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!
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.
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!!!
@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
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!
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.
@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)
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.
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!
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!
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
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@
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.
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.
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.
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.
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.
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.
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.
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
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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.
dam now i want to go to MIT
Who doesn't?
@@shi_shii_ 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!
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.
ruclips.net/video/vFDMaHQ4kW8/видео.html ...💐
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!!!
his chalk is so big
it's sidewalk chalk, writes better.
brother this is MIT
It’s girthy
“Chalk”
mmmm
@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
Beautifully explained,
I liked your teaching.
These are actually very instructive excercises.
ruclips.net/video/vFDMaHQ4kW8/видео.html ...💐
Thank you I have to study at home because of the corona virus. This came in so clutch!!
Happy to help!
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!
Thanks a lot MIT. I've finally understood the concept!!!
This is one of the best explanations for this topic.
You don’t just cancel the r in 10:45. You solve for quadratic
What a simple explanation which everyone could understand
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.
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 ..
@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)
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.
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!
Finally, it makes perfect sense
I have an advanced calc exam on Tuesday AND YOU ARE A LIFE SAVER ILY
First time i watch a MIT class and i understand everything
To the instructor, thank you.
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!
yeah, he didn't realize it was 1/r^3 when he re-wrote it
limits explanation was awesome --which obviously is the heart and soul of the prob...lol in C in forgot to put square root.........
Mindblowing explanation! Thanks!
Explained beautifully........
Thank you, just thank you.
@marcuswauson you're evaluating 1/2 sin theta from 0-Pi/4.
@maplestorypl He wrote it correctly. dA becomes r dr dtheta so the integral does become 1 / r^2.
What can be his age in 2011🤔
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
Brillient.... Absolutly great
Well done young man, really nailed it
Finally understood the concept
nice and concise. thank you david
the -(1/r) can be changed to (1/r) if you swap the limits of integration, which is what he did here.
Best ! Like IT ! Simply ,straight forward and lucid :D
Thank you David!! This was extremely helpful !
thank you so much David ❤️
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
Really helpful, thank you!
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.
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.
Very nicely demonstrated. I appreciate it!
Nicely explained. But getting out of the board feels very awkward.
this helped me, thanks so much!
Best lecture . Thanks for your kindly helping
Good job bro. You really explain well.
I think somewhere in the middle of a.), you accidentally replaced 1/r^3 with 1/r^2 ...
Watch it again its correct
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.
at 10:33, you should not cancel the variable r because you miss the answer r = 0.
Best online classroom
is this the same as multiple integrals? im just starting out and im very confused...
ruclips.net/video/vFDMaHQ4kW8/видео.html ...💐
I loved it! Thanks so much!!!
So how do you solve this if e^-x^2 is multiplied by Cos(bx) with respect to x
He is a genius...
MIT depends heavily on the chalk industry
thanks very helpful examples
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.
How did he know 1/(x^2+y^2)^1/2 was r?
How did he know x^2+y^2=r^2?
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?
loveddd ittt!!! thankyouuu so much david sir
Than you man, great video, simple and nice explanation :D
Thank you professor You cleared all my doubts
🙏🙏 Dhanyawad and Namaskar
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.
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.
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)
for the last problem, it would be easy to switch the coordinate instead of x give y and so on.
Welcome back
shouldn´t teta go from -π/2 to π/2?
Very clear teaching!
Final answer for c) ??
Excellent no one able to tell how limits of polar are going on.🙏
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.
youre stupid
You’re replying to another comment aren’t you?
Thank you thank you thank you I appreciate it a lot. Nice instructor very helpful
ARCTAN(y/x) when y==x is π/4 for the rest of us
SUPER!
Great Video!!! Greetings from GT!!!
Awesome thank you so much mit
why in example c teta goes only to Pi/2 = 90 and not to pi =180
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
Great video! Really helpful.
Thanks a ton!
Great videos, David. Thanks kindly!
Excellent work
Good base ,so love you sir!
how could you just assume the lower limit oangle of the parabola to be 0
Very well explained!
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.
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
That was really helpful. Super clear!
I am an old man now. seems I can understand this better than my young age. feel like to go back to school.
Perfect
Worth the watch! Very helpful!
Oh my gosh.. Our IIT is world's best in terms of proffessor knowledge
bro iit nahi VO MIT ka he .
aur MIT no 1 he world me
Thank you so much kind sir.
ruclips.net/video/vFDMaHQ4kW8/видео.html ...💐
great video mate, really helpful!!