Integration in polar coordinates | MIT 18.02SC Multivariable Calculus, Fall 2010
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- Опубліковано 29 лип 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
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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!!
dude, you`re awesome, i was able to do my homework thanks to you
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!
dam now i want to go to MIT
Who doesn't?
@@shi_shii_ You?
This is one of the best explanations for this topic.
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!
Thanks a lot MIT. I've finally understood the concept!!!
Mindblowing explanation! Thanks!
Well done young man, really nailed it
Thank you I have to study at home because of the corona virus. This came in so clutch!!
Happy to help!
Thank you David!! This was extremely helpful !
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.
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his chalk is so big
it's sidewalk chalk, writes better.
brother this is MIT
It’s girthy
“Chalk”
mmmm
So how do you solve this if e^-x^2 is multiplied by Cos(bx) with respect to x
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!
@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
thank you so much David ❤️
These are actually very instructive excercises.
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I have an advanced calc exam on Tuesday AND YOU ARE A LIFE SAVER ILY
Very nicely demonstrated. I appreciate it!
this helped me, thanks so much!
Great videos, David. Thanks kindly!
Really helpful, thank you!
I loved it! Thanks so much!!!
What a simple explanation which everyone could understand
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?
Best ! Like IT ! Simply ,straight forward and lucid :D
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?
Thank you thank you thank you I appreciate it a lot. Nice instructor very helpful
To the instructor, thank you.
Best lecture . Thanks for your kindly helping
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.
That was really helpful. Super clear!
Thank you, just thank you.
Than you man, great video, simple and nice explanation :D
Good job bro. You really explain well.
loveddd ittt!!! thankyouuu so much david sir
nice and concise. thank you david
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 ..
Finally, it makes perfect sense
Worth the watch! Very helpful!
Brillient.... Absolutly great
Explained beautifully........
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
@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)
Good base ,so love you sir!
Great video! Really helpful.
Awesome thank you so much mit
great video mate, really helpful!!
Very well explained!
Can any one tell me that integration in polar coordinates and double integration in polar coordinates are same????
Finally understood the concept
Very clear teaching!
why in example c teta goes only to Pi/2 = 90 and not to pi =180
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
@marcuswauson you're evaluating 1/2 sin theta from 0-Pi/4.
Great Video!!! Greetings from GT!!!
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
Una explicacion bastante acertada.
is this the same as multiple integrals? im just starting out and im very confused...
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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
Thanks a ton!
Thank you very much for video
Excellent work
U really helped me :)
Thanx
limits explanation was awesome --which obviously is the heart and soul of the prob...lol in C in forgot to put square root.........
great video!
how could you just assume the lower limit oangle of the parabola to be 0
Perfect
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?
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Welcome back
Great video
Thank you so much kind sir.
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Very well explained,,,,
shouldn´t teta go from -π/2 to π/2?
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!
@maplestorypl He wrote it correctly. dA becomes r dr dtheta so the integral does become 1 / r^2.
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.
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.
SUPER!
What can be his age in 2011🤔
How did you know which point ot use to find r? in the first example you used x=1 and x=2 to find r =sec and 2sec. but in the second and third example you used the y=x and y=x^2 rather than using the x=0 and x=1, just confusing stuff like that makes me get different points of integration
i also was very confused by that. he tried to be clever in the first example rather than just using a consistent method. if you take x=1 as the lower bound function and swap in the polar coordinates, you get r*cos(theta)=1. solving for r gets you 1/cos(theta) i.e. sec(theta). Same for the upper bound of x=2.
Exactly. Did anyone figure out a general way to do this? :/
Nevermind. I found the solution. He didn't explain this but here it is:
The on which theeta is changing is the one we use to find the upper limit for r. Just simply equate that curve by replacing x and y with rsin(0) and rCos(0) respectively.
In the first part, theeta was changing on x=1 so he used that to find the r's limits. In the second and third, you can see what i mean.
Thank u !!!
the -(1/r) can be changed to (1/r) if you swap the limits of integration, which is what he did here.
Excellent no one able to tell how limits of polar are going on.🙏
Final answer for c) ??
how is dxdy equal to rdrdθ
to prove this , we use jacobian method. And we all know that from cartessian to polar coordinate , x=cos@ and y=sin@ then dxdy change to drd@ and apply this to jacobian matrix (dx/dr dx/d@ , dy/dr dy/d@) then we get rdrd@
or... dA = (dr) (ds)? = (dr) d(rθ) = r (dr) (dθ) = (dx) (dy)
use Jacobian
There is a simpler method to understand that, dxdy is actually a very small square. So its equal to rdrdtheta
Look up the Jacobian for polar coordinates
thank you
Thanks a lot !!!
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Best online classroom
Thank u 4 dis video
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.
in c , what if it was a circle or half circle with its origin in (1,1),
like (x-1)^2 + (y-1)^2 =1
it must be the whole circle due to its positive origin
THANK YOUUUUUUUUUUUUU