An aesthetic double integral.

[@demenion3521]

one small "trick" that you could use for the determinant of the transformation: as you have already drawn the new and old region, you can easily see that the volume/area of the region in the new coordinates is the same as the volume in the old coordinates (pi²/2 each) so that the determinant is simply 1

[@DeanCalhoun]

I think he was just walking through the process a bit slower so those unfamiliar with it could have a better understanding. Just from a glance it’s easy to see that transformation results in a jacobian of 1

[@PunzL]

Haven't encountered the jacobian before. Is it something similar to the eigenvalue? Like a multi-dimensional extension of the eigenvalue in this case the scalar of the area of the triangular region?

[@demenion3521]

@@DeanCalhoun i think it's nice to have both the technical tools to calculate jacobian determinants as well as the geometrical understanding about them

[@demenion3521]

@@PunzL you can view the determinant of a matrix as something like a condensed version of eigenvalues (as it's equal to the product of the eigenvalues). but geometrically speaking the determinant is the volume factor when you transform one polyhedron into another (as one can see by transforming a unit cube which is aligned with the principal axes of the matrix which all just get scaled by the eigenvalues). the sign of the determinant just tells you whether the matrix inverts the orientation of the polyhedron. the jacobian determinant is like a special case of the determinant for the matrix which changes one set of integration variables to another. you can think of the absolute value of the jacobian determinant times the new volume element as having the same volume as the old volume element (if you view the volume element as something like a tiny cube)

[@adamjennifer6437]

If you like more interesting Math solutions, just watch this impressive Math channel kzbin.info/door/ZDkxpcvd-T1uR65Feuj5Yg

[@filipristovski88]

Congrats on the 100k Michael, really well deserved

[@andrewmichel2525]

Congrats on 100k subs, Michael! I can't think of a content creator who deserves it more. Easily my favorite mathematics channel on youtube.

[@adamjennifer6437]

If you like more interesting Math solutions, just watch this impressive Math channel kzbin.info/door/ZDkxpcvd-T1uR65Feuj5Yg

[@Icenri]

I like that the last part works the same as if we were using the complex definition of the sine function: getting the Ln(1/2) out and then cancelling the exponentials.

[@captainintegral]

Yes! 100k. Congratulations, sir!

[@astrosky9464]

Congratulations with 100k!!!!! (Great video by the way)

[@jonathanweihing9650]

100k Congratulations from Germany. Nice videos!!!

[@mohammedsalouani7672]

Congratulations for the 100k subscribers !!!! Amazing

[@GroundThing]

The "trick" that you mentioned around 9:15 I don't think is as far outside of reasonability as you claim, since it's basically king's rule, i.e. integral of f(x)dx from a to b = integral of f(a+b-x)dx from a to b, which I don't know if it's taught in every calculus course, but it's far from uncommon.

[@DeanCalhoun]

Congrats on 100k Michael! Love this channel!

[@eduardoschiavon5652]

Great video as always Mr. Penn! I noticed that at 8:50 if you'd substituted (x,y) for (1/2u+1/2v,-1/2u+1/2v) you would've already gotten the integrals split up into two.

[@aumshreeraval8249]

Congrats on 100K🎉

[@sergeigrigorev2180]

Congratulations for 100k subs!

[@themathsgeek8528]

Congrats for the 100k, Prof Penn!! Keep going :)

[@WiseSquash]

at the beginning I really believed this was gonna be a very easy problem... haha congratz on the 100k, Michael!

[@armanrasouli2779]

CONGRATS!!! 100K

[@KyleDB150]

So at 7:36 why are you able to say that "ln(sin(v)) does not depend on u" even though v is a function of u via v=x-y and u=x ? Is it because you transformed variables using the Jacobian, so you can now treat u and v as independent variables instead of functions of each other?

[@markusbrachert]

Short answer: Yes.

[@KyleDB150]

@@markusbrachert cool XP I can see how the Jacobian lets you account for inter-dependence of your new variables, and if its just 1 then you can now treat them as independent, imma go learn about it now lol

[@bebarshossny5148]

Awesome man I'd love to see more multivariable calc from you

[@mikeschieffer2644]

00:59 What the heck was that?!

[@williamadams137]

A flying cockroach?

[@ישראלישראלי-ע1ב]

I didn't get how you are allowed to assume that pi-tetha equals tetha (at 11:35) outside of the sin (it was at the same equation, I mean how you can assume that the tetha from the left integral is the same as the tetha from the right integral) ?

[@azmath2059]

Check this out, it may help kzbin.info/www/bejne/eZC5aWh9f7Wgm5I

[@azmath2059]

Hi Michael. After you split the integral in two, you make a change in variables on both integrals incorporating θ, but they are not the same. You then combine the integrals under the new variable θ. I cant see how you can do this considering that the original variable , v , is changed differently for the two integrals?

[@azmath2059]

I've just discovered an alternative way of solving int(xlnsinx) here at kzbin.info/www/bejne/eZC5aWh9f7Wgm5I. I understand this method better.

[@BikeArea]

I second that. 😮

[@rounaksinha5309]

Hii sir Please make a video on proving the properties of floor function(greatest integer function).Nowhere on the internet is this available.

[@IoT_]

Prof. Penn, How many substitutions do you want to have? Prof. Penn: Yes. P.S. Congratulazione with 100k subs

[@Saki630]

anxiety integral. Please tell me there is an easier way using symmetries to point out that future terms were going to cancel out.

[@winky32174]

I'd love to see a more complex version of this, where the Jacobian is not 1.

[@leif1075]

Are u kidding? How could it possibly get more complex than this?

[@adamjennifer6437]

If you like more interesting Math solutions, just watch this impressive Math channel kzbin.info/door/ZDkxpcvd-T1uR65Feuj5Yg

[@leif1075]

@@bengrant5670 Right thanks for clarifying. I'm wondering if this couldnt be solved with a straightforward u substitution like u equals sine x or something...then it becomes the integral of ln u/( sqr root of 1-u^2) and I think that's one of those special e natutal number functions you could just look up the solution or do a second substitution after maybe?

[@skylardeslypere9909]

@@leif1075 I'm very late to the party. But, what we're doing in this video is essentially the "u-substitution" of double integrals. You're integrating over an area instead of an interval (or subset of IR, really), so we need to define 2 new variables. The way Michael wrote it down is a bit less intuitive, but it's essentially applying the substitution { u = x { v = x-y

[@leif1075]

Those manipulations at the end are so convoluted i dont think anyone would ever think to solve this way..can't you do a u substitution like u equals sine of theta..I think that's more intuitive and would work...

[@yuseifudo6075]

It wouldn't

[@mathsamtube2741]

Congrats for being silver button owner ;)

[@trevthecellist]

How can you cancel the different thetas? One theta represents “v” while the other represents “pi-v”. If you subtract theta-theta aren’t you subtracting v-(pi-v)? Maybe I’m missing something about the function being periodic but any clarification would be appreciated :) Edit: I kept watching and saw all the change of variables to the SAME variable. Maybe my calc is just REALLY rusty but this just seems so wrong to me. Plz help

[@demenion3521]

it's just a matter of notation. he could just as well call the new variable x=pi-theta. the name of the integration variable is arbitrary, but it's easier to see which integrals cancel if you have the same names for the variables in both integrals. as an example, consider int(x, 0

[@leif1075]

Are we finding the area ofbthe whole triangle or the area of that function ln sine of x minus y , the part within the triangle..I'm assuming..

[@demenion3521]

@@leif1075 well, we are integrating a function over a region. in the 2D case you can view this as calculating the volume under the graph

[@azmath2059]

I agree, I can't see how that can be done considering there are two representations of theta after which he combines them?

[@azmath2059]

Check this out, it may help kzbin.info/www/bejne/eZC5aWh9f7Wgm5I

[@aakashhaque9805]

He hit 100K. I didn't even notice

[@prollysine]

From Budapest BME villamosmérnöki kar eljárása /Feynman ?/. Int.(Int.)ln(sin(x-y))dA = immaginarius része az Int.(Int)ln(exp(x-y))dA megoldásnak. Tisztelt tanár úr, régen volt már, tisztelettel kérem, segítene e módszer szerinti megoldásban ? Üdvözlettel Ágoston Budapestről !

[@d4slaimless]

So we calculated the area and the result is negative value? (ln(1/2)

[@yuseifudo6075]

Yes

[@jd123-]

anyone could help me ?! I have a hyperbola , I have to find the conic section , I know that 2× conjugate axis of the hyperbola is equal to 3×Focal Length

[@SKARTHIKSELVAN]

I am new to two variable substitution in double integral. Can you recommend some books?.

[@manucitomx]

Wow! I feel like I should do a backflip.

[@janzentwong8094]

00:59 I thought there was a bug crawling on my screen

[@themathsgeek8528]

One video without ''Good Place To Stop'

[@kianushmaleki]

Interesting.

[@tomatrix7525]

Is it fully legit how he defined theta differently in the two integrals around 10:00 ? I’m guessing so, but how is this?

[@DavidSavinainen]

Theta is just a dummy variable. It doesn’t matter what it’s called, since the two integrals are separate. You couldn’t do such a thing if you had a double integral, but since they are completely disconnected from each other, it’s not a problem.

[@adamjennifer6437]

If you like more interesting Math solutions, just watch this impressive Math channel kzbin.info/door/ZDkxpcvd-T1uR65Feuj5Yg

[@azmath2059]

@@DavidSavinainen But then he combines the two separate integrals together using theta which is defined differently?

[@nombreusering7979]

Where is A Gooad Place To Stop?

[@AlbusVacuus]

Why was the determinant just 1 instead of minus 1?

[@popescuervin7893]

absolute value

[@rogerlie4176]

When making a change of variables in a multivariable integral, the absolute value of the determinant is used.

[@leif1075]

Why not just integrate over the triangle in x and y as is, seems just as easy..

[@rogerlie4176]

@@leif1075 No, since then you need to find a primitive function for ln(sin(x-y)), which you won't.

[@leif1075]

@@rogerlie4176 what's a primitive function?

[@fredfrancium]

Hey Michael, I do not know, you should slove a CRAZY Math Problem for reaching 100k 😁. We are waiting for it. 😀.

[@ByakuyaKuchiki006]

100k subs 🎉🎉🎉🎉🎉🎉🎉🎉🎉🎉🎉🎉

[@PerseoRapax]

nice

[@Mathcambo]

Good teacher

[@Tiqerboy]

I treat mathematics as a tool to get the job done. I just checked my toolbox and I don't have the tools necessary to solve that problem, so I'm going to have to watch your video to see how it is done. To be honest, I never got far enough in mathematics to solve something like that, and I never would have come up with all those substitutions either, which I suppose comes from experience. That gets around from finding the integral of a log(sin(x)) which can only be done using numerical methods (as far as I know).

[@samegawa_sharkskin]

100k subs!! :D

[@zarifjubair707]

Is this suitable for school level?

[@ByakuyaKuchiki006]

Nope

[@OuroborosVengeance]

Nope. You need a few years into college to get this

[@raystinger6261]

Kinda annoyed he kept the ln(1/2) instead of -ln(2). It's not wrong, but... Anyway, good video!

[@aleksapupovac]

Me 2

[@Nickesponja]

Me (1/2)^(-1)

[@angelogandolfo4174]

Me rt4.

[@danv8718]

00:58 - 01:01... UFO warning!

[@edwardlulofs444]

The integral was, advertisements every 3 minutes was not fun. Is KZbin running out of money or does KZbin just need all of the money in the universe?

[@danv8718]

I've pretty much stopped watching KZbin on my tablet or phone because of this. The ad blocker does a good job on the pc so far, but if it ever stops working, that's it.

[@andreemery4964]

@@danv8718 I've downloaded KZbin Vanced on my android, which lets you watch youtube without ads (though it doesn't appear to be able to block the occasional KZbin Red/Music pop up ads)

[@danv8718]

@@andreemery4964 Thanks for the tip, I'll give it a try!

[@JSSTyger]

Right after sex.. "and thats a good place to stop."

[@zza7195]

No body: Me: being disturbed by a non-symetric hood

[@anushrao882]

Hmm yes that's definitely a good place...

[@infinityinf1]

That's a good place to what?

[@rickzhang1796]

This must be witchcraft...

[@mantkcmantk8717]

Too tricky...