Integration in polar coordinates | MIT 18.02SC Multivariable Calculus, Fall 2010

[@rach3834]

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.

[@m.donnediego587]

I like how you subtitle what you speak, it's helpful for me.

[@wssz112]

dam now i want to go to MIT

[@shi_shii_]

Who doesn't?

[@martinrosol7719]

​@@shi_shii_ You?

[@ianmoseley9910]

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!

[@rthelionheart]

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.

[@beoptimistic5853]

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[@VocalBeast]

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!!!

[@benn7139]

his chalk is so big

[@vtace1]

it's sidewalk chalk, writes better.

[@Tech4GamingIndia]

brother this is MIT

[@PunkHippie1971]

It’s girthy

[@captainobvious1415]

“Chalk”

[@sambalder9343]

mmmm

[@AldenRyno]

@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

[@muhammadzeeshankhan7251]

Beautifully explained, I liked your teaching.

[@morgard211]

These are actually very instructive excercises.

[@beoptimistic5853]

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[@ashtongaeta2581]

Thank you I have to study at home because of the corona virus. This came in so clutch!!

[@mitocw]

Happy to help!

[@DeprecatedAPI]

This was really awesome, thanks! Helped me in my end sems!!

[@oliverbrace4505]

dude, you`re awesome, i was able to do my homework thanks to you

[@anikethsridhargund6671]

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!

[@suruchisolanki3603]

Thanks a lot MIT. I've finally understood the concept!!!

[@shreyassahu6930]

This is one of the best explanations for this topic.

[@meerismailali6082]

You don’t just cancel the r in 10:45. You solve for quadratic

[@rkumaresh]

What a simple explanation which everyone could understand

[@alexandresauve7560]

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.

[@sairajbhosale3984]

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 ..

[@Jameel263]

@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)

[@MCSPT117]

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.

[@23StudiosSports]

That helps sp much you have no idea, you are a legend.

[@st.johntsuno-wayne2489]

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!

[@Paulovrish7334]

Finally, it makes perfect sense

[@LICKSandWINKS]

I have an advanced calc exam on Tuesday AND YOU ARE A LIFE SAVER ILY

[@edwinsebastianperezrodrigu8659]

First time i watch a MIT class and i understand everything

[@vrendus522]

To the instructor, thank you.

[@NickZachPattyWack]

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!

[@DoggoWillink]

yeah, he didn't realize it was 1/r^3 when he re-wrote it

[@anuraagkaravadi1640]

limits explanation was awesome --which obviously is the heart and soul of the prob...lol in C in forgot to put square root.........

[@HimanshuShekhar13s]

Mindblowing explanation! Thanks!

[@quratulainfatima109]

Explained beautifully........

[@GradientSoln-En]

Thank you, just thank you.

[@gnauhandy]

@marcuswauson you're evaluating 1/2 sin theta from 0-Pi/4.

[@jbonn5]

@maplestorypl He wrote it correctly. dA becomes r dr dtheta so the integral does become 1 / r^2.

[@bhaktimd9299]

What can be his age in 2011🤔

[@joeferreira-qr7iq]

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

[@usamafarooqi7292]

Brillient.... Absolutly great

[@MayankSharma-qd9ny]

Finally understood the concept

[@vychuck]

Well done young man, really nailed it

[@platinumk]

nice and concise. thank you david

[@joshie92]

the -(1/r) can be changed to (1/r) if you swap the limits of integration, which is what he did here.

[@Dhavalc2011]

Best ! Like IT ! Simply ,straight forward and lucid :D

[@WolfBoyBenRawr]

Thank you David!! This was extremely helpful !

[@AbhayArsekar]

thank you so much David ❤️

[@gabrieltmapondera9697]

how is dxdy equal to rdrdθ

[@adellewilliams3266]

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@

[@molkgfmf5699]

or... dA = (dr) (ds)? = (dr) d(rθ) = r (dr) (dθ) = (dx) (dy)

[@AkashYadav-mr4hg]

use Jacobian

[@JamilKhan-hk1wl]

There is a simpler method to understand that, dxdy is actually a very small square. So its equal to rdrdtheta

[@АзХашми]

Look up the Jacobian for polar coordinates

[@vinitamaharaj5738]

Really helpful, thank you!

[@marcuswauson]

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.

[@rain74925]

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.

[@erikumble]

In the introduction video of the TAs, David was a graduate student. Of course, that was 10 years ago.

[@QuantumDisciple7]

Very nicely demonstrated. I appreciate it!

[@AhmadKhan-hb3je]

Nicely explained. But getting out of the board feels very awkward.

[@HomoSapiensMember]

this helped me, thanks so much!

[@hargeysasomaliland4374]

Best lecture . Thanks for your kindly helping

[@VikasSingh-cv2fu]

Good job bro. You really explain well.

[@pedroff_1]

I think somewhere in the middle of a.), you accidentally replaced 1/r^3 with 1/r^2 ...

[@sumeepriya5454]

Watch it again its correct

[@ChucksSEADnDEAD]

It was subtle but you have to watch out for the 1/r^3 r dr dtheta. The r in the nominator position crosses out one of the rs in the denominator leaving 1/r^2 dr dtheta.

[@syoushiro1837]

at 10:33, you should not cancel the variable r because you miss the answer r = 0.

[@sumitgupta6905]

Best online classroom

[@riya6549]

is this the same as multiple integrals? im just starting out and im very confused...

[@beoptimistic5853]

kzbin.info/www/bejne/rHenfpR-hpmejZo ...💐

[@русскийпартизан-ь6п]

I loved it! Thanks so much!!!

[@femiairboy94]

So how do you solve this if e^-x^2 is multiplied by Cos(bx) with respect to x

[@ERLAANKABABU23BME706]

He is a genius...

[@nelsonlopez9587]

MIT depends heavily on the chalk industry

[@moasfco11]

thanks very helpful examples

[@anjaneyasharma322]

Students note double or triple integration does not give correct results. Follow simple integration. The question is why count something twice as it happens in double triple integration.

[@the_eternal_student]

How did he know 1/(x^2+y^2)^1/2 was r? How did he know x^2+y^2=r^2?

[@BlazeCyndaquil]

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?

[@ananyarathore4678]

loveddd ittt!!! thankyouuu so much david sir

[@pudingstorm1]

Than you man, great video, simple and nice explanation :D

[@prakharuttarpradesh3517]

Thank you professor You cleared all my doubts 🙏🙏 Dhanyawad and Namaskar

[@rationalLeft]

I think there is a little mistake in the end of letter a). The result should be sqrt (2)/ 4 - 1/2, instead of sqrt (2)/4. Thank you for the excelent explanation.

[@waltvanamstel6807]

you have to integrate the 1/2*cos(θ), which gives you 1/2*sin(θ) integrating from 0 to pi/4. So sqrt(2)/4 is indeed correct.

[@samuelminea5520]

rationalLeft He had to integrate cos(a)-1/2*cos(a)=1/2*cos(a) which has the antiderivative equal to 1/2*sin(a) from 0 to pi/4 = sqrt(2)/(4)

[@freakingik2781]

Welcome back

[@mustaphabouchaqour7770]

for the last problem, it would be easy to switch the coordinate instead of x give y and so on.

[@MarikoIchigo]

shouldn´t teta go from -π/2 to π/2?

[@thecivilizedguy8425]

Very clear teaching!

[@darprahimi9692]

Final answer for c) ??

[@deepakkumarchandel2444]

Excellent no one able to tell how limits of polar are going on.🙏

[@johnpwnsyou]

No. There are several reasons why. First, r is a length, which logically can't be negative. Second, If you look at the formula, r=(x^2+y^2)^(1/2), you'll realize that there's no way for it to be negative since it's components are all squared.

[@valkarez1137]

youre stupid

[@sreekarg9553]

You’re replying to another comment aren’t you?

[@ricardinhovorkes4876]

Thank you thank you thank you I appreciate it a lot. Nice instructor very helpful

[@celsiusfahrenheit1176]

ARCTAN(y/x) when y==x is π/4 for the rest of us

[@kamaraju1551]

SUPER!

[@zeldadu09]

Great Video!!! Greetings from GT!!!

[@Dineshkumar-xv4xz]

Awesome thank you so much mit

[@dant9944]

why in example c teta goes only to Pi/2 = 90 and not to pi =180

[@noahmckeever9059]

In part a) why do you go from 1/r^3 to 1/r^2?

[@AlexVX_]

he multiplied 1/r^3 by the r from "r dr d theta", leaving 1/r^2

[@Lagos3sgte]

Great video! Really helpful.

[@anannyauberoi7984]

Thanks a ton!

[@FirstGradeCalculus]

Great videos, David. Thanks kindly!

[@matthewskatuta1305]

Excellent work

[@krishnaghorai8146]

Good base ,so love you sir!

[@aditkalyani7966]

Very well explained!

[@animeshpathak3921]

how could you just assume the lower limit oangle of the parabola to be 0

[@HereToday32]

how do you know the maximum line is pi/4?

[@jeffreychang2293]

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.

[@hasanabs]

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

[@_tasneem7378]

That was really helpful. Super clear!

[@l1mmg0t]

I am an old man now. seems I can understand this better than my young age. feel like to go back to school.

[@grovestreet9165]

Perfect

[@beercity123]

Worth the watch! Very helpful!

[@sajathsalim261]

Oh my gosh.. Our IIT is world's best in terms of proffessor knowledge

[@MohitSharma-wt9ex]

bro iit nahi VO MIT ka he . aur MIT no 1 he world me

[@nitesjung1274]

Thank you so much kind sir.

[@beoptimistic5853]

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[@sumballaboi]

great video mate, really helpful!!