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!!!
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.
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!
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!
@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!
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@
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
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! 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.
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.
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.
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.
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?
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.
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.
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.
If you look at the initial bounds of x; they go from 0 to sqrt(2y-y^2). So think of the integral as slices across x from the horizontal x=0 to the x value determined by the equation. Thus overall, this means that the region R is on the right side of the x=0 horizontal line (y axis) and thus the right side of the circle which has bounds 0 to pi/2.
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.
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.
In part a) how did he know that theta is pi/4?? I read in the comments someone wrote that y=x is half of the first quadrant and the first quadrant is pi/2 so half of that is pi/4... but what if I don't know that y=x is half of the first quadrant?
+Nafa1 My thought is to draw an arbitrary line with equation: x=# that will intersect the line, y=x. This would give you a right triangle, so you can find where they intersect for the y value, then solve for theta.
+Nafa1 you can also substitute y with rsin(theta) and x with rcos(theta). then you get rsin(theta)=rcos(theta), now solve for theta , r's cancel , sin/cos = tan(theta) = 1 Do inverse tan and solve for angle theta and you get same answer.
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!!!
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 ...💐
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!
dam now i want to go to MIT
Who doesn't?
@@shi_shii_ You?
Thank you I have to study at home because of the corona virus. This came in so clutch!!
Happy to help!
his chalk is so big
it's sidewalk chalk, writes better.
brother this is MIT
It’s girthy
“Chalk”
mmmm
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 ..
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!
dude, you`re awesome, i was able to do my homework thanks to you
These are actually very instructive excercises.
ruclips.net/video/vFDMaHQ4kW8/видео.html ...💐
@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
This was really awesome, thanks! Helped me in my end sems!!
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!
I have an advanced calc exam on Tuesday AND YOU ARE A LIFE SAVER ILY
Beautifully explained,
I liked your teaching.
You don’t just cancel the r in 10:45. You solve for quadratic
This is one of the best explanations for this topic.
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.
Thanks a lot MIT. I've finally understood the concept!!!
First time i watch a MIT class and i understand everything
What a simple explanation which everyone could understand
@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!
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
So how do you solve this if e^-x^2 is multiplied by Cos(bx) with respect to x
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
Mindblowing explanation! Thanks!
How does he go from integral of 1/r^3 to the integral of 1/r^2 in one movement... 6:30-7:10 ??
The r in rdrd∅ cancels one of the R's in (1/r³)
@@jeskomargargot3454 Thank you sir! I watched this before the chapter was introduced in my course, makes sense now.
Well done young man, really nailed it
is this the same as multiple integrals? im just starting out and im very confused...
ruclips.net/video/vFDMaHQ4kW8/видео.html ...💐
thank you so much David ❤️
Finally, it makes perfect sense
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
Thank you David!! This was extremely helpful !
the -(1/r) can be changed to (1/r) if you swap the limits of integration, which is what he did here.
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
@maplestorypl He wrote it correctly. dA becomes r dr dtheta so the integral does become 1 / r^2.
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.........
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
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.
at 10:33, you should not cancel the variable r because you miss the answer r = 0.
6:31 how do we know that dydx becomes rdrdθ?
we change from cartessian to polar coordinate , to prove this use jacobian matrix then you will get the answer
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.
Best ! Like IT ! Simply ,straight forward and lucid :D
Finally understood the concept
loveddd ittt!!! thankyouuu so much david sir
What can be his age in 2011🤔
Explained beautifully........
Thank you professor You cleared all my doubts
🙏🙏 Dhanyawad and Namaskar
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.
Good job bro. You really explain well.
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.
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?
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?
shouldn´t teta go from -π/2 to π/2?
Thank you, just thank you.
To the instructor, thank you.
Nicely explained. But getting out of the board feels very awkward.
Very nicely demonstrated. I appreciate it!
Brillient.... Absolutly great
Can any one tell me that integration in polar coordinates and double integration in polar coordinates are same????
@marcuswauson you're evaluating 1/2 sin theta from 0-Pi/4.
Final answer for c) ??
nice and concise. thank you david
Best online classroom
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.
explain more on how you find those limits. pie over 4. how did you find it
ruclips.net/video/vFDMaHQ4kW8/видео.html ...💐
Best lecture . Thanks for your kindly helping
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?
ruclips.net/video/vFDMaHQ4kW8/видео.html ...💐
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?
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)
Great videos, David. Thanks kindly!
How to explain why dydx becomes rdrdθ ?
Professor Leonard has a full explanation about it.
Denis Auroux multivariable calculus lect 16, 17, 18, 19
I am an old man now. seems I can understand this better than my young age. feel like to go back to school.
IMAGINE
Playing Chalk Fight with this MIT chalks😂😂😁🙈
this helped me, thanks so much!
Sir plz help to make this in polar:double integral y=1 to 2,x=0 to 1 (1/x^2+y^2)dxdy
MIT depends heavily on the chalk industry
At 7:05 he messed up and put 1/r^2 not 1/r^3
He is correct only
He simplified r/r^3
This video helps me for my calculus 3 paper tomorrow!! Thanks a lot 💖
why in example c teta goes only to Pi/2 = 90 and not to pi =180
How did he know 1/(x^2+y^2)^1/2 was r?
How did he know x^2+y^2=r^2?
sqrt(2)/4 = "...elementary kind of integral" ... really?
oh god, what have I been learning this whole time >.
8:14 : incorrrect calculation, it is actually sqrt(2)/4 -1
Than you man, great video, simple and nice explanation :D
Excellent no one able to tell how limits of polar are going on.🙏
Welcome back
ARCTAN(y/x) when y==x is π/4 for the rest of us
Worth the watch! Very helpful!
for the last one the int.val is not just from 0 to pi?
If you look at the initial bounds of x; they go from 0 to sqrt(2y-y^2). So think of the integral as slices across x from the horizontal x=0 to the x value determined by the equation. Thus overall, this means that the region R is on the right side of the x=0 horizontal line (y axis) and thus the right side of the circle which has bounds 0 to pi/2.
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.
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
In part a) how did he know that theta is pi/4?? I read in the comments someone wrote that y=x is half of the first quadrant and the first quadrant is pi/2 so half of that is pi/4... but what if I don't know that y=x is half of the first quadrant?
+Nafa1 My thought is to draw an arbitrary line with equation: x=# that will intersect the line, y=x. This would give you a right triangle, so you can find where they intersect for the y value, then solve for theta.
+Nafa1 you can also substitute y with rsin(theta) and x with rcos(theta). then you get rsin(theta)=rcos(theta), now solve for theta , r's cancel , sin/cos = tan(theta) = 1
Do inverse tan and solve for angle theta and you get same answer.
Thank you thank you thank you I appreciate it a lot. Nice instructor very helpful
Excellent work
how could you just assume the lower limit oangle of the parabola to be 0