Double Integration in Polar Coordinates | Example & Derivation
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- เผยแพร่เมื่อ 7 ธ.ค. 2019
- While Cartesian coordinates are great and all, some regions and some integrands are way nicer when described using polar coordinates. In this video we play around with polar coordinates, derive the formula for double integration in polar coordinates, and see an example. The derivation is much like it was in Cartesian coordinates - a limit of a sum of little volumes - but the geometry of the little volumes change in polar coordinates as they no longer have rectangular bases.
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Not only are your formula derivations presented extremely logically but your diagrams show exactly where the formulas come into play. The latter is what makes your explanations so much more cogent than others I've seen. It's one thing being able to follow where a formula comes from, making the abstraction to intuit how that formula applies to examples other than that which was used to derive it is the subsequent "jump" where most students falter in their understanding. From videos like yours and those from a select few other youtubers, I've learned to love calculus for the fact that its one of the only branches of math where it's possible to get a complete intuitive understanding of virtually concept without the rigorous mathematical proofs.
I'm so happy you've been loving calculus recently:)
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I wish you were my teacher, you have explained it perfectly :)
Beautifully clear and concise. Thanks.
Man ! Dr Trefor, I am an engineer who studied math, and you're the best one who can break anything big to small, lol Thank you also for the polar curves which BTW are important in Antenna studies, like the smith chart, Thank you so much Dr, I will definitely follow your videos as they are an immense pleasure to watch. Still have Schaum McGraw hill books and Piskounov books as well, lol but also a CAS calculator to help with all this when one doesn't have time.
amazing teaching! thanks, Dr. Bazett
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Why I love and appreciate your videos is because you give a clear visual picture.
Thank you! Glad they are helping:)
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Your videos give a conceptual based geometrical analysis of every topic which is interesting for me
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These Visualizations really help a lot
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First port of call for an intuitive grasp of mathematics.
thank you sir....
Thank you for excellent explanation , I also watched your integration video in 3-D graphing (makes it easy to understand), wish I had access to 3-D graphs when I took calculus (1970's) Thanks again............Jay
Sir, I am really curious to know how did you managed comprehending such difficult topics without 3d graphs ??
Must say we are lucky, I cant even imagine my life without 2d and 3d graph calcualtors .
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I have my first sem engineering end sem tomorrow your video series was great for a quick revision thanks a lot sir.
Good luck!
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Thanks again Dr. Trefor! Could you please explain where the 'k sub r' goes when you're converting from the summation to the integral?
Thanks sir
Thanks man
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I really really love the way you teach. It is so great. One thing I would like much more of is to see which books you recommend and also love. I see a few books that you like through the link you give about Math Books you love. But I really want to see much more. A bit like the Math Sorcerer but your personal ones even those that are not really study books. You have a unique style and I love it because it makes me understand where it all comes from.
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Nice one haha more examples please
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Excellent as always. I was confused by the f(r, theta) = r function though. This is just z=sqrt(x^2+y^2) in Cartesian coordinates right? So essentially you found the volume under this curve bounded by the cardioid of course?
thank you professor for your wonderful effort. I have a question about how you came up with the r_k idea. I really liked how the proof pieced together and I would like to learn how to think like that or how these solutions are thought up. can you give me pointers what to study or where to look to know how to think up these assumptions and how to verify or nullify them?
I might be wrong but at 5:08 isn't rk +deltar/2 suppose to be the left side of the wedge since an increase in theta results in a counter clockwise rotaion and rk-deltar/2 should then be on the right side of the wedge?
In calculus 1, it is stated that a function is a one-to-one function if the vertical line test intersects the function at most once. How does this change when dealing with multivariable functions?
But what about spherical transform what is its formula
Does the Jacobian always use polar cords, or is it sometimes equal to something besides the scaling factor of r?
Sir, from which software i make this wonderful sketch . 🔥
excuse me professor, but when the angel is from 0 to 2pi should the r be from 1 to 1-sin(&) due to the equation???
Sir , I am not understand why this area change by changing the value of theta...???🙏
W Teacher
sir basiclly double integral is used to find area .... then y we use it in finding area bound by a curve
Why the limit I eg is from 0 to 1- sintheta
Good. But one things wasn't very clear to me. Does one r represent a region in the polar plane and the other r a surface above it ?
Are there two ways of doing this like how you did in the cartesian system? Can we integrate by d-theta first and then dr?
It's possible, but you are far more likely to have curves written r(theta) than as theta(r) so it isn't really that helpful to reverse.
Somehow this feels like I used the Jacobian determinant without actually computing for it
When describing dA you made the choice to have r_k be the average radius of the wedge. I tried it defining r_k as the smaller radius and got dA=1/2*dTheta((r_k+dr)^2-r_k^2) and this expands to dA=r*dr*dTheta+1/2*dr*dr*dTheta. Can we say the second term is significantly smaller than the first term because it has another dr (which goes to 0 in the limit)? It feels similar to the way we defined differentiability for multivariable functions that df=...+epsilon1*dx+epsilon2*dy and the epsilons went to 0 quickly. Thanks so much for answering questions btw!
thanks for solving my doubt...
Delta r /2 ,how to get that ?
At 10:02, where has "dr" gone? Of course, I get that r^3/3 is integral of r^2 which is from (r * (r dr dTh)) but why is "dr" disappear here?
I got lost at the 5 minute mark. If I understood you correctly, the volume of any region is the height times area of the base. If I understood you correctly, you divide the volume into infinitely small volumes and you multiply the number of these values by the limit of infinity. Is this correct? Thank you for your help. I had trouble with Calculus, so please bear with me.
@@DrTrefor Thank you. Your response is very helpful. I will review your lecture again and again. I hope you don't mind if I ask you some more questions about it.
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@Trefor Bazett Addicted to your videos! In cartesian coordinates, there is a way of visualising the integration where you take the area under the curve between two x coordinates at 'one end' and imagining that area between those two x coordinates swept across y axis between your y limits (how you visualised this in your previous video where you calculated the integral two ways). Does this visualisation apply to polar coordinates - where you can imagine an area being swept circularly? It seems like it ought to be because areas further away from the origin are scaled by r. Am I right in thinking that?
I get most of the video and I think I don't have much of a problem with the rk, but I have 1 problem though
When we do the approximation in the Cartesian region we take a certain point, say (xk, yk), that lies at least *in* the small regions we're gonna approximate the volume with, but when you picked here while rk may be in the small region how do we know its theta is *also* in that small region ?
It is the same thing. We are choosing rk and thetak to be some point inside that region...as in we know it is true because it is up to choose it
@@DrTrefor I see, but do we even need to know whether or not that theta k is insinde the small region? I rethought it, on the assumption that the point inside f, f(rk, theta k), is different from the rk we chose, they would reach the same r which is *r* but the theta of the point we chose isn't used so we may not need its theta at all
if its already in polar dont you need the extra r?
Nope, the r comes when we convert from Cartesian, a generic one already in polar doesn’t require an r
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The title is misspelled unless that was intentional.
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@@DrTrefor I appreciate the reply!
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