Integration in polar coordinates | MIT 18.02SC Multivariable Calculus, Fall 2010
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- čas přidán 5. 08. 2024
- Integration in polar coordinates
Instructor: David Jordan
View the complete course: ocw.mit.edu/18-02SCF10
License: Creative Commons BY-NC-SA
More information at ocw.mit.edu/terms
More courses at ocw.mit.edu
This was honestly so so so helpful. From the bottom of my heart, thank you thank you THANK YOU so much. We need more tutors like you.
I like how you subtitle what you speak, it's helpful for me.
Man thank you so much, literally the first time i actual properly understood polar coordinate transformations despite trying to wrap my head around it for like 3 months. Appreciate you!!!
Beautifully explained,
I liked your teaching.
This was really awesome, thanks! Helped me in my end sems!!
Thanks a lot, I was struggling to solve a problem related usage of polar coordinates in my assignment, but now i solved it in just 5min. thanks once again!
makes me realise how much I have forgotten in the 40+ years since uni. Expressing the integral in that graphic form was not something we were taught and it really helps me to grasp the whole concept; would have grokked so much more if we had been!
dude, you`re awesome, i was able to do my homework thanks to you
This is one of the best explanations for this topic.
Thanks a lot MIT. I've finally understood the concept!!!
In my school they simply say: evaluate the following integral just as given in part (a) above. It is entirely ALL UP TO YOU how you approach it. Of course one method makes life a whole lot easier than say compared to another one. It is expected that you know which method to use accordingly.
czcams.com/video/vFDMaHQ4kW8/video.html ...💐
Thank you I have to study at home because of the corona virus. This came in so clutch!!
Happy to help!
Well done young man, really nailed it
Thank you David!! This was extremely helpful !
dam now i want to go to MIT
Who doesn't?
@@shi_shii_ You?
Mindblowing explanation! Thanks!
So how do you solve this if e^-x^2 is multiplied by Cos(bx) with respect to x
@marcuswauson
The integrand is [cos(theta)-(1/2)*cos(theta)] but when integrated, it becomes [sin(theta)-(1/2)*sin(theta)] from zero to pi/4. When evaluated equals sqrt(2)/2 - sqrt2)/4 = sqrt(2)/4
only in Calc BC trying to solve an argument with a friend when i found this… video made the concepts simple to understand and was extremely informative!
I have an advanced calc exam on Tuesday AND YOU ARE A LIFE SAVER ILY
Very nicely demonstrated. I appreciate it!
This was so helpful to solve questions.
My professor just solve 3-4 easy questions and left us with such questions
Thanks a lot sir ..
thank you so much David ❤️
Best ! Like IT ! Simply ,straight forward and lucid :D
What a simple explanation which everyone could understand
this helped me, thanks so much!
Worth the watch! Very helpful!
To the instructor, thank you.
Great videos, David. Thanks kindly!
I loved it! Thanks so much!!!
This is a graphical approach, which is fine, but can this be done directly by looking at the Cartesian bounds of integration i.e without Graphing the regions?
Really helpful, thank you!
That was really helpful. Super clear!
Finally, it makes perfect sense
Than you man, great video, simple and nice explanation :D
These are actually very instructive excercises.
czcams.com/video/vFDMaHQ4kW8/video.html ...💐
Does the last example diverge? Because the the internal integration integral[r=0,r=2sin(theta)] r^-2 dr = -1/r | [r=0,r=2sin(theta)] = -(1/2sin(theta)-1/0) = diverge
Good job bro. You really explain well.
great video mate, really helpful!!
Thank you, just thank you.
On the first one, shouldn't the result have been sqrt(2)/4 - 1/2?
Wouldn't you have to subtract off the (1 - 1/2) from the lower bound of theta?
nice and concise. thank you david
his chalk is so big
it's sidewalk chalk, writes better.
brother this is MIT
It’s girthy
“Chalk”
mmmm
loveddd ittt!!! thankyouuu so much david sir
Great video! Really helpful.
Best lecture . Thanks for your kindly helping
@PeaceUdo
Question a and b) The upper bound for y is y=x.
The line y = x is always at a 45 degree (pi/4) angle with the x axis.
If you dont get why, then for example lets say y = x = n (as y=x)
then
tan θ = n/n
tan θ = 1
therefore θ=45 degree (pi/4)
Thank you thank you thank you I appreciate it a lot. Nice instructor very helpful
Brillient.... Absolutly great
Explained beautifully........
Great Video!!! Greetings from GT!!!
Very well explained!
yes; and this is how we deal with it. find the range of angles for which r is negative, split these segments away from your first integral and put a negative sign in front in order to find the area. example r=cos(theta) from 0 to 2pi gives negative r from pi/2 to 3pi/2. the area enclose by r=cos(theta) = integral of rcos(thta) from 3pi/2 to pi/2 plus (-) integral cos(thta) from pi/2 to 3pi/2. integrating the original function from o to 2pi gives 0 if you ignore the fact r goes negative.
You don’t just cancel the r in 10:45. You solve for quadratic
First time i watch a MIT class and i understand everything
Good base ,so love you sir!
Very clear teaching!
Finally understood the concept
Welcome back
Can any one tell me that integration in polar coordinates and double integration in polar coordinates are same????
Awesome thank you so much mit
why in example c teta goes only to Pi/2 = 90 and not to pi =180
shouldn´t teta go from -π/2 to π/2?
@marcuswauson you're evaluating 1/2 sin theta from 0-Pi/4.
Great video
Thanks a ton!
great video!
Excellent work
Perfect
thanks very helpful examples
the one thing i think that is missing when converting is that the F function at the end is not a function of x, y but r, theta
you cant just plug in f given a function of x,y when in polar form
SUPER!
thank you
how do you know the maximum line is pi/4?
Look at a unit circle. You see that y = x is half of quadrant one and quadrant one is from 0 to pi/2. and half of that is pi/4 so you get theta = pi/4 as your max theta value.
Thank u !!!
Very well explained,,,,
is this the same as multiple integrals? im just starting out and im very confused...
czcams.com/video/vFDMaHQ4kW8/video.html ...💐
Not only do the brilliant students of MIT get excellent student teachers, but I can't understand majority of what my teacher says through his thick accent...and no my school is not well known even in the city that it resides in. Thanks David and MIT!
Sir plz help to make this in polar:double integral y=1 to 2,x=0 to 1 (1/x^2+y^2)dxdy
U really helped me :)
Thanx
Amazing! Thanx
Thank you very much for video
Best online classroom
Hi I need more practice to understand how to identify the sector which is to be integrated. Do you have a preceding video that demonstrates this?
czcams.com/video/vFDMaHQ4kW8/video.html ...💐
Helpful!
how could you just assume the lower limit oangle of the parabola to be 0
Una explicacion bastante acertada.
Final answer for c) ??
He is a genius...
limits explanation was awesome --which obviously is the heart and soul of the prob...lol in C in forgot to put square root.........
Thank man...
@maplestorypl He wrote it correctly. dA becomes r dr dtheta so the integral does become 1 / r^2.
In part a) why do you go from 1/r^3 to 1/r^2?
he multiplied 1/r^3 by the r from "r dr d theta", leaving 1/r^2
Thank u 4 dis video
Yes yes this video is good
What can be his age in 2011🤔
THANK YOUUUUUUUUUUUUU
thanks
Very helpful
How did he get rdrd(Theta) from dydx?