Can You Solve This Tricky Section of a Quarter Circle? | 2 Awesome Methods
ฝัง
- เผยแพร่เมื่อ 8 ก.พ. 2025
- Think you can solve this tricky section of a quarter circle? In this video, we explore two awesome methods to find the area of this fascinating shape.
We’ll start with a geometric approach, using visual reasoning to break down the problem step-by-step. Then, we’ll dive into an integral-based method, applying trigonometry and analytical geometry for a deeper mathematical perspective.
Which method will you prefer? Watch the video, try it yourself, and let us know in the comments!
🔔 Don't forget to like, subscribe, and hit the notification bell for more exciting math puzzles!
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Beautiful!!
I have no idea what was happened.
Good job.
Great problem 😃
Thank you! Glad you liked it!
@ThePhantomoftheMath Actually I wanted to say that why don't you try to make some shorts out of these long videos?
It will really help grow the channel as you could reach more viewers 📈👍
@@BhumanBudhiraja I'm not very big fan of shorts...but...yeah maybe I'll do that...I will definitely consider your advice on that. Thanks!
Thank you ❤
Geometric solution is lovely, THX
Thank you! I'm glad you like it!
Great sir, really great!!!
Ty sir!
At 1:10, designate the point where the arc intersects the line through CD as point E. Construct OA and OB. We note that sector OAE is 1/6 of the circle's area and OBE 1/12. We can find the red area by computing the area of sector OAE and deducting the areas of ΔOAC and the black area bounded by BD, DE and arc BE. That black area is the area of sector OBE less the area of ΔOBD. So the red area is area (sector OAE - ΔOAC - (sector OBE - ΔOBD)) = sector OAE - sector OBE - ΔOAC + ΔOBD. As found in the video, ΔOAC and ΔOBD are congruent, so they cancel out. Sector OAE - sector OBE is sector OAB, or 1/12 the area of the circle, = (1/12)πr² = (1/12)π(6²) = 3π, as The Phantom of the Math also found.
Very nice!
@@ThePhantomoftheMath Thank you!
I just set up a triangle and a rectangle of the red area, using standard trig and geometry. Got those areas. Then added the amount from the segment area of sector AOB. ( the space between the chord and the arc AB of sector AOB . As I say, I added up everything and the total came to 9.39. Pretty damn close to suggested solution. Short and sweet.
Great job!!!
I did the same thing. Simple is better
Great method. good job
If you end up using trig sub, why not just do a polar integral from the start?
That's a great question! I figured that most people would take the polar approach, so I wanted to go a bit unconventional and explore another method. That's why I mentioned in the video that there are many different ways to solve this problem-just to show the versatility of approaches in math.
Nice video
Thank you! Glad you like it!
9.425
Draw a line from the circle's center to the outer side of the largest height of red shaded to form a 60-30-90 right triangle AOC.
It is 60-30-90 since the quarter circle is divided into 30 degrees, and hence the angle to the left is 60 degrees ( 90 -30)
The dimensions of this triangle, hence, are 6 (radius), 3, and 5.196 ( x, 2x, x sqrt 3)
Draw a 30-degree angle from the lowest height of the red shaded to the circle's center to form another 30-60- 90 right triangle, thus forming also triangles BDO, COP, AOP, a trapezoid, BPCD
and a sector ABP
Notice that the shaded region is the trapezoid BPCD, and the sector, ABP. Hence, the sum of their combined area = area of the shaded region
Let's start with the trapezoid, which is inside triangle BDO.
line OB = 6, Hence line OD = 5.196 and line BD =3 since BDO is a 30-60-90 triangle
Hence, the area of BDO = 7.794
Triangle COP's bases are 3 (OC) and 1.732 since it is also a 30-60-90 triangle.
Hence, it area = 2.598
Hence, the area of the (shaded) trapezoid = 5.196 (7.794 - 2.598)
Calculating the area of the remaining shaded section ABP:
To Calcuate the area of the remaining shaded sector, ABP (30 degrees) = area of circle/ 12 ( since the circle is 360 degrees and 30
degrees is 1/12 of 360. Hence, area = area of circle /12 = 113.1/12 = 9.425
9.425 minus the triangle AOP will give the remaining area of shaded
Triangle OAP's area = area of triangle AOC - OPC = 5.196 * 3 *1/l2 - 3* 1.732 * 1/2 = 7.794 - 2.598 =5.196
Hence, the area of remaing shaded = 9.425 - 5.196 = 4.229
Hence, the area of total shaded = 4.229 + 5.196= 9.425 Answer
3pi = 9.4247, approximately 9.425
Well done!!!
@@ThePhantomoftheMath Thanks
i did it by using trignometry by extending it to semi circle and then dividing the resulted area by 2.
integral olmadan da çözülür yay uzunlukları eşit ise açılarda 30,60 ve 90 olur açı özelliklerinden alanlar bulunabilir ve kısaca hesap yapılırsa alan 3pi bulunur.
çok uzun süre almış sadece 1 dakika süreniz var imtihanda
You could have also used the substitution cos^2u=(1+cos2u)/2 in the integral method .
Absolutely! That would also work like a charm!
I followed the second (integral) method but calculated each of the 3 slices (both total area and the 3 sub-areas).
Left area = 18 (π/6 + √3/4) = 3π + 9√3/2. Red (middle) area = 18 (π/3 - π/6) = 18 x π/6 = 3π, and the right area = 18 (π/6 - √3/4) = 3π - 9√3/2.
Total area from adding all of them = 9π which matches if you integrate from π/2 to 0 (so the indefinite integral (sin 2T/2 - T) must be accurate, which also matches quarter circle area = π (6/2)^2 = 9π, thereby confirming the middle (red) area correctly equal to 3π.
You are underrated.
Thank you so much! That means a lot to me! Just subscribed to your channel as well!
For the integral of cos^2(x), if you don’t have the reduction formula memorized you can use integration by parts
I = ∫cosx*cosx dx
I = cosx*sinx + ∫sin^2(x) dx
I = cosx*sinx + ∫1 - cos^2(x) dx
I = cosx*sinx + ∫1 dx - I
2*I = cosx*sinx + x
I = (cosx*sinx + x)/2
The vertical lines are equal to 3 and 3*sqrt(3) (both are part of a 30 60 90 triangle). The base is 3*sqrt(3)-3. The area is the trapezoid which area is (3*sqrt(3)+3)*(3*sqrt(3)-3) / 2 = 9. The remain part of the area is the area of a sector with angle 30° minus the triangle formed by two radiuses having the same 30° angle. The area of the sector is 6 *6 * π / 12. The area of the triangle is 6 * 3 / 2 = 9. So the curved region is 3π - 9. Adding the trapezoid area which is 9 I get 3π - 9 + 9 = 3π
My answer:3[π+2√3-4] sq.unit.
Very nice method! I like it!
Don't forget to re-compute the limits of your integral to account for the u-substitution. Leaving the limits with respect to x in your u-substituted integral is technically incorrect. I only bring this up since it's easy to forget to un-sub and just use the x-based limits, and has the added bonus of not even needing to un-substitute. 😊
Yea that part definitely was confusing / worst practice
Second solution is sick 😂
I know I know 😅
@@ThePhantomoftheMath LOL. "Sick" is ambiguous in that usage.
It can mean "the very worst"; and it can mean "the very best".
What a great response.
@@pietergeerkens6324 😅😅😅
@@pietergeerkens6324 That's why I answered for both options...to be safe 😂😂😂
The area equality or substitution in the first method is very elegant! That's how such a problem is meant to be solved. The second method is awful! So it's a good contrast. The lazy one does well to think creatively before begining work on the first idea that pops up. First chess master Steinitz said that the difficult thing with chess is, when one has come up with a good move, to keep looking for a better one.
I don't find very elegant the first method, because it is only valid by dividing the quadrant into three equal parts.
For other angles, doesn't work.
Second method is awful, I agree, but is generic for any angle
See my generic method in the comments
I love chess!!! That's an awesome quote!
x = 3√3, 3
Area = ∫[3,3√3] √(36 - x^2)dx = 3π
1) semicircle 2) first circle piece 12pi-9*sq(3) 3) secont circle piece 6pi-9*sq(3) 4) fcp-scp=6pi 5) 6pi/2=3pi From Greece -- no good English
Everything understood, no "good English" required for math friend!
6×6=36×3.14159268÷12=9.4247
Position of points A and B:
h₁= x₂ = R.cos30° = 3√3 cm
h₂= x₁ = R.cos60° = 3 cm
x = x₂ - x₁= 3√3-3 = 2,19615cm
Area of right trapezoid:
A₁ = ½(h₁+h₂).x = ½(3√3+3)*2,19615
A₁ = 9 cm²
Area of circular segment:
A₂ = ½R²(α-sinα) = ½6²(30°-sin30°)
A₂ = 0,4248 cm² = 3π-9
Red shaded area:
A= A₁+A₂= 9,4248cm² = 3π (Solved √)
This area is equal to 30° angular sector area:
A = ½αR² = ½*π/6*6²= 3π cm².
I couldn't see it before. The solution could have been simpler !!!!!
That also works like a charm! Nice!
I was so lost on the 2nd solution
9:37 how does x = 6 sin u?
Why 6?
Why Sine?
What does the "u" represent?
I don't get it. Please help.
When we face an integral like ∫ √(36−𝑥^2)𝑑𝑥, it resembles the form √(𝑎^2-𝑥^2). So, in these cases, using a trigonometric substitution can simplify the expression. Here’s the reasoning:
1. Why 6? The number 6 comes from the fact that 36 is a perfect square: 36=6^2. In trigonometric substitutions for expressions like 𝑎^2−𝑥^2, we typically choose 𝑥=𝑎 sin 𝑢 (or occasionally
𝑥=𝑎 cos 𝑢 ) because it fits well with the Pythagorean identity (sin 𝑢)^2+(cos 𝑢)^2 = 1. In this case, 𝑎=6, so I set 𝑥=6 sin 𝑢.
2. Why sine? The choice of sin 𝑢 (rather than cosine or another function) ensures that 𝑥 stays within the interval [−6,6], which matches the bounds of the square root function 36−𝑥^2 because sin 𝑢 ranges between -1 and 1. This substitution simplifies the square root and keeps the integral solvable with trigonometric identities.
3. What does 𝑢 represent? The variable 𝑢 is a new angle we introduce for the substitution. It’s just a placeholder that helps convert our variable 𝑥
in terms of trigonometric functions. After making the substitution, we can simplify the integral and later back-substitute to return to the original variable 𝑥.
So, when we set 𝑥=6 sin 𝑢:
We replace 𝑑𝑥 with the derivative of 𝑥 with respect to 𝑢, which will involve cos 𝑢.
The square root term √(36−𝑥^2) becomes
√(36−36sin^2 𝑢) = 6 cos u, leading to a more straightforward integral in terms of 𝑢.
This substitution is a common approach in integrals involving square roots of quadratic expressions and is part of a method called "trigonometric substitution." Let me know if you have more questions about any of these steps!
Cheers, and thanks for watching!
👍
A = A₁- A₂
A = ½[½R²(α-sinα) - ½R²(β-sinβ)]
A = ¼R²(120°-sin120°)-(60°-sin60°)
A = ¼6²(⅔π-√3/2-⅓π+√3/2)
A = 3π cm² ( Solved √ )
My solution is subtracting the two circular segments, EASIER !!! and generic.
Video solution (first method) only works when dividing the arc in three equal parts !!!
@@marioalb9726 Since the problem IS dividing into 3 equal parts, solution 1 is valid in this case, and I wanted to avoid trigonometry, since method 2 is little bit obnoxious...if you know what I mean...
And, yes: Your solution is valid for every angle! Nice one! 👍
Just add a triangle on the left side, and then remove the triangle with the same area on the right side.
Then you are left with a third of this quarter circle.
The 2nd. method is a very mess !! I prefer to use the 1st.one. Thanks
10:04
I am pretty sure it has something to do with the integral.
Area=3π
We can use an Integral to solve this.
01) sin(30º) = 1/2
02) sin(60º) = sqrt(3)/2
03) Interval of the Integral [3sqrt(3) ; 3]
04) Equation of the Circle : X^2 + Y^2 = 36
05) Y^2 = 36 - X^2 ; Y = +/- sqrt(36 - X^2)
06) INT sqrt(36 -X^2) dX between 3sqrt(3) and 3.
07) Area = 9,425
Lets call the bottom right corner of the quarter circle F.
The dark area with curved edge BDF = The Arc of OBF - The triangle OBD
Area with curved edge ACF = The arc OAF - triangle OAC
OAF = 2 x OBF
therefore;
Area with curved edge ACF = 2xOBF - OAC
Red shaded area. = ACF - BDF
2xOBF - OAC - (OBF - OBD) = 2xOBF - OAC - OBF + OBD = OBF - OAC + OBD
OBD = OAC
therefore;
Red Shaded Area = OBF
Not really the most convenient way to use integration to compute this area. Computing the path integral of (x dy - y dx)/2 along the boundary of the region using the obvious parameterizations of each segment yields much simpler integrals to evaluate. Indeed, once you’ve set up those four (really three) integrals, you might even notice that they directly correspond to the geometric decomposition in the first half of this video. Even if you don’t spot that, you should be able to see that two of them cancel and one of the remaining integrands is uniformly zero, leaving a single simple integral to evaluate.
Well I get a strange answer of 3π
Knowing the 3 equal arc lengths means you can know some angles from the center of the circle to each vertex. The angles will be multiples of 30°.
Solution:
M = center of the quarter circle,
A = right corner point of the quarter circle,
B = upper corner point of the quarter circle,
C = lower point of the red stripe on the left,
D = lower point of the red stripe on the right,
E = lower third point,
F = upper third point.
r = radius of the quarter circle = 6.
Triangle MCF is congruent to triangle MDE (Ssw) because
1. MF = ME = r = 6,
2. Angle CFM = 90°-60° = 30° = angle DEM,
3. Angle FCM = angle EDM = 90°.
Black area =
= circular section MFB + triangle MCF + circular section MDE - triangle MDE
= circular section MFB + circular section MDE [triangle MCF = triangle MDE]
= π*6²*30°/360°+π*6²*30°/360° [both circular sections with 30° angle]
= π*36/12+π*36/12 = 6π
Red area = quarter circle - black area = π*6²/4-6π = 9π-6π = 3π ≈ 9.4248
Excellent job! Well done!
Solution using integral calculus:
The lower limit of the integral is r*cos(60°) = 6*1/2 = 3.
The upper limit of the integral is r*cos(30°) = 6*√3/2 = 3*√3.
The equation of the circle with the center at the origin is x²+y² = r² = 6². ⟹
y = +√(r²-x²) = +√(6²-x²) = +√(36-x²). So the integral that represents the red area is:
3*√3 3*√3 3*√3
Red area = ∫√(36-x²)*dx = ∫√[36*(1-x²/36)]*dx = 6*∫√[1-(x/6)²]*dx =
3 3 3
---------------------
Substitution: u = x/6 du = 1/6*dx dx = 6*du
lower limit = 3/6 = 1/2 upper limit = 3*√3/6 = √3/2
----------------------
√3/2
= 36*∫√(1-u²)*du =
1/2
----------------------
Substitution: u = sin(t) t = arcsin(u) du = cos(t)*dt
lower limit = arcsin(1/2) = π/6 upper limit = arcsin(√3/2) = π/3
---------------------
π/3 π/3
= 36*∫√[1-sin²(t)]*cos(t)*dt = 36*∫cos(t)*cos(t)*dt =
π/6 π/6
Solution of the indefinite integral ∫cos(t)*cos(t)*dt with partial Integration:
∫cos(t)*cos(t)*dt =
--------------------------------------
Solution by partial integration:
Partial integration can be derived from the product rule of differential calculus. The product rule of differential calculus says:
(u*v)’ = u’*v+u*v’ |-u’*v ⟹
(u*v)’-u’*v = u*v’ ⟹
u*v’ = (u*v)’-u’*v |∫() ⟹
∫u*v’*dx = u*v-∫u’*v*dx
------------------------------------
∫cos( t)*cos(t)*dt = cos(t)*sin(t)+∫sin(t)*sin(t)*dt |+∫cos(t)*cos(t)*dt ⟹
2*∫cos(t)*cos(t)*dt = cos(t)*sin(t)+∫sin²(t)*dt+∫cos²(t)*dt |/2 ⟹
∫cos(t)*cos(t)*dt = {cos(t)*sin(t)+∫[sin²(t)+cos²(t)]*dt}/2 ⟹
∫cos(t)*cos(t )*dt = {cos(t)*sin(t)+∫1*dt}/2 ⟹
∫cos(t)*cos(t)*dt = [cos(t)*sin(t)+t]/2+C ⟹
π/3 π/3
So the red area = 36*∫cos(t)*cos(t)*dt = 18*[cos(t)*sin(t)+t]
π/6 π/6
= 18*[cos(π/3)*sin(π/3)+π/3-cos(π/6)*sin(π/6)-π/6]
= 18*[1/2*√3/2+π/3-√3/2*1/2-π/6] = 18*[π/3-π/6] = 3π ≈ 9.4248
@@gelbkehlchen Very very nice!!! Great job 👍👍👍
@@ThePhantomoftheMath Thank you!
Sorry for my previous answer which was wrong.Now I found the correct answer which is equal to your answer,but my method is different.ok.Thanks for a a good question.
Never mind about that, when it comes to math, we all make mistakes. Thank you for commenting, and for watching!
The area should be 9 pi not 3 pi, you made a mistake in converting degree to radian, it's theta / 180 * pi, not divided by 360, that was overkill on the integral method too 😅, but can you tell me where the cos(arcsin(x)) formula comes from cause I never saw it before.
I calculated the area geometrically by dividing it into a trapezium + a sector with angle (30) it was faster for me and simpler this way and still gets the 9 pi in the end
Hi! Thanks for watching and taking the time to comment. I see where you're coming from, but the area being 9π can't be right. Here's why:
If the radius of the quarter circle is 6, the area of the entire quarter circle would be:
(1/4) * π * r² = (1/4) * π * 36 = 9π.
So, if the whole quarter circle’s area is 9π, it wouldn’t make sense for a smaller segment inside that quarter circle to have the same area! Maybe there was a small miscalculation somewhere.
As for your other question about the cos(arcsin(x)) formula:
Think about the definitions of sine and cosine in a right triangle. If y = arcsin(x), then:
sin(y) = opposite / hypotenuse = x / 1.
We can imagine a right triangle with an angle y, an opposite side of length x, and a hypotenuse of 1. Using the Pythagorean theorem, the adjacent side would be:
sqrt(1 - x²).
So:
cos(y) = adjacent / hypotenuse = sqrt(1 - x²) / 1 = sqrt(1 - x²).
Thus, cos(arcsin(x)) = sqrt(1 - x²). I hope this makes it clearer!
I also appreciate your method of dividing the shape into a trapezium and a sector; it’s always cool to see different approaches. Keep up the great work!
@@ThePhantomoftheMath I revised the problem again and turns out I forgot one half while doing it the first time, thanks for taking the time to answer patiently to all my questions, shows how pationate and great of a person you truly are, and I hope your channel becomes one of the top channels on YT as well
@@Z-eng0 No worries at all! It happens to the best of us-math can be tricky sometimes. 😊 Thank you so much for your kind words! It really means a lot to me. I'm just happy to share my passion for math and engage with awesome people like you.
I hope you stick around for more puzzles and challenges! With support like yours, I’m sure the channel will keep growing. Thanks again, and feel free to ask anytime if something isn’t clear or if you have new ideas to share! 🙌
I calculated the (area of) circle half-segment beyond AC, then subtracted the circle half-segment beyond BD. Half-segment beyond AC is the circle sector below OA minus right triangle OAC. Same for half-segment BD, sector below OB, triangle OBD. Sector below OA = 2/3 of the quarter circle's area, and sector below OB is 1/3. Relative sides of 90-60-30 triangles known by heart.
Excellent thinking!
Why make it so complicated?
I did solve it by trigonometry
HI. I completely understand your question! I guessed that most people would take the polar approach, so I wanted to go a bit unconventional and explore another method. That's why I mentioned in the video that there are many different ways to solve this problem-just to show the versatility of approaches in math. Ana a lot of people asked me to do integrals, so I implemented that into this problem. But your trigonometric approach would definitely be much easier...as well as the integral with polar coordinates. But like I said, I wanted to make "high contrast" between 2 methods.
You lost me @ 9:48.
I threw up @ 11:30.
I can feel your pain!!!