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.
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
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.
(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
@@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
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
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.
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!
I am 2 years old and i already learn calculus🤓 you make it look easier😇
4 года назад
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.
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 ruclips.net/video/I1vRL-5e2lQ/видео.html
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.
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...
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 ruclips.net/video/I1vRL-5e2lQ/видео.html
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.
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?
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.
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.
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
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???
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!
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 ruclips.net/video/I1vRL-5e2lQ/видео.html
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.
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.
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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.
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
I prefer the reliable Wolfram Alpha method. It applies to almost every integral you throw at it.
Weak sauce.
Which method is that?
@@ninjawayxd6211 it’s an online calculator that gives the answer for you lol
The force is strong on this one
Believe in math, not Wolframalpha!
BlackPen RedPen *BluePen*
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.
It is actually good exercise to practice substitution method. Not that hard. Maybe I will make a video about it :).
Here you go the video that I promised :): ruclips.net/video/hhi9iaK8-g8/видео.html
Bermatematika.com You should! I subbed to you.
I did it using the substitution method
(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
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!
lololol.
wait high school students learn this:)?
@@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
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
I feel like this guy can never stop holding his microphone, it's just a part of his thing now
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.
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!
That Smile when you realize that you did it again.
YOU REALLY KNOW YOUR THING
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?
The Polar method is like one of the earliest things taught in multivariable calculus
I prefer the Toblerone method.
I am 2 years old and i already learn calculus🤓 you make it look easier😇
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.
Double integral, Triple coulours
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
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 ruclips.net/video/I1vRL-5e2lQ/видео.html
Why am I watching math at 1 am? I guess I can claim this as studying
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.
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
Toblerone = The Best
Andi Tafel what is toblerone please?
Filip Kochan the best method
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...
The method you applied at the beginning...I call it 'clumsy integral' whenever I encounter it lol
*GREEN'S THEOREM INTENSIFIES*
nobody is talking about the GIANT TOBLERON CHOCOLATE BAR AT THE END ??
With the first method it was so complicated that I ended up with the wrong answer
At 1:05, why would you ADD the exponent, and then divide the exponent by 3? I've never seen this before.
In the second method you should use absolute value.
the toblerone in the last tho
that was so coooolll thank you for this amazing video
The third way blew my mind! Thank you!
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 ruclips.net/video/I1vRL-5e2lQ/видео.html
amazing video bro!
from null to awesome.... i love second and thrid method....tq
To continue in first way, use u=9-x^2, that's also easy!
Polar method is suitable for this problem
Pls do more polar coordinates integration videos! They're so cool
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.
Okay. So I made a video with the working out for the first way: ruclips.net/video/svWkm8s2ABQ/видео.html
Thanx bro... u taught us very well
18:45 what is TOBLERONE?? :D
x((9-x^2)^3/2)/3 disappears because one is positive and one negative
Polar coard is the best
Calc genius! Wow!
That coffee cup tho
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?
thank you
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.
Polar way made it so easy.
do you have playlist for this? double integral and triple integral
I was so confused where the r came from when we switch dydx to polar form😭 thanks for giving me so much peace😄❤️❤️
Can you do a triple integral please? Triangle integration? THANKS !
polar coordinates are my choice
Thank you so much
Wow thank you explain very well and I take advantage of you.
He is the best.
Im way too drunk to understand this, but im still watching lol
I did the integral and it still took me a long time I had to do two integrals.
Polar coordinate
ruclips.net/video/BEBB3HRPl1E/видео.html
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.
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
the first one seems unnecessarily cruel xD anyhow... fun video!! :)
Polar is easy. Did it in my head!
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???
The integral computes the volume between the given region and the given f(x,y), not the area of the region
Da secund one is very cool
Do you have a video on sketching the integration domain for a double integral?
I like this!! Thank You
oh nice!
polar coordinate is best for me
Thank you very much...
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
Can you make a triple integrals?
just a question, can u reverse the order of the integtation signs? would that give the same answer?
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!
Published on my birthday 😍
Rash Scientist happy belated birthday
Polar!
trig sub looks intimidating but actually is pretty simple if you go forward with it, obviously the other methods can be considered better though😅
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 ruclips.net/video/I1vRL-5e2lQ/видео.html
I did it without changing the order of integration or coordenate system... I have to do it 3 times to get the result DX
I thought the polar coordinate was easier than the first two methods
The polar form solution almost seems like cheating.
Polar is the easier method
Awesome!
Amazing!
im still confused why is it the theta is -pi/2 instead of 3pi/2 huhuhu
Why in polar coordinate dxdy is equal to rdr(theta)
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.
What is this Jacobian? Can anyone explain?
You can see it in the following video from Dr. Peyam: ruclips.net/video/MIxTvKXG1jY/видео.htmlm55s
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.
I spent way too much time trying to solve it the first way :(
Why couldn't i see it before 😭😭😭😭
great video which helped me a lot. I think you lost a bit of steam near the end!
I bet he didn’t do the second method all the way
Great
thanks
I want to exercise Limited
Thank youu❤❤
polar world will save the world
I just didn't understand how the value of dxdy was found.
It is calculated by the Jacobian. I believe Dr. Pyyam has a video on it.
Easy explanation without Jacobian: mathforum.org/library/drmath/view/74707.html
EVERY TIME I INTERACT WITH YOUR VIDEO SIR, I GET UNDERSTAND EVERYTHING ABOUT THAT PARTY OF THE COURSE
BUT THE CHEN LU!
Wait how did blackpenredpen got r dr dtheta ?
I did it using the first way but got 354/5 or 70.8
Are you a wizard?
It all ads up!
danke