r - s$ ( or $p = | r - s |$ since you want the length of p). The p = 0 occurs when $\theta$ reaches ${\pi\over 2} + \arcsin ({d\over 2r})$; that's when the bottom end of the blue line is on the circle as well.

Note only the + sign in your solution of the quadratic equation for s applies because s > 0. So $s = \sqrt{r^2-d^2\sin^2\theta} - d\cos\theta$

In your word document you want to integrate $$
\int_0^{\pi/2} r + d\cos\theta - \sqrt{d^2\cos^2\theta + r^2 - d^2 } \ d\theta$$ but you evaluate $$
\int_0^{\pi/2} - \sqrt{d^2\cos^2\theta - r^2 + d^2 } \ d\theta$$could that be the problem ?

By the way, $d^2\cos^2\theta + r^2 - d^2 = r^2 - d^2\sin^2\theta$ ; as long as $d \le r$ taking the square root should be OK.

This is nice step that: $d^2\cos^2\theta + r^2 - d^2 = r^2 - d^2\sin^2\theta$ , I overlooked it at all.
Nothing wrong with $$ \int_0^{\pi/2} r + d\cos\theta - \sqrt{d^2\cos^2\theta + r^2 - d^2 } \ d\theta$$ because I know how to evaluate the the first two terms but I stuck at the third one.

Now how to evaluate

$$\int_0^{\pi/2} \sqrt{r^2 - d^2\sin^2\theta } \ d\theta$$

 

[LCKurtz]

This is nice step that: $d^2\cos^2\theta + r^2 - d^2 = r^2 - d^2\sin^2\theta$ , I overlooked it at all.
Nothing wrong with $$ \int_0^{\pi/2} r + d\cos\theta - \sqrt{d^2\cos^2\theta + r^2 - d^2 } \ d\theta$$ because I know how to evaluate the the first two terms but I stuck at the third one.

Now how to evaluate

$$\int_0^{\pi/2} \sqrt{r^2 - d^2\sin^2\theta } \ d\theta$$

I don't think that is an elementary integral. Maple gives an answer in terms of Elliptic functions.

 

[BvU]

So are we doing OK now Adel ?

 

[Adel Makram]

So are we doing OK now Adel ?

But still I don`t know how to evaluate
$$\int_0^{\pi/2} \sqrt{r^2 - d^2\sin^2\theta } \ d\theta$$
WolframAlfa didn`t give me a solution.

 

[LCKurtz]

But still I don`t know how to evaluate
$$\int_0^{\pi/2} \sqrt{r^2 - d^2\sin^2\theta } \ d\theta$$
WolframAlfa didn`t give me a solution.

Maple gives this:

where

 

[Adel Makram]

I am not familiar with elliptic integrals so I don not know what z and k in this solution are related to my problem.

For example, $$ \int_0^{\pi/2} r + d\cos\theta - \sqrt{d^2\cos^2\theta + r^2 - d^2 } \ d\theta = {\pi/2} r + d - f(d, r)$$. So how to represent f(d,r)? If it is not elementary function so how to approximate it to an elementary one or at least represent it as a series?

 

[BvU]

Approximate with elementary functions, represent as series: Is that really going to help you ? When I plot $p(\theta)$ for r = 3 and d = 2, I get something like

so you still want to pay attention to the point where the bottom end of d crosses the circle ( I assume you want |p| ). I have no hope of doing anything analytical with this one, so I would turn to numerics without much hesitation.

--

 

[Adel Makram]

So if we taking the area under the curve, p(theta), it should represent the desired integration, right!.

Remember Buffon needle problem, a needle with a length 2d and the distance between the vertical lines is L crosses those lines with a probability = $$ 2d/ \pi L$$, One of smart solution to this problem is shown in wikipedia where it evaluates the integral $$\int_0^{(\pi/2} 2d cos\theta d\theta$$ as a nominator divided on $$\int_0^{\pi/2} L d\theta$$ as a denominator.

So what if I want to calculate the probability of the needle of length 2d crossing the circle circumference with a radius r.

I proposed to consider the length p, as it is done in this post, instead of $$d cos\theta$$ in the classical problem. Intuitively, the probability of crossing the circle is larger than the probability of crossing vertical lines which is consistent with $$p> dcos\theta$$.

Lastly, I am not sure if considering the bottom point of d when it crosses the circle will have a logic ground in calculating the probability. The reason is that when $$\theta > \pi/2$$, the lower half of the needle under the diameter of the circle will be considered instead of the upper half because of the symmetry. So, we always considering the half of the needle that makes $$\theta<\pi/2$$ with the diameter of the circle.

 

[BvU]

So if we taking the area under the curve, p(theta), it should represent the desired integration, right!.

Not $p(\theta)$ but the absolute value of $p(\theta)$. After all, in your original post, you wanted to integrate "the length shown in red" .

Remember Buffon needle problem, a needle with a length 2d and the distance between the vertical lines is L crosses those lines with a probability = $$ 2d/ \pi L$$, One of smart solution to this problem is shown in wikipedia where it evaluates the integral $$\int_0^{(\pi/2} 2d cos\theta d\theta$$ as a nominator divided on $$\int_0^{\pi/2} L d\theta$$ as a denominator.

So what if I want to calculate the probability of the needle of length 2d crossing the circle circumference with a radius r.

I proposed to consider the length p, as it is done in this post, instead of $$d cos\theta$$ in the classical problem. Intuitively, the probability of crossing the circle is larger than the probability of crossing vertical lines which is consistent with $$p> dcos\theta$$.

Lastly, I am not sure if considering the bottom point of d when it crosses the circle will have a logic ground in calculating the probability. The reason is that when $$\theta > \pi/2$$, the lower half of the needle under the diameter of the circle will be considered instead of the upper half because of the symmetry. So, we always considering the half of the needle that makes $$\theta<\pi/2$$ with the diameter of the circle.

You are losing me completely. My perception was that your d is fixed and you wanted p as a function of theta. So your blue line has the top end point on the circle and the bottom end point on the horizontal axis. Nowhere is there a probability coming in.

"So what if I want to calculate the probability of the needle of length 2d crossing the circle circumference with a radius r." Is not a well defined problem. Is the circle in a football field or on a chess board ?


--

 

[Adel Makram]

This is nice step that: $d^2\cos^2\theta + r^2 - d^2 = r^2 - d^2\sin^2\theta$ , I overlooked it at all.
Nothing wrong with $$ \int_0^{\pi/2} r + d\cos\theta - \sqrt{d^2\cos^2\theta + r^2 - d^2 } \ d\theta$$ because I know how to evaluate the the first two terms but I stuck at the third one.

Now how to evaluate

$$\int_0^{\pi/2} \sqrt{r^2 - d^2\sin^2\theta } \ d\theta$$

I am not familiar with elliptic integrals so I don not know what z and k in this solution are related to my problem.

For example, $$ \int_0^{\pi/2} r + d\cos\theta - \sqrt{d^2\cos^2\theta + r^2 - d^2 } \ d\theta = {\pi/2} r + d - f(d, r)$$. So how to represent f(d,r)? If it is not elementary function so how to approximate it to an elementary one or at least represent it as a series?

Not $p(\theta)$ but the absolute value of $p(\theta)$. After all, in your original post, you wanted to integrate "the length shown in red" .

You are losing me completely. My perception was that your d is fixed and you wanted p as a function of theta. So your blue line has the top end point on the circle and the bottom end point on the horizontal axis. Nowhere is there a probability coming in.

"So what if I want to calculate the probability of the needle of length 2d crossing the circle circumference with a radius r." Is not a well defined problem. Is the circle in a football field or on a chess board ?--

d is still fixed and represents half length of the needle.
I followed the same reasoning of the classical Buffon needle. In classical version, the needle crosses the vertical line when the projection of the needle on the horizontal axis is not larger than $$2dcos\theta$$ . In other words, the needle needs to be displaced away from the nearest vertical line by $$2dcos\theta$$ until its upper end touch it.
In the circle version with a very large radius, the problem is then reduced to the classical one omitting very small number of times where the upper tip of the needle crosses the circle circumference. When the circle is getting smaller in radius, the needle needs to be displaced to the left side away from the circumference by a length p in order not to touch the circumference.

N.B: 2d is used in the classical version and d in the circle version but this should not be a source of confusion as long as the horizontal line between the vertical lines is used in the classical and only the radius (half the diameter) is used in the circle version.

 

[Ray Vickson]

Maple gives this:
View attachment 87079
where
View attachment 87080View attachment 87079

Maple gives this:
View attachment 87079
where
View attachment 87080View attachment 87079

I get a much simpler-looking result, using Maple 14:
f:=sqrt(r^2 - d^2 * sin(t)^2):
J:=Int(f,t=0..Pi/2):
VJ:=value(J) assuming d>0,r>d;
VJ:=r*EllipticE(d/r)