What is the shaded area??

[@ahzong3544]

7:29 "Let's do it with integration by parts, and let's do that just because I've got this nice, slick way of calculating this integral with integration by parts that I'd like to show everyone." How can I not subscribe when I hear this

[@davidbrisbane7206]

Indeed. If Michael can find a harder way of doing something, he usually does 😂🤣😂🤣.

[@ethang8250]

Brilliant; not too hard, not too easy and everything linked up nicely. Thank you for doing these.

[@ixian98]

in the integral part: shouldn’t answer be [1-sqrt(3)/4 - Pi/6]? not plus Pi/6?

[@nullplan01]

I'm getting the same result as you, so I think that's correct.

[@tomatrix7525]

Yep, that.’s just a typical Michael Penn error. He does all the hard work and gets a basic error haha. He just needed to distribute the negative ofc.

[@azzteke]

@@tomatrix7525 OFC???

[@romajimamulo]

At 10:25 , you forgot the to fully distribute the minus sign

[@divyanshaggarwal6243]

* at 10:25

[@romajimamulo]

@@divyanshaggarwal6243 thanks for getting the exact time stamp

[@physjim]

i did it a little bit differently, but arrive at the same result, i calculated the integral from 0 to 1/2 of sqrt(1-x^2)-sqrt(1-(x-1)^2)dx which is appr. 0.17, this is the area of a segment of the circle. Then that area plus the blue area are equal to 1-pi/4 (the area of the square - 1/4 the area of one circle) and then blue area= 1- pi/4 - the integral

[@artsmith1347]

Again, you have provided a good drawing for the problem: lines that are dark and thick, with helpful colors and notations, with no smudges on a clean board. And then the solutions were well organized, clearly presented, and interesting. The arrows at 03:07 were a nice touch. Well done, as always.

[@aaryunik]

Hi! A big fan of your videos @Michael Penn Keep 'em coming!

[@angelpatino805]

9:58 I think you meant to say cos(pi/6) = sqrt(3)/2 and that sin(pi/6) = 1/2 :)

[@mathwithjanine]

Great video! You always explain things so well! :)

[@Deegius]

Surprising how often simple geometry easily beats the complex and time consuming calculus. Love th videos tho.

[@keep7smiling]

Depends. I happened to know the cos^2 trick so it was pretty straight forward for me. I would have had to think a while and look things up to come up with the geometry solution. (I used integration A LOT for my studies and barely ever came in touch with circle-sectors to be fair)

[@PRIYANSH_SUTHAR]

Geometry is a good old friend.

[@luisoncpp]

This reminds me to a classic problem from the ACM UVa(competitive programming), named "Is this integration?", with the ID 10209. Even when it's a competitive programming problem, the solution is just classic geometry and all the programming part is read the length of the side and printing the output. That problem goes a little bit farther: it draws 4 unitary circles, each one in the corner of an unitary square and asks for the area in the middle (the intersections of the 4 circles).

[@goodplacetostop2973]

15:18 Loading 95.4%...

[@blackpenredpen]

Loading 95.6%... I have the champagne ready already!

[@rehmmyteon5016]

@@blackpenredpen Fun fact my math teacher gave this exact problem to me a few weeks ago

[@ramaprasadghosh717]

Let a be the common radius of these two cicrles. Then their meet point forms an equilateral trriangle with the two end point of the base line. So area formed by right arc with left end point is πa"a/6. So shaded area is area of the square - (area formed by right arc with left end of base + area formed by left arc with right end of base- area of the equilateral triangle ) = a*a( 1-π/3 +√3/4)

[@leickrobinson5186]

What’s the non-sequiturial “root 2” at 13:18? 🤔

[@nelsblair2667]

(Root 3) over 2 is the height of the triangle in base times height. “Root 2” may have been an abbreviation?

[@heisenbergheisenberg8751]

At 5:46 you could've wrote √(a)^2 - (x)^2 as 1/2.x√(a)^2 - (x)^2 + a^2/2.sin^-1x/a it's a formula from integration itself(where"a"is constant and "x" is variable)

[@heisenbergheisenberg8751]

@@angelmendez-rivera351 i totaly agree with you sir but I'm just saying and I'm not a crammer here myself

[@dushyanthabandarapalipana5492]

Thank you!

[@punditgi]

More math fun!

[@VerSalieri]

Never stop making videos...please.

[@mathematicsmi]

Very interesting

[@leonardomontalvoosores501]

Nice video, continue doing more videos

[@mathlessons4116]

Good as always

[@tomatrix7525]

Great video.

[@imobusters4049]

Please include this problem in one of your videos. Let x,y,z be reals in the interval [4,40] such that: x+y+z=62 xyz=2880 find all such triples (x,y,z)

[@Ni999]

The geometric solution for integral of the circle segment is pretty easy, just areas of the triangle (½bh) and pie slice (½θr²). ∫ √(r²-x²) dx ½arcsin(x/r)r² + ½x√(r²-x²) + c Not criticizing anything, I just like the method. It stares at you from the integral, no need to waste time with substitutions.

[@Ni999]

@@angelmendez-rivera351 I don't have the formula memorized. Edit - except for the areas of a triangle and a pie slice, those I have memorized. And the formula for a circle.

[@Ni999]

@@angelmendez-rivera351 Speaking of time, I took honors calculus in college probably before you were born. The prof in the first year held an eraser in her left hand and chalk in right. Solutions took one line about two feet wide and woe to you if you if you didn't pay attention or weren't prepared to follow the solution. There was no pausing. Before solving any Integral, one was required to stand up and verbally describe how the function in question would appear in real life or what it might physically be describing. Only after passing peer review would one be allowed to take to the board and demonstrate a proposed solution. We were part of the shakeout of the text we used by Lewis and Kaplan - and the text is still used today. You cast understanding purely in terms of following some rote methods that you approve of as The Way and also The True Path To Learning Because You Say So. Of course it involves memorizing general strategic approaches and developing experience so one becomes proficient at tackling problems - but God help anyone avoid your condescension for suggesting that visualization and mapping ought to be considered as valid tools in the struggle. It's also important to note that you seem to have an opinion worthy of a snide remark about how much free time I have. Let me remove the guesswork for you - I'll be dead long before you. Today, I survived another cardiac episode. I treasure my time and I use what I have left as I please. I hoped my use of it here might have encouraged some student to think outside of the straitjacket if they were having trouble. But of course, that's not good enough for you. So let me tell you exactly how much time I have - if I have to die before having to listen to your pretentious condescension again, then it won't be a moment too soon.

[@Jack_Callcott_AU]

Error at 10:22. It should be -pi/6.

[@gleydsonqueiroz6397]

Nicee!! But instead of using integration by parts, you could do cos²θ = (1 + cos 2θ)/2

[@NavyBlueMan]

Michael, here's a similar question. It's also one you can solve geometrically and with calculus. It's hard to put into non-ambiguous words, so I coded up a visualisation here - editor.p5js.org/Big_Dog/full/M__EHXViL For a given x (with the origin at the center of the circle, x representing the distance of the vertical lines from the origin), find the area of the black shaded region, and hence find the optimal value of x to maximise the shaded region.

[@weirdfrog1196]

Great problem! I obtained 2/sqrt(4+pi2) as the optimal x.

[@artsmith1347]

Wouldn't it require trig to evaluate the area at an arbitrary value of x? If I did it correctly, the value of 'x' is related (not equal) to interesting number, but it doesn't happen at a "nice" angle. I don't see how geometry alone could be used to maximize the shaded area.

[@Yuuki3uwu]

7:05, since theta is pi/6 , then why is 5pi/6 not accepted?

[@nathanisbored]

sinθ = sin(π - θ) sinθ = sin(θ + 2nπ) therefore: sinθ = sin(π - (θ + 2nπ)) so: if sinθ₁ = 0, then θ₁ = π - (arcsin(0) + 2nπ) = π - 2nπ if sinθ₂ = 1/2, then θ₂ = π - (arcsin(1/2) + 2nπ) = 5π/6 + 2nπ therefore 5π/6 is a valid value for θ₂ when n = 0 but that means you also need θ₁ = π to match the n value if you want to be thorough, then (π - 2nπ) works for the lower bound, but you would need to use (5π/6 + 2nπ) for the upper bound. any value for n will give you the same value in the end for the integral, because the n values match.

[@jotave49]

Dude last night I had a dream in which dr Penn was my calc II professor and I failed the final lol.

[@get2113]

Put the origin in center of top line segments. Then turn. Figure upside down. Easy after that.

[@divyanshtripathi7867]

Sir,at 8:30 else doing integration by parts you can change cos^2ø to (1+cos2ø)/2 and do the integration easily this will save the time at last thank you 😊 for such good videos

[@dingo_dude]

the easiest way to calculate the integral of cos^2(x) is just to remember that cos^2(x) = 1/2 (1 + cos(2x))

[@manavaggarwal2714]

From which book you have taken this problem if any?

[@notananimenerd1333]

Presh talwalker

[@leohsu8560]

Believe it or not, elementary students in Taiwan are asked to do this kinda problem without having to integrate anything. Cool approach though!

[@tonyhaddad1394]

10:28 you have make a littile mistake 1_(sqrt(3)/4+pi/6)=1_sqrt(3)/4 _pi/6

[@txikitofandango]

Didn't you already have to know the triangle was equilateral to know the height of the triangle? If not, then how else would you know the height was root(3)/2?

[@txikitofandango]

You at least need the 60-degree base angle

[@someguy3335]

U know this triangle is equilateral, because one of the sides is the side of the square and the other 2 are the radius of the circle

[@txikitofandango]

@@someguy3335 he hadn’t shown that yet when he gave the height of the triangle

[@txikitofandango]

Once you know the height, use Pythagorean theorem to show the triangle is equilateral. But you probably should go the other direction

[@shivansh668]

This problem is also on the channel "Mind Your Decisions" too 😁

[@takyc7883]

I cant believe i didnt identify that easy geometric method.

[@aamierulharith5294]

I didn't split it into two halves because that easier... you'll get the blue area as 1 - equilateral triangle - 2(30° sector of radius 1)

[@lordstevenson9619]

Nice coincidence the answer was simply, r-θ-(cos(θ)/2). r=1 simplifies everything

[@insouciantFox]

I think the power-reducing formulae are the least-taught and most-useful identities.

[@insouciantFox]

@@angelmendez-rivera351 Yeah, I'm sure HE knows them, but my point is that come across problems that require it all the time and there are people who make their lives very difficult because they don't know the identities. That's all I meant by it.

[@danielaguiar9306]

with a double integral, this should be easy

[@Tiqerboy]

Your geometry problems are easy. Presh on the other hand sure hands us some doozies, but I'm sure you can arm wrestle the solutions out of him. Your number theory problems on the other hand ..... I seldom get those.

[@邓兴-u6s]

👍️💯️

[@theguythatmakesyoumad3834]

You have loud neighbours

[@darreljones8645]

To four decimal places, the correct answer is 0.0434.

[@manishmarda6636]

Simple geometric method is so easy, why to use the other one

[@CM63_France]

Hi, For fun: 4 "ok, great", 1 "but let's go ahead and", 1 "so let's go ahead and".

[@stewartcopeland4950]

At __ : __ coming soon...

[@kingpin1199]

Sir these kind of problems are kind of trivial as they are basically just mindless computation. Please select deeper problems, we really enjoy those

[@srijanbhowmick9570]

Bruh

[@someguy3335]

Its called "variety"

[@kingpin1199]

@@someguy3335 Next Up : 1+ 2 in 10 different ways in the name of variety

[@someguy3335]

Call me crazy, but i think this is the right kind of problem to get school kids etc. interested. The second way they can easily understand and the first way might make them curious.

[@artsmith1347]

I liked this one.