a double integral with 3 ways, original way, change of order, use polar coordinate, blackpenredpen, math for fun, blackpenredpen.com/bprplive, / blackpenredpen , blackpenredpen@gmail.com
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.
@@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
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 ua-cam.com/video/I1vRL-5e2lQ/v-deo.html
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!
(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
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 ua-cam.com/video/I1vRL-5e2lQ/v-deo.html
from null to awesome.... i love second and thrid method....tq
4 роки тому
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.
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...
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?
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.
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.
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
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 ua-cam.com/video/I1vRL-5e2lQ/v-deo.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.
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 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 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!
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 :): ua-cam.com/video/hhi9iaK8-g8/v-deo.html
Bermatematika.com You should! I subbed to you.
I did it using the substitution method
BlackPen RedPen *BluePen*
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
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 ua-cam.com/video/I1vRL-5e2lQ/v-deo.html
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!
(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 prefer the Toblerone method.
Thanx bro... u taught us very well
that was so coooolll thank you for this amazing video
YOU REALLY KNOW YOUR THING
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 ua-cam.com/video/I1vRL-5e2lQ/v-deo.html
The Polar method is like one of the earliest things taught in multivariable calculus
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
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?
That Smile when you realize that you did it again.
Calc genius! Wow!
Pls do more polar coordinates integration videos! They're so cool
Wow thank you explain very well and I take advantage of you.
I like this!! Thank You
Gold as usual
Jordan Rogers thank you!
I am 2 years old and i already learn calculus🤓 you make it look easier😇
from null to awesome.... i love second and thrid method....tq
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
do you have playlist for this? double integral and triple integral
Thank you very much...
thank you
Amazing!
Toblerone = The Best
Andi Tafel what is toblerone please?
Filip Kochan the best method
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.
amazing video bro!
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...
Awesome!
He is the best.
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?
Do you have a video on sketching the integration domain for a double integral?
The method you applied at the beginning...I call it 'clumsy integral' whenever I encounter it lol
In the second method you should use absolute value.
nobody is talking about the GIANT TOBLERON CHOCOLATE BAR AT THE END ??
I was so confused where the r came from when we switch dydx to polar form😭 thanks for giving me so much peace😄❤️❤️
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.
To continue in first way, use u=9-x^2, that's also easy!
the first one seems unnecessarily cruel xD anyhow... fun video!! :)
Can you do a triple integral please? Triangle integration? THANKS !
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: ua-cam.com/video/svWkm8s2ABQ/v-deo.html
Great
Thank youu❤❤
That coffee cup tho
x((9-x^2)^3/2)/3 disappears because one is positive and one negative
Why am I watching math at 1 am? I guess I can claim this as studying
thanks
I did the integral and it still took me a long time I had to do two integrals.
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
At 1:05, why would you ADD the exponent, and then divide the exponent by 3? I've never seen this before.
oh nice!
polar coordinate is best for me
Polar method is suitable for this problem
*GREEN'S THEOREM INTENSIFIES*
Im way too drunk to understand this, but im still watching lol
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!
Polar coordinate
ua-cam.com/video/BEBB3HRPl1E/v-deo.html
Da secund one is very cool
Can you make a triple integrals?
great video which helped me a lot. I think you lost a bit of steam near the end!
Polar coard is the best
the toblerone in the last tho
Polar is easy. Did it in my head!
Polar way made it so easy.
polar coordinates are my choice
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 ua-cam.com/video/I1vRL-5e2lQ/v-deo.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
With the first method it was so complicated that I ended up with the wrong answer
YOu are awesome!!!
Published on my birthday 😍
Rash Scientist happy belated birthday
danke
Polar!
18:45 what is TOBLERONE?? :D
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.
Yeeeeeeeeeeeeeeeeessssssssssssssssssss
What is this Jacobian? Can anyone explain?
You can see it in the following video from Dr. Peyam: ua-cam.com/video/MIxTvKXG1jY/v-deo.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.
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
Why couldn't i see it before 😭😭😭😭
im still confused why is it the theta is -pi/2 instead of 3pi/2 huhuhu
I spent way too much time trying to solve it the first way :(
Lol I paused it and did trig sub, it works but its hella work 🥴
It all ads up!
❤
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.
Hi!
BUT THE CHEN LU!
EVERY TIME I INTERACT WITH YOUR VIDEO SIR, I GET UNDERSTAND EVERYTHING ABOUT THAT PARTY OF THE COURSE
Are you a wizard?
I want to exercise Limited
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
I did it using the first way but got 354/5 or 70.8
Polar is the easier method
Wait how did blackpenredpen got r dr dtheta ?