For anybody wondering why 1/2 is added at 3:53, the constant of 1/2 comes from the fact that the region is symmetric and 2D. When you're averaging values along a dimension that involves a parabolic distribution of mass, the factor of 1/2 accounts for how the average "weight" is distributed across the parabola. In essence, you are averaging out the mass distribution to get your center of mass.
literally a miracle worker, taking my calc 2 final today and this is the last concept I need to understand. your quick and comprehensive explanations carried me through this class- cannot thank you enough.
@@blackpenredpen Wait wait wait... should it be xydx and xydy? We are basically meassuring amount of change coordinate within area? That way we are limiting rate of change to coordinate and area... right? X bar i get, it's basically searching the x coordinate, but y bar? Shouldn't it be y = 1-x^2 that implies yx is y d/dx dx of sqrt(1-y) which is y/2sqrt(1-y) from 0 to 1?
Bless you, sir. You teach thoroughly and you teach fast, enough so that I was able to pick up the core concepts in about two minutes while running close to a deadline. I'll be coming back.
4:33 isn't the procedure when finding centroid of a region between two curves is to actually subtract the squares of the functions, not to square the difference of the functions?
Your diffeq videos look scary, I'm taking it parallel with Calc 3 in the fall. I'm told it only uses calc2 concepts but man those equations look intimidating. Hope you have a nice playlist setup for diffeq :)?
Ashwin Santhanam I usually do that in class with my students. Because I want to make it into a discussion along my proof. Not sure if I will have the time to do it in a video during these days tho since finals are coming.
Maybe this year? I haven't seen this before either. If there's a connection to probability theory, I'd like to hear that too. Got some stats classes coming up. I looked for other videos on UA-cam, but none of them go over the intuition.
I appeared in Calc 2 for me. I did have a unit in Calc 3 for applications of double integrals, in which we covered Moments and center of mass. It is not the same though.
Pretty nice video and excelent explanation. Thank you for your job. But, what if we had to find the centroid of a function with two variables? Is there some kind of hint or similarity with single vlariable function?
I'm sure this is too late, but this problem was specifically looking for the centroid of the region that is in the first quadrant and is bounded above by the function. To get the right bound, you just find the x-intercept in the first quadrant. To get the left bound, you just find the y intercept.
graet efforts from great man But personally i wish soon you do a New video exclusivly on cintroid for bigginer like me Many guys like me faces hard obstakle in the basics of cintroid problems solvig I look for that very soon Couse out time is roo short comparing with the final exam Please if you see my comment dont ignore it I need to understand the pricibes of how to face a problem & how to solve it easily And with details Step by step
Dudes.. the area of the region is basically one-fourth of the area of a circle of radius 1 unit.. pi/4 is not equal to 2/3. How did you get that 2/3... Dem. I'm confused..
Does y=1-x² even have anything to do with a circle? The equation of a circle with center in the origin and radius one is x²+y²=1, if we solve for it for y, then, assuming y is positive, we get y=√(1-x²), not y=1-x². His answer is correct.
@blackpenredpen Sir, as per my calc book, it says that for y bar we could use the same formula for x bar except now its the integral with the limits of y rather than x and y*[y2-y1] as the integrand of the upper integral. I think that also works. Thanks a lot though !
Ahhhhhhhh ..... The Area calculation is wrong because the equation for the quarter circle should be y = Sqrt of 1 - x^2 NOT just y = 1 - x^2..... Checking this is easy. This is 1/4 of the area of a circle with radius = 1. So A = pi r^2 = pi 1^2 = pi. The quarter area is thus pi/4 NOT 2/3...............BUT the solution is correct if the shape is not a Circle but is a Parabola. It just looks like a circle so the problem should have stated it was a Parabola.
i found this man when i was searching help for calculus and now um doing engineering mechanics "statics "he still saves my life again
..
exactly same here, I found him while I was having maths exam in class 10, then in 12, now i'm in first year CSE and he is still helping me...
@brianramz6681
Wish I found it earlier in the semester tbh... He was _great_ for calc two!
For anybody wondering why 1/2 is added at 3:53, the constant of 1/2 comes from the fact that the region is symmetric and 2D. When you're averaging values along a dimension that involves a parabolic distribution of mass, the factor of 1/2 accounts for how the average "weight" is distributed across the parabola. In essence, you are averaging out the mass distribution to get your center of mass.
literally a miracle worker, taking my calc 2 final today and this is the last concept I need to understand. your quick and comprehensive explanations carried me through this class- cannot thank you enough.
Thanks! Best of luck to you on your final!
update: I got a 98! :)
@@blackpenredpen Wait wait wait... should it be xydx and xydy? We are basically meassuring amount of change coordinate within area? That way we are limiting rate of change to coordinate and area... right? X bar i get, it's basically searching the x coordinate, but y bar? Shouldn't it be y = 1-x^2 that implies yx is y d/dx dx of sqrt(1-y) which is y/2sqrt(1-y) from 0 to 1?
Bless you, sir. You teach thoroughly and you teach fast, enough so that I was able to pick up the core concepts in about two minutes while running close to a deadline. I'll be coming back.
woah it's woodfur00! epic
You explained that in such a straight forward and easy way. It helped so much!
holy crap i finally understand. this man explained in 7 minutes what my professor discussed in 2 hrs
I found him while I was having maths exam in class 10, then in 12, now i'm in first year CSE and he is still helping me...
This explanation is absolutely excellent my friend thank you so much, you are a brilliant teacher.
only helpful video on this site that is concise and under 10 mins and explains everything perfectly
Wow, my calculus 2 teacher couldn't even give us those basic functions... Thanks for the video.
Where do I find the x bar and y bar equations and how they are created?
Thank you sir. You will probably save my mechanical quiz.
Edit: You did. 25/25
4:33 isn't the procedure when finding centroid of a region between two curves is to actually subtract the squares of the functions, not to square the difference of the functions?
I like the way you change pen
thank you that's so helpful
3:50 why do you multiply by 1/2
because math tells you to, do not argue with the math gods (joking lol)
why you multiplied by 1/2 @3:53 min ?
i also don't know
Bundle of thanks you've made my day...♥
So clear, so good. Thanks man!
Thank you, sir!
It was awesome.loved this
Thank you so much sir
I literally understand this, but my professor wants us to use her method and it's so hard
Amazing video, thank you very much!!♥
Bruh, you are the best.
3:53 whyyy?? tho WHYYY!!!!???
ua-cam.com/video/MHRN2lj15LE/v-deo.html
@@PrincessPea1 thnx bro, that guy explained just what i was looking for
@@PrincessPea1 Thank you!
Why?
Since you're working out co-ordinates, should the x bar and y bar values not satisfy the function y = 1 - x^2?
The centroid doesn't have to lie on the function. The centroid of a circle doesn't lie on its perimeter
@@danielmencl2764 the centroid never lies on the curve or the shape according to geometry
Great Video mate ! Can u please help me understand how the equations for centroid(x',y') have been derived ?
Anantha s rao I may not have the time recently to do that video since I have tons on my to do list. Sorry.
I am sure if u search on YT u can find some on it
Your diffeq videos look scary, I'm taking it parallel with Calc 3 in the fall. I'm told it only uses calc2 concepts but man those equations look intimidating. Hope you have a nice playlist setup for diffeq :)?
Anantha s rao yup. U can go visit my site www.blackpenredpen.com
ua-cam.com/video/MHRN2lj15LE/v-deo.html
can u make a video for the proof of centroid formula
Ashwin Santhanam I usually do that in class with my students. Because I want to make it into a discussion along my proof. Not sure if I will have the time to do it in a video during these days tho since finals are coming.
blackpenredpen I hope you have time. This seems really interesting, but it wasn't covered in my calc 2 class.
Maybe this year? I haven't seen this before either. If there's a connection to probability theory, I'd like to hear that too. Got some stats classes coming up. I looked for other videos on UA-cam, but none of them go over the intuition.
Would this appear in Calculus 3 by chance?
I appeared in Calc 2 for me. I did have a unit in Calc 3 for applications of double integrals, in which we covered Moments and center of mass. It is not the same though.
I don't think this is in my Calc2 for engineers course. I'm like wtf is a centroid..
Centroid is really just the center of mass.
Caarve on a 2d object I imagine so kinda like the nucleus?
I did not learn this in my Cal 2 class.
Simple...f3f33f3 👏
Pretty nice video and excelent explanation. Thank you for your job. But, what if we had to find the centroid of a function with two variables? Is there some kind of hint or similarity with single vlariable function?
Hell yes this is the exact example I needed first try.
Sir you didn't give the reason that why we have to square and divide equation by 2, when we are integrating for y bar.
I wonder if you can get the same answer by doing the same as x bar in terms of y
Hello Akash, I believe the reason is because it’s an integral rule!
thank you sir
i didnt understand (actually dont know) what the centroid is? could you please explain it??
Theo Papa imagine that's a plate, u can place ur finger under that point and u can balance it. It's also called the center of mass
ok thank you very much!!
i thought he was holding a glass of wine
Sir where did the 0-1 limits came from ?
Thank you for the answer
I'm sure this is too late, but this problem was specifically looking for the centroid of the region that is in the first quadrant and is bounded above by the function. To get the right bound, you just find the x-intercept in the first quadrant. To get the left bound, you just find the y intercept.
you have humor noice
thank you
graet efforts from great man
But personally i wish soon you do a New video exclusivly on cintroid for bigginer like me
Many guys like me faces hard obstakle in the basics of cintroid problems solvig
I look for that very soon
Couse out time is roo short comparing with the final exam
Please if you see my comment dont ignore it
I need to understand the pricibes of how to face a problem & how to solve it easily
And with details
Step by step
Why you multipled y with half?
cant u also determine the moment of inertia?
I think there are a wrong because the circle equation is (x^2 + y^2 = r^2). Thus ( y = ( r^2 _ x^2) ^1/2).
yes, thank you!
You are amazing. I wish I was your friend.
I really like to follow you
Based teacher
Yes, I was looking for the derivation of the centroid formulas, too ...
yeah theres clearly a double integral somewhere, derivation is better
Integration applications please
this can also be done using double integerals. great video.
sir= why y x 1 / 2 ? --- thank u sir - amarjit= india
Thank you, my lord, and savior 🙏
How 1/2 of y bar??
لان هو بياخد بعد ال centrid من الشريحة اللي اخد عندها تكامل
Dudes.. the area of the region is basically one-fourth of the area of a circle of radius 1 unit.. pi/4 is not equal to 2/3. How did you get that 2/3... Dem. I'm confused..
You are correct ! ! .... He made a careless error. See my post on April 3, 2020 above.
Does y=1-x² even have anything to do with a circle? The equation of a circle with center in the origin and radius one is x²+y²=1, if we solve for it for y, then, assuming y is positive, we get y=√(1-x²), not y=1-x². His answer is correct.
@blackpenredpen Sir, as per my calc book, it says that for y bar we could use the same formula for x bar except now its the integral with the limits of y rather than x and y*[y2-y1] as the integrand of the upper integral. I think that also works. Thanks a lot though !
Ahhhhhhhh ..... The Area calculation is wrong because the equation for the quarter circle should be y = Sqrt of 1 - x^2 NOT just y = 1 - x^2..... Checking this is easy. This is 1/4 of the area of a circle with radius = 1. So A = pi r^2 = pi 1^2 = pi. The quarter area is thus pi/4 NOT 2/3...............BUT the solution is correct if the shape is not a Circle but is a Parabola. It just looks like a circle so the problem should have stated it was a Parabola.
Isn't that supposed to be a parabola?
Why is the area not doubled?
what is ur qualification academics?
I understood this lesson in 7 minutes what i did not understand in 40 minutes from organic chemistry tutor
Sorry OCT😢
how to find centroid of mass in:R3
👍
kiss
Why when i use double integration the y bar = ⅕ not ⅖