enter image description here
Assume we got a scenario in the picture. I want to calculate the area between two chords with angles theta1 and theta2. Could anyone suggest me how to calculate it based on the angles and distance defined in the picture?
Thank you very much!
[Math] area between two parallel chords
geometry
Related Solutions
I don't care to make the calculation for you, but I think(?) this is the region you intend:
Addendum.
Here is an idea to avoid computing a complex integral.
Let me assume that the slice "fits" in one octant, as I drew above
(as opposed, e.g., to exiting through the bottom of the octant).
First, imagine that you wanted, not the area in an octant, but the area in a halfspace,
demarcated by one of the coordinate planes.
Then, from a side view, we have this:
Your two planes are $p_1$ and $p_2$, with $p_1$ through the sphere center.
The halfspace boundary is coordinate plane $q$.
Now note that the difference between the half of the $2 \pi h$ area of the slice
and the truncated area is just the area of two spherical triangles (one illustrated,
one in back), whose angles
are all known.
So I believe the area in your octant differs from a quarter of $2 \pi h$ by two spherical triangles, one at either end of the octant. The dimensions of both triangles are known from the orientation of the planes $p_1$ and $p_2$.
Well, I guess I'll take a stab at this. This is definitely a calculus problem. To take the area between two curves, you want to take the integral of the greater function minus the lesser function. For the yellow area, the greater function is $y=1$. The lesser function will take some manipulation. The formula for the circle is:
$(x-4)^2+(y-4)^2=16$
$(y-4)^2=16-(x-4)^2$
$y-4=-\sqrt{16-(x-4)^2}$
$y=4-\sqrt{16-(x-4)^2}$
So our integral is $\int^3_2[1-(4-\sqrt{16-(x-4)^2}]dx$=$\int^3_2-3dx+\int^3_2\sqrt{16-(x-4)^2}dx$. The first integral is $-3x$ evaluated from 2 to 3, or in other words, $-3(3)-[-3(2)]=-9+6=-3$.
For the second half of that integral, we'll use the info from Andreas's comment. We'll perform a change of variable
$u=x-4,du=dx$
$\int^3_2\sqrt{16-(x-4)^2}dx=\int^{-1}_{-2}\sqrt{16-u^2}du=\frac12[u\sqrt{16-u^2}+16sin^{-1}\frac u4]^{-1}_{-2}=\frac12[(x-4)\sqrt{16-(x-4)^2}+16sin^{-1}\frac{x-4}4]^3_2$
That solves the yellow area. For the other 2, you'll want to know where the 2 functions cross.
$(x-4)^2+(1-4)^2=16$
$(x-4)^2=7$
$x=4-\sqrt7$
For the green area, the 2 functions are the same, but it's evaluated from $4-\sqrt7$ to 2. For the orange area, the greater function is $y=2$, but the lesser function changes. It should be easy to see, though, that the right half is a rectangle. The left half is integrated from 1 to $4-\sqrt7$. Also, the first half of the integral has changed from $\int-3dx$ to $\int-2dx=-2x$. So the total orange area is $2x-\frac12[(x-4)\sqrt{16-(x-4)^2}+16sin^{-1}\frac{x-4}4]$ evaluated from 1 to $4-\sqrt7$ plus $1[2-(4-\sqrt7)]$, or $\sqrt7-2$.
Best Answer
Note, all angles here are in RADIANS not degrees.
Given: $O$ denotes the center of the circle, $ R $ is the radius of the circle, $\theta_1 $ is the angle in the upper triangle opposite to the upper chord , $ \theta_2$ is the angle of the lower triangle opposite to the lower chord
Let $A$ be the leftmost point of intersection b/w the lower chord and the circle, $B$ be the rightmost P.O.I, $C$ be the leftmost P.O.I for the upper and $D$ be the rightmost P.O.I for the upper:
The area we are looking for is the Area of the Green shaded Region, Let $A_{green}$ be the Area we are looking for.
$A_{green} = A_{⌓AGB} - A_{⌓CGD}$:
First find $A_{⌓AGB}$
$ A_{⌓AGB} = A_{⌔OAB} - A_{△OAB}$ :
And use the same formula to find $A_{⌓CGD}$
$ A_{⌓CGD} = A_{⌔OCD} - A_{△OCD}$:
Thus, $A_{green} = A_{⌓AGB} - A_{⌓CGD} = A_{⌔OAB} - A_{△OAB} - (A_{⌔OCD} - A_{△OCD})$
$ = A_{⌔OAB} - A_{△OAB} - A_{⌔OCD} + A_{△OCD}$
$$ A_{⌔OAB} = \pi R^2 \big(\frac{\theta_2}{2\pi}) = \frac{R^2(\theta_2)}{2}$$ $$ A_{⌔OCD} = \pi R^2 \big(\frac{\theta_1}{2\pi}) = \frac{R^2(\theta_1)}{2}$$
You can use trigonometry to find the area of the triangles, $A_{△OAB}$ and $ A_{△OCD}$ :
$$ A_{△OAB} = R^2(\sin(\frac{\theta_2}{2})\cos(\frac{\theta_2}{2})) $$ $$ A_{△OCD} = R^2(\sin(\frac{\theta_1}{2})\cos(\frac{\theta_1}{2}))$$
and $A_{green} = A_{⌔OAB} - A_{△OAB} - A_{⌔OCD} + A_{△OCD}$
$$ = \frac{R^2(\theta_2)}{2} - \frac{R^2(\theta_1)}{2} - R^2(\sin(\frac{\theta_2}{2})\cos(\frac{\theta_2}{2})) + R^2(\sin(\frac{\theta_1}{2})\cos(\frac{\theta_1}{2})) $$
$$ = \frac{R^2(\theta_2-\theta_1)}{2} + R^2[(\sin(\frac{\theta_1}{2})\cos(\frac{\theta_1}{2}))-(\sin(\frac{\theta_2}{2})\cos(\frac{\theta_2}{2}))] $$
Recall the compound angle formulas: $$ \sin \left(A+B\right)=\sin A\cos B+\sin B\cos A $$ $$ \sin \left(A-B\right)=\sin A\cos B-\sin B\cos A $$ $$ \sin(2A) = \sin(A+A) =\sin A\cos A+\sin A\cos A = 2\sin A\cos A $$ $$ \frac{\sin(2A)}{2} = \sin A\cos A $$
Thus, $$ (\sin(\frac{\theta_1}{2})\cos(\frac{\theta_1}{2}))-(\sin(\frac{\theta_2}{2})\cos(\frac{\theta_2}{2})) = \frac{\sin(2(\frac{\theta_1}{2}))}{2} - \frac{\sin(2(\frac{\theta_2}{2}))}{2} $$ $$ = \frac{\sin(\theta_1)}{2} - \frac{\sin(\theta_2)}{2} = \frac{\sin(\theta_1) - \sin(\theta_2)}{2} $$
Therefore
$$ A_{green} = \frac{R^2[\theta_2 -\sin(\theta_2) -\theta_1 + \sin(\theta_1)]}{2} $$