a double integral, 3 ways

[@DjVortex-w]

The double integral is essentially calculating the volume of the origin-centered half-cylinder, which is capped by the surface x^3+xy^2. It baffles my mind how this volume can be a rational number, given that a circle is involved.

[@simenjorissen5357]

If the height of the cylinder is 1/π then the volume will be r², if r is integer, the volume will not only be rational but also an integer and a perfect square

[@seanfraser3125]

Polar is the method I used immediately. I’m very curious to see a full version of the first method just to know how ridiculous it is by comparison. Not enough to do it myself though.

[@Bermatematika]

It is actually good exercise to practice substitution method. Not that hard. Maybe I will make a video about it :).

[@Bermatematika]

Here you go the video that I promised :): kzbin.info/www/bejne/npnMapyXgJ1gnZo

[@brooksgunn5235]

Bermatematika.com You should! I subbed to you.

[@falkinable]

I did it using the substitution method

[@JohnAlejo99]

BlackPen RedPen *BluePen*

[@winnablebtw459]

I prefer the reliable Wolfram Alpha method. It applies to almost every integral you throw at it.

[@MarkMcDaniel]

Weak sauce.

[@ninjawayxd6211]

Which method is that?

[@2muchnrg268]

@@ninjawayxd6211 it’s an online calculator that gives the answer for you lol

[@hectorbrizuelavega9214]

The force is strong on this one

[@Supercatzs]

Believe in math, not Wolframalpha!

[@unknown6000]

I'm only a high school student so I had no idea about the third method so I just tried the first one right away. What a tedious process that was!

[@yash1152]

lololol.

[@thaovu-yi5ts]

wait high school students learn this:)?

[@epikherolol8189]

​@@thaovu-yi5tsWe don't but it's pretty self explanatory that we gotta do the inside integral first. It's kinda like those 10yr old algebra questions where u use bodmas and do inside out ig But yeah being a highschool student myself I only knew how to do the first method and i got stuck afterwards

[@randomname9291]

(2x/3)(sqrt(9-x^2)^3) is actually relatively simple to integrate, as it fits the formula of the integral of f’(x)*f^n(x) Where f(x)=9-x^2 And n=1.5 Therefore all it is is (1/3)*(((9-x^2)^2.5)/2.5) Idk how to integrate the other part though as my integration knowledge is very limited

[@retired5548]

the second way is so much clearer, however i cannot help but try the first method as well edit: well it was intimidating to integrate at first, but wasn't so bad in the end

[@lmao4982]

I feel like this guy can never stop holding his microphone, it's just a part of his thing now

[@colt4667]

I just LOVE it when you solve a problem two or three different ways and you get the same answer each time! Ain't mathematics grand?

[@tungboychak2295]

YOU REALLY KNOW YOUR THING

[@ralfbodemann1542]

I prefer the Toblerone method.

[@anegativecoconut4940]

That Smile when you realize that you did it again.

[@eugeneimbangyorteza]

The Polar method is like one of the earliest things taught in multivariable calculus

[@andi_tafel]

Toblerone = The Best

[@filip-kochan]

Andi Tafel what is toblerone please?

[@TrueGamerWoo]

Filip Kochan the best method

[@duncanw9901]

Double integral, Triple coulours

[@Bayerwaldler]

Your video reminded me of the time when I was a teaching fellow more than thirty years ago. 14:01 At that point, since we're integrating first w.r.t. r and *then* w.r.t. theta, I wouldn't have depicted semi circles ranging from r=0 to r=3 but rays with angles ranging from theta = -Pi/2 to Pi/2. I would have also shown the other order of integration too which is just as easy to do. Then of course the semi circles would have come into play! Great video nonetheless. ... I wish I could have communicated as well as you!

[@gergananikolovagery5058]

Very nice! I just learned about the polar coordinates method at uni and I like your explanation best. Seems much easier! haha I love it

[@ArifSolvesIt]

using polar coordinates is unfortunately not always the best way. Hence, if you use the polar coordinates to calculate the area of an ellipse, the integral you need to solve turns out to be more difficult to handle than the one you solve using Cartesian coordinates; you can see kzbin.info/www/bejne/f2LZg39jasplorM

[@paulnokleberg5188]

Great video as always. There's actually a 4th way also, since changing the order of integration in polar form also works. I wrote all 4 methods out in detail and got 486/5 each time.

[@xxshogunflames]

The third way blew my mind! Thank you!

[@ArifSolvesIt]

using polar coordinates is unfortunately not always the best way. Hence, if you use the polar coordinates to calculate the area of an ellipse, the integral you need to solve turns out to be more difficult to handle than the one you solve using Cartesian coordinates; you can see kzbin.info/www/bejne/f2LZg39jasplorM

[@fletcherk3296]

Day before my calculus exam and i think you may have just saved me from losing a good amount of marks lol!! Thank you! Great explanation

[@DanNguyen-oc3xr]

Why am I watching math at 1 am? I guess I can claim this as studying

[@freeze2win697]

I am 2 years old and i already learn calculus🤓 you make it look easier😇

[@bensnodgrass6548]

Surely to integrate with respect to y where there are x's you have to assume that the two functions are independent? Like if you wrote x as a function of y (not treating it as constant) it would look different and you would get a different answer. But then later he connects them by saying x^2 + y^2 = 9...

[@v0igr409]

nobody is talking about the GIANT TOBLERON CHOCOLATE BAR AT THE END ??

[@thevirus7020]

thank you

[@vai_-cn9br]

The method you applied at the beginning...I call it 'clumsy integral' whenever I encounter it lol

i pause this video at 0:50 and i want to solve this integral by original way by myself , it take along time and very complex, then when i solved it i continous see this video, that amazing way to solve it, 2 way is so good.

[@CubeMontster17]

Pls do more polar coordinates integration videos! They're so cool

[@nadia-sy8cn]

that was so coooolll thank you for this amazing video

[@yash1152]

4:08 > _"represents bottom part of circle"_ holly molly, i entirely forgot that and was thinking about root of inverted parabola. and by the way, never noticed this connection before too: root of a parabola gives a semi-circle. awesome.

[@danmart1879]

Calc genius! Wow!

[@DanielAsada-m4u]

Thank you so much

[@mimiphan9582]

He is the best.

[@guilhermepimenta_prodabel]

In the second method you should use absolute value.

[@arifahimran5773]

from null to awesome.... i love second and thrid method....tq

[@GhostyOcean]

*GREEN'S THEOREM INTENSIFIES*

[@dankylesiao4755]

the toblerone in the last tho

[@rkumaresh]

Polar method is suitable for this problem

[@lovejotsingh7018]

Thanx bro... u taught us very well

[@kuantumalex7937]

Polar coard is the best

[@persekarva6444]

Im way too drunk to understand this, but im still watching lol

[@josammarenye4021]

I was so confused where the r came from when we switch dydx to polar form😭 thanks for giving me so much peace😄❤️❤️

[@acehabib5008]

Can you do a triple integral please? Triangle integration? THANKS !

[@innocentmhlanga9117]

Thank you very much...

[@Tarheb]

x((9-x^2)^3/2)/3 disappears because one is positive and one negative

[@Tomcat721]

I like this!! Thank You

[@stydras3380]

the first one seems unnecessarily cruel xD anyhow... fun video!! :)

[@vainqueurndangi6848]

amazing video bro!

[@appybane8481]

To continue in first way, use u=9-x^2, that's also easy!

[@3manthing]

polar coordinates are my choice

[@user-vm6qx2tu3j]

Published on my birthday 😍

[@victorkkariuki]

Rash Scientist happy belated birthday

[@cameronspalding9792]

With the first method it was so complicated that I ended up with the wrong answer

[@lostwizard]

I did it the first way and messed up the numerical calculations the first time through. It looks really scary after substituting in the Y values. However, a little fiddling around and using u = 9 - x^2 gives a relatively nice second integration. There is an x^2 that doesn't immediately disappear from the substitution but it's easy enough to represent x^2 in the u world. Sure, it's not as nice as the other two methods since the square roots don't disappear. However, with the converted integration limits, you end up substituting a 9 into the square roots so the actual calculation is straight forward enough.

[@lostwizard]

Okay. So I made a video with the working out for the first way: kzbin.info/www/bejne/qae6nKBuqJd0eLM

[@زهرةالنرجس-ن2ل2ب]

Wow thank you explain very well and I take advantage of you.

[@omerangi4695]

That coffee cup tho

[@pappaflammyboi5799]

I know this is a dumb question, but I gotta ask it. But before I do, I understand how you did the double integral all 3 ways. Not too bad. Now here's my question: Once you find out it's a circle of radius 3 from theta = -pi/2 to pi/2, and you're interested in finding the area, which is what this integral is doing, why not just apply the function A = 1/2*pi*r^2 where r = 3. Thing is, it's not the same answer... what went wrong???

[@lukandrate9866]

The integral computes the volume between the given region and the given f(x,y), not the area of the region

[@karljoyeux5148]

Da secund one is very cool

[@mcwulf25]

Polar is easy. Did it in my head!

[@paulbooer7171]

Awesome!

[@Timorftw]

Polar way made it so easy.

[@dellaameliaa27]

do you have playlist for this? double integral and triple integral

[@andualemfetene9237]

oh nice! polar coordinate is best for me

[@sayanpakira8]

Great

[@Patapom3]

Amazing!

[@fireemblem2770]

At 1:05, why would you ADD the exponent, and then divide the exponent by 3? I've never seen this before.

[@copperfield42]

I did it without changing the order of integration or coordenate system... I have to do it 3 times to get the result DX

[@sushruttadwalkar7701]

trig sub looks intimidating but actually is pretty simple if you go forward with it, obviously the other methods can be considered better though😅

[@ArifSolvesIt]

using polar coordinates is unfortunately not always the best way. Hence, if you use the polar coordinates to calculate the area of an ellipse, the integral you need to solve turns out to be more difficult to handle than the one you solve using Cartesian coordinates; you can see kzbin.info/www/bejne/f2LZg39jasplorM

[@jonshonjohn4756]

the first way of doing it is not THAT hard, you can make the change of variable 9-x^2=t and it becomes quite easy from there

[@Kurtlane]

I tried the same arrangement, but with function x^2+y^2 (instead of x^3+xy^2). Following method 3 (polar), I got (81 / 4) * pi. But if this is half a circle, then its area should be pi * r^2 / 2, and if r = 3, it should be (9 / 2) * pi. What did I do wrong, or maybe the whole thing is not really the area of half a circle? Please explain. Thanks.

[@cicciobombo7496]

x^2+y^2 in 3d is not a plain circle, it is a parabola rotated on itself in the y axis, so what you calcultae with this double integral is the volume under this shape, very different from the area of a circle :D

[@anicetoaniceto1020]

I did the integral and it still took me a long time I had to do two integrals.

[@wkingston1248]

Back in my day we didn't have these new fangled 'polar coordinates' we did some good old fashion integration. It builds character unlike the youth with their fancy tricks.

[@jarikosonen4079]

How can be sure dy integrate from -3 to +3, not from +3 to -3? And theta from -pi/2 to pi/2 instead of pi/2 to -pi/2... Maybe always integration from smaller coordinate to larger cooridate? It looks quite certain though that final integration result is positive... If dx was from 3 to 0, would you use theta pi/2 to -pi/2 or r from 3 to 0? This case does it matter which one of the variables would integrate in the negative direction?

[@thomasblackwell9507]

Polar!

[@JaskoonerSingh]

great video which helped me a lot. I think you lost a bit of steam near the end!

[@marianesaliba2594]

Thank youu❤❤

[@lou.104]

Do you have a video on sketching the integration domain for a double integral?

[@andreasvalen8802]

thanks

[@mathmathician8250]

Can you make a triple integrals?

[@emmanuelontiveros8446]

Polar is the easier method

[@mariahannaherickasingson9827]

im still confused why is it the theta is -pi/2 instead of 3pi/2 huhuhu

[@ゴテンクス-q8q]

Yeeeeeeeeeeeeeeeeessssssssssssssssssss

[@sdbstar8515]

Why couldn't i see it before 😭😭😭😭

[@aaronargottelopez3488]

danke

[@yash1152]

18:45 what is TOBLERONE?? :D

[@margintoosmall925]

I spent way too much time trying to solve it the first way :(

[@chesteezy5197]

BUT THE CHEN LU!

[@j0j083]

I did it using the first way but got 354/5 or 70.8

[@JohnnyPerson1]

YOu are awesome!!!

[@pharundps9434]

I want to exercise Limited

[@kirillous]

Lol I paused it and did trig sub, it works but its hella work 🥴

[@ev4_gaming]

just a question, can u reverse the order of the integtation signs? would that give the same answer?

[@achyuthramachandran2189]

There's a whole method of evaluating double integrals by changing the order of integration. However, you have to change the bounds between which they are evaluated as well. You can't simply switch dy and dx and the integral bounds in the front. Hope that helps!

[@lornacy]

❤

[@Kurtlane]

What is this Jacobian? Can anyone explain?

[@botondosvath2331]

You can see it in the following video from Dr. Peyam: kzbin.info/www/bejne/g3rbhamBjaxkoLsm55s

[@alanhiguera3484]

Kurtlane it is a matrix of the partial derivatives of the change of coordinates. in this case, x=rcos(theta) and y=rsin(theta) are the change of coordinates, you takes the partial derivatives of both with respect to r and theta, and you take the determinant of the matrix which gives r. its essentially the multidimensional analogue to dealing with the differential du in u-substitution in the single variable case.

[@emontrailers]

polar world will save the world

[@aninditasarkar688]

Why in polar coordinate dxdy is equal to rdr(theta)

[@KingRustee]

Essentially dxdy or dydx is a small change in x multiplied by a small change in y to give a small rectangular change in area. To create this same rectangle in polar coordinates, you take a small change in the radius (dr) and multiply it with a small change in the arc (rdθ) to give rdrdθ.

[@OonHan]

Hi!

[@mimiphan9582]

Wait how did blackpenredpen got r dr dtheta ?

[@vatsalgp]

I just didn't understand how the value of dxdy was found.

[@kgshbteambeasto_o963]

It is calculated by the Jacobian. I believe Dr. Pyyam has a video on it.

[@cicciobombo7496]

Easy explanation without Jacobian: mathforum.org/library/drmath/view/74707.html

[@cameronspalding9792]

It all ads up!

[@tungboychak2295]

EVERY TIME I INTERACT WITH YOUR VIDEO SIR, I GET UNDERSTAND EVERYTHING ABOUT THAT PARTY OF THE COURSE