Integration in polar coordinates | MIT 18.02SC Multivariable Calculus, Fall 2010
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- เผยแพร่เมื่อ 1 ก.ค. 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!!!
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!
Beautifully explained,
I liked your teaching.
his chalk is so big
it's sidewalk chalk, writes better.
brother this is MIT
It’s girthy
“Chalk”
mmmm
This was really awesome, thanks! Helped me in my end sems!!
@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
dude, you`re awesome, i was able to do my homework thanks to you
dam now i want to go to MIT
Who doesn't?
@@shi_shii_ You?
This is one of the best explanations for this topic.
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!
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.
th-cam.com/video/vFDMaHQ4kW8/w-d-xo.html ...💐
I have an advanced calc exam on Tuesday AND YOU ARE A LIFE SAVER ILY
These are actually very instructive excercises.
th-cam.com/video/vFDMaHQ4kW8/w-d-xo.html ...💐
@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)
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.
Mindblowing explanation! Thanks!
What a simple explanation which everyone could understand
Well done young man, really nailed it
Thank you David!! This was extremely helpful !
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!
You don’t just cancel the r in 10:45. You solve for quadratic
Thank you I have to study at home because of the corona virus. This came in so clutch!!
Happy to help!
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 ..
Very nicely demonstrated. I appreciate it!
thank you so much David ❤️
To the instructor, thank you.
Best ! Like IT ! Simply ,straight forward and lucid :D
First time i watch a MIT class and i understand everything
Finally, it makes perfect sense
@maplestorypl He wrote it correctly. dA becomes r dr dtheta so the integral does become 1 / r^2.
this helped me, thanks so much!
@marcuswauson you're evaluating 1/2 sin theta from 0-Pi/4.
Great videos, David. Thanks kindly!
Worth the watch! Very helpful!
the -(1/r) can be changed to (1/r) if you swap the limits of integration, which is what he did here.
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?
I loved it! Thanks so much!!!
Good job bro. You really explain well.
Thank you thank you thank you I appreciate it a lot. Nice instructor very helpful
That was really helpful. Super clear!
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!
nice and concise. thank you david
loveddd ittt!!! thankyouuu so much david sir
Finally understood the concept
Really helpful, thank you!
Thank you, just thank you.
Than you man, great video, simple and nice explanation :D
Best lecture . Thanks for your kindly helping
So how do you solve this if e^-x^2 is multiplied by Cos(bx) with respect to x
Explained beautifully........
Brillient.... Absolutly great
limits explanation was awesome --which obviously is the heart and soul of the prob...lol in C in forgot to put square root.........
yeah, he didn't realize it was 1/r^3 when he re-wrote it
great video mate, really helpful!!
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
Because he has (1/r^3) r drdΘ, if you combine that r with (1/r^3) it would have a negtive power if you put it over one, (r has a power of one), therefore it becomes (1/r^(3-1)... and becomes 1/r^2. Hope this helps.
That helps sp much you have no idea, you are a legend.
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?
Great video! Really helpful.
Great Video!!! Greetings from GT!!!
Very well explained!
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
Welcome back
Very clear teaching!
Good base ,so love you sir!
Excellent no one able to tell how limits of polar are going on.🙏
shouldn´t teta go from -π/2 to π/2?
Awesome thank you so much mit
Excellent work
Best online classroom
Can any one tell me that integration in polar coordinates and double integration in polar coordinates are same????
Thanks! I think you articulate it really well. That's awsome! Wish you're my prof.
Not to be disrespectful, but are you an undergraduate? You look like a peer.
In the introduction video of the TAs, David was a graduate student. Of course, that was 10 years ago.
why in example c teta goes only to Pi/2 = 90 and not to pi =180
Sir plz help to make this in polar:double integral y=1 to 2,x=0 to 1 (1/x^2+y^2)dxdy
thanks very helpful examples
Nicely explained. But getting out of the board feels very awkward.
great video!
Great video
What can be his age in 2011🤔
Thank you professor You cleared all my doubts
🙏🙏 Dhanyawad and Namaskar
cos of 0 is not 0, cos of 0 is 1, your answer to a should be (squareroot of 2) minus 2 all over 4. As cos of 0 is one, and one minus a half should give you squareroot of 2 over four minus a half.
Final answer for c) ??
for the last problem, it would be easy to switch the coordinate instead of x give y and so on.
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?
th-cam.com/video/vFDMaHQ4kW8/w-d-xo.html ...💐
Thanks a ton!
Perfect
explain more on how you find those limits. pie over 4. how did you find it
th-cam.com/video/vFDMaHQ4kW8/w-d-xo.html ...💐
Very well explained,,,,
I am an old man now. seems I can understand this better than my young age. feel like to go back to school.
Thank you very much for video
Una explicacion bastante acertada.
SUPER!
This video helps me for my calculus 3 paper tomorrow!! Thanks a lot 💖
how could you just assume the lower limit oangle of the parabola to be 0
ARCTAN(y/x) when y==x is π/4 for the rest of us
at 10:33, you should not cancel the variable r because you miss the answer r = 0.
MIT depends heavily on the chalk industry
is this the same as multiple integrals? im just starting out and im very confused...
th-cam.com/video/vFDMaHQ4kW8/w-d-xo.html ...💐