This professor is amazing no one else was actually explaining the set up. I was very confused as to why the dx was being integrated just at 2, this is the first video to explain it. Amazing!
I've been trying to understand finding the bounds of both double and triple integrals for this whole semester and you've really helped me out. Thank you so much!
This was so helpful. And like you said, if you understand this then everything else going forward is a lot easier. I especially liked learning that once you integrate once, it becomes just a sing variable function. Thank you!
@@daniel_an im sorry if my question is stupid the way u explain is really plain and simple which is what students need but lets say i start my boundaries for dy the other way around... I was just curious why it will be negative?I might think it a bit better and reask :)
@@daniel_an i ll give u an example of some problem I encountered . lets say R is the sector of the circle of radius 2 centered at the origin above x-axis and below y=x firstly I said: SSdxdy= the double integral from 0->1 (of dy(outer integral)) ,integral from square of 2-y^2 to x ( inner integral ) which gave me a wrong number of 1-square root of 2(negative). then I changed my inner integral boundaries ( just turned them around ) and gave me square of 2 -1 which is positive. I hope I was clear xD sry for longpost
Searched the whole internet for this sort of clear explanation, thank you.
facts 💯💯
The paint brush analogy really helps to understand what to put for the bounds depending on the order!
This professor is amazing no one else was actually explaining the set up. I was very confused as to why the dx was being integrated just at 2, this is the first video to explain it. Amazing!
I've been trying to understand finding the bounds of both double and triple integrals for this whole semester and you've really helped me out. Thank you so much!
Very concise and apt explanation of whats going on, especially adding up the blocks of numbers part. Thank you so much for this!
Thank you so much I finally understand bounds. Setting up the blocks and adding them was a great analogy of what double integrals were doing.
I can see now I wasn't the only one who looked up on how to determine the bounds of the double integral. Thanks for such a clear explanation!
This was so helpful. And like you said, if you understand this then everything else going forward is a lot easier. I especially liked learning that once you integrate once, it becomes just a sing variable function. Thank you!
Very concise and simple to understand ! that painting analogy cleared any doubts i had. 💯💯💯
Searched for this video everywhere!!! Love from India!!!
vera level nenga vera level enga super excellent keep doing senpei
Thanks you, I finally managed to learn how to properly find the bounds a couple of days before my exam
Já fizeste o exame?
This was extremely useful! Thank you.
That analogy was really helpful
Thank you :) searched everywhere for that painting analogy
you are an amazing legend
Thanks, very patient and clear explanation
Your explanation is amazing, thank you so much, God bless you!
very good explanation with the paint analogy, thank you
this was the best explanation ever thank you
Great video, great work, great explanation! A Very helpful video!!!❤️
Thanks, your explanation was really intuitive
Wonderful!!Simply wonderful!
thank you, didn't know the trick before 14:00
thank you very much
Fabulous explanation sir. Thanks a lot
thanks for explaining how it like flattened it out really helped me
Very quick and concise, thank you
Great work man, thankyou
thank you so much cleared so much up for me
Great explanation! I'm able to pike it up quickly
You're a Rockstar.
thank you so much for this!!!
That's exactely what i was looking for
phenomenal video sir! my professors should take a look at this and maybe learn a thing or two...
Very good explanation
Very Helpful
Thank you Sir! :)
very very useful thank you!
Thank you so muchhh
FUCKING LOVE YOU, YOU'RE A WONDERFUL BEING! THANK YOU SO MUCH!
THANK YOU
thank you sooo much.
Thank you so much
Thanks)
Best one
AMAZINGG
what happens if i start my boundaries from y=4-2x to 0 ?
Then the resulting integral will be negative of what I have in the video
@@daniel_an im sorry if my question is stupid the way u explain is really plain and simple which is what students need but lets say i start my boundaries for dy the other way around... I was just curious why it will be negative?I might think it a bit better and reask :)
@@aggelosspirou8815 It's just like single integral. Integral of x from 1 to 4 would be negative of the integral of x from 4 to 1. Right?
@@daniel_an i ll give u an example of some problem I encountered . lets say R is the sector of the circle of radius 2 centered at the origin above x-axis and below y=x
firstly I said: SSdxdy= the double integral from 0->1 (of dy(outer integral)) ,integral from square of 2-y^2 to x ( inner integral ) which gave me a wrong number of 1-square root of 2(negative). then I changed my inner integral boundaries ( just turned them around ) and gave me square of 2 -1 which is positive. I hope I was clear xD sry for longpost
@@aggelosspirou8815 sqrt(2-x^2) should be sqrt(4-x^2). Also, try changing to polar coordinates. That should make it easier.
hory shet i nevar knew it was so easa
I learned here more than in 4 lectures I had about this at my Uni.