I am Alex who send you the question,thank you for answering and doing a video for my problem❤❤.You explained very well and now I understand it.Have a nice day❤(sorry for my bad english)
I can't seem to find videos in your channel about line integrals and surface integrals I would love to see you explaining them and doing more of them if you have time (if you have already done them tell me please). Thanks, love your channel!!!!
I swear, I've first found your channel a few years ago when you had like 10k subs and you have almost a million now. I remember some people would just leave some stupid hate comments everywhere in your comment section and I was like, why would people do that? Glad to see your channel becoming one of the biggest educational channels out there.
I dont even have this subject but u explain so clearly and doesnt skip any small details makes me learn a lot. I might pass my failed calculus 1 this summer.
omg your so amazing, i love how you explain things in such a pleasant way that makes math feel intuitive. Could you please do an example where the circles are not centered at the origin and when there is no pleasant symmetry to abuse? I seem to really struggle to set boundaries for the integrals. thanks
Just reading down the comments, it seems a few are upset by the 5 X pi -26 resulting in a negative value. Firstly, have a look at tutorial.math.lamar.edu/Classes/CalcIII/DoubleIntegrals.aspx and see that the double integral can be interpreted as a volume. Secondly, consider that a negative answer in an integration is nothing to be unexpected. Go to a simple single variable integration as an example, say integrate sin x between -3pi/4 and pi/4. Sketch that function. Compute the integral and achieve a negative answer. Geometrically, we expect it. So with a double integral, and a geometric interpretation of a volume, well that volume may have elements above and below the dA plane. So a negative value is legitimate.
Blackpenredpen so much love. Can you do a video explaining why you get the extra ‘r’ term when you switch to polar coordinates from Cartesian one. I heard it was something to do with a Jacobian Discriminant
Hey quick question. Don't polar coordinates go from -pi to pi so the limits on the theta would be -pi/2 to -pi for the third quadrant? Or does it make no difference
@@stephenbeck7222 No it wouldn't. When in polar coordinates, you normalise the angles such that you go anti-clockwise only from 0 to positive pi and clockwise from 0 down to -pi. If something would go above pi, for example 3pi/2, you would change that to -pi/2. Therefore that would only get the third quadrant because it's going clockwise -pi/2 to -pi. I'm sure this is probably something to do with semantics? More than the results being wrong I think but I was just curious.
Hey! I had a quick question about this problem. The final answer is approximately -10.292. Isn't area always positive? or am i missing a critical piece of information
If we tried using geometry formula, will that work also? All we have to do is to get big quarter circular area minus the small quarter circular area making the total area to be negative since it’s under the x-axis.
There is a short presentation that geometrically answers the above quite nicely - ua-cam.com/video/luAx7dUVM5w/v-deo.html . Of course multiplying the sides of a rectangle give you area, so dx.dy = dA . Think of it also as like dx in units of metres and dy, also in units of metres, gives dA in units of metres squared. In the polar coord system, consider the near rectangle which looks like dr.d(theta). The radial side of the "rectangle" can have the dimension dr of units of metres, but the curved side has a length of r.d(theta) - not d(theta) on its own - because the length of a curve is simply r.theta. Look at the units involved. d(theta) is an angle, not a length, so dr. d(theta) would give metre-radians and not the needed metres squared for dA. r.theta is also in units of metres-radians, but the radians are effectively dimensionless, so the overall measure is metres alone. Likewise, r.d(theta) has an overall unit of metres as well - accepting that theta is dimensionless. Put the whole lot together dimensionally and you have dr (in metres) X r.d(theta) (in metres) = dA (in metres squared). Swap terms around and r.dr.d(theta)=dA.
Double and triple integrals are so fun, but I wish there was a way to do higher dimensional integrals. Like a quadruple integral would be so cool! But, alas, there is no way to define this higher integral :(
Seems that the result of this integration is a volume not an area. This was not explained clearly. Integrating y2 + 3x (a ‘z’ value) over an area in x,y should result in a volume? Seems that some of the volume is above the x,y plane +ve and some below -ve and the result is the difference between the two volumes. Is this right?
That is correct, the double integral represents a volume as per the excellent "Paul's Online Notes" tutorial.math.lamar.edu/Classes/CalcIII/DoubleIntegrals.aspx .
Why should I search the double integral but not do it geometric f.e. I have two circles with S1= pi*R^2=9*pi and S2= pi. Than we have S2-S1=8*pi.8*pi/4=2*pi What role plays the y^2-3x ?
I may be wrong but I think y^2 - 3x defines a surface in 3 dimensions (i.e. z = y^2 - 3x) and what the double integral represents is the volume of the shape made between the shaded area D and the surface.
Can you help me with this problem? I didn't get it totally. Thanks in advance @blackpenredpen! 😁 Problem: The surface area of a rectangular box without top is said to be 108 ft². Find the greatest possible volume.
Somebody please help me with a refresher. I took calculus but it’s been a long time. What is the question here asking? I know that a integral gives us the area under a curve (and above the x-axis) but what are we finding when we find this double integral?
Is It too rude to ask for greens theorem,stokes theorem,gauss divergence theorem ,line, surface and volume integrals 😂😂😂😂 ? Jokes aside , love your videos
Its the graph of a plane in R3. The integral ur evaluating is the volume under that plane (y² + 3x) under a region D which is the region between the two circles. Its like the normal integral where you calculate the area under a function y = f(x) in an interval from a to b. Now you go up one dimension and you evaluate the volume under a graph z = f(x,y) in a region D
Hey man, great problem! With it we can “prove” all positive integers are equals, which is absurd. But, I think the solution is quite simple: we cannot conlude that m = n from 1^m = 1^n; here is why. 1^m = e^log(1^m) = e^(m*log1) 1^n = e^log(1^n) = e^(n*log1) 1^m = 1^n if and only if e^(m*log1) = e^(n*log1). To conclude that e^m = e^n (which is m = n), we should raise every member to the power of 1/log1, which is 1/0, which is impossible. I hope I am right :)
If DumbGuy = me, I don't understand. The way I understand circles, it's A=pi*r^2. Let r=1 and R=3 A1=pi*1^2 => 3.1415... A2=pi*3^2 => 28.2743... Shaded area = (28.2743 - 3.1416) / 4 = 6.2832... His answer => (5*pi-26) = -10.2920... whaaa...?
Well, u're not at all a dumb... See, that shaded region you see in the xy-plane is only the domain we are concerned to integrate our 2-variable function on. This 2-variable function represents a 2d curved surface in 3d space. So, we are asked to perform an integration on the curved surface OVER THAT SHADED REGION and not on the shaded region itself. What you did, for 2d integrals it's just like saying the answer is 2 when we are asked to integrate x^5 wrt x from x=6 to 8(since you just found the region of the domain), which does not make sense. Hope this helped. Have a great day.
when you put x=r*cos(f) and y=r*sin(f) in x^2+y^2=1 you get r^2=1 so its r=1 and r=-1, and you know that radius cant be negative so only solution is r=1
I am Alex who send you the question,thank you for answering and doing a video for my problem❤❤.You explained very well and now I understand it.Have a nice day❤(sorry for my bad english)
Alex 7 yay!!! Don’t worry. Your English is great!
Bad english? If you know the difference between adjectives and adverbs you are a great English speaker!
You speak better english than people who actually live here!! (Assuming you don't live in the US...)
@@blackpenredpen yes!
Yes @blackpenredpen
u really should do more multivariable calculus
I can't seem to find videos in your channel about line integrals and surface integrals I would love to see you explaining them and doing more of them if you have time (if you have already done them tell me please). Thanks, love your channel!!!!
really craving for the double integral playlist from you!
TRIPLE INTEGRALS
Its so adorable how blackpenredpen gets excited for every math question that he does. Keep doing what you doing :D.
Loved the Doraemon music in the beginning
Nice video btw
I swear, I've first found your channel a few years ago when you had like 10k subs and you have almost a million now. I remember some people would just leave some stupid hate comments everywhere in your comment section and I was like, why would people do that? Glad to see your channel becoming one of the biggest educational channels out there.
lol I was really focusing on the process then suddenly Peyam shows up screaming USE THE CHAN LU ... that kinda scared the crap outta me lol
Here I am now, in 2022...
Learning Calculus just like Alex three years ago.
Thanks for the vid, you explained everything great🙂
I want somebody to smile at me like blackpenredpen smiles at his math questions :)
I dont even have this subject but u explain so clearly and doesnt skip any small details makes me learn a lot. I might pass my failed calculus 1 this summer.
Your smiling face is a mercy, Mr. Bprp. Have a nice day.
omg your so amazing, i love how you explain things in such a pleasant way that makes math feel intuitive.
Could you please do an example where the circles are not centered at the origin and when there is no pleasant symmetry to abuse? I seem to really struggle to set boundaries for the integrals.
thanks
why the hell is this unlisted? only 1.1k views!
I haven't done a double integral in 35 years......Great review!!
Please do 100 double integrals 😁
Oh, my.
No, triple intregal!!!
no, 50 for equality.
Very neat, I never mastered double integrals at school
Use the Chen Lu!!! 😄😄😄
Dr Peyam go sho!!!
big up to you bro im following you from morocco
I like how you use Doraemon music like we’re just kids even tho we’re doing cal III questions
Amazing I can’t wait to be able to do this
Could you do a video explaining some topics just after passing high school topics that would be taught to us in colleges please
Just reading down the comments, it seems a few are upset by the 5 X pi -26 resulting in a negative value.
Firstly, have a look at tutorial.math.lamar.edu/Classes/CalcIII/DoubleIntegrals.aspx and see that the double integral can be interpreted as a volume.
Secondly, consider that a negative answer in an integration is nothing to be unexpected. Go to a simple single variable integration as an example, say integrate sin x between -3pi/4 and pi/4. Sketch that function. Compute the integral and achieve a negative answer. Geometrically, we expect it.
So with a double integral, and a geometric interpretation of a volume, well that volume may have elements above and below the dA plane. So a negative value is legitimate.
Blackpenredpen so much love. Can you do a video explaining why you get the extra ‘r’ term when you switch to polar coordinates from Cartesian one. I heard it was something to do with a Jacobian Discriminant
Could you do a video on cycloid,And how to calculate its area through calculus .Cheers
Very helpful video.
Fantastic vid! Happy Pi day! Thanks. I would like to know the source of the video of the "chen lu"
It is Dr. Peyam. Peyam misheard his heavily accented professor say Chain Rule as Chen Lu.
Hey quick question. Don't polar coordinates go from -pi to pi so the limits on the theta would be -pi/2 to -pi for the third quadrant? Or does it make no difference
Pi/2 to -pi would be getting the 1st, 4th, and 3rd quadrant areas. There is no general range requirement for using polar coordinates.
@@stephenbeck7222 No it wouldn't. When in polar coordinates, you normalise the angles such that you go anti-clockwise only from 0 to positive pi and clockwise from 0 down to -pi. If something would go above pi, for example 3pi/2, you would change that to -pi/2. Therefore that would only get the third quadrant because it's going clockwise -pi/2 to -pi. I'm sure this is probably something to do with semantics? More than the results being wrong I think but I was just curious.
it would be -pi to -pi/2 and it doesnt make a difference
@@Andrei-rp3dz it always goes anti-clockwise
Can you prove please why should we multiply by the jacobian in change of variables please
Do for double integral for bounded regions e.g substitution
Dear Blackpenredpen,
Can you please calculate the temperature gradiation of a cylindrical cooling fin?
Tell about methods of getting pi digits pls
So good!
Hey! I had a quick question about this problem. The final answer is approximately -10.292. Isn't area always positive? or am i missing a critical piece of information
Me too
thanks so much. am greatly helped
If we tried using geometry formula, will that work also? All we have to do is to get big quarter circular area minus the small quarter circular area making the total area to be negative since it’s under the x-axis.
In right side, why dA=dx.dy than become dA=rdrd@? Can you explain it more?
Search about the Jacobian determinant for when you change variables in double integrals
@@joao_pedro_c in the polar case i believe theres a geometric proof too, because i was taught that before i was taught about jacobians
There is a short presentation that geometrically answers the above quite nicely - ua-cam.com/video/luAx7dUVM5w/v-deo.html . Of course multiplying the sides of a rectangle give you area, so dx.dy = dA . Think of it also as like dx in units of metres and dy, also in units of metres, gives dA in units of metres squared. In the polar coord system, consider the near rectangle which looks like dr.d(theta). The radial side of the "rectangle" can have the dimension dr of units of metres, but the curved side has a length of r.d(theta) - not d(theta) on its own - because the length of a curve is simply r.theta. Look at the units involved. d(theta) is an angle, not a length, so dr. d(theta) would give metre-radians and not the needed metres squared for dA. r.theta is also in units of metres-radians, but the radians are effectively dimensionless, so the overall measure is metres alone. Likewise, r.d(theta) has an overall unit of metres as well - accepting that theta is dimensionless. Put the whole lot together dimensionally and you have dr (in metres) X r.d(theta) (in metres) = dA (in metres squared). Swap terms around and r.dr.d(theta)=dA.
Wonderful
lol, random Dr Peyam cameo
Double and triple integrals are so fun, but I wish there was a way to do higher dimensional integrals. Like a quadruple integral would be so cool! But, alas, there is no way to define this higher integral :(
Do more multiple integrals
Hollow, he can do a triple intregal?
So since 5*pi-26
This integral is not equal to the (signed) area, so this is not confirmation of a correct answer.
its the integral over the signed area
imagine the function y^2 + 3x is what we are integrating and we are looking at it from above
how do i find the fucking radius from the graph it looks like half a donut ( 3rd and 4th quadrant)
Hey bro , do you have a playlist of triple integrals ?
Great video. But confused because the answer seems to indicate a negative area??
Why dA=rdrd(theta)?
why the extra r in dA? Like why it is not just drdtheta!?
5 pi minus 26 is a negative area... may be 26 minus 5 pi right? thank you
hello sir.. where are u from? ty
Why don't you say 2 of the pi over 2s for pi so I can read in a clearer way starting from 0 to pi?
Hey bprp, can you calculate the sum of 1/(k^2+1) from k=1 to infinity?
Seems that the result of this integration is a volume not an area.
This was not explained clearly.
Integrating y2 + 3x (a ‘z’ value) over an area in x,y should result in a volume?
Seems that some of the volume is above the x,y plane +ve and some below -ve and the result is the difference between the two volumes. Is this right?
That is correct, the double integral represents a volume as per the excellent "Paul's Online Notes" tutorial.math.lamar.edu/Classes/CalcIII/DoubleIntegrals.aspx .
Why should I search the double integral but not do it geometric f.e. I have two circles with S1= pi*R^2=9*pi and S2= pi. Than we have S2-S1=8*pi.8*pi/4=2*pi What role plays the y^2-3x ?
I may be wrong but I think y^2 - 3x defines a surface in 3 dimensions (i.e. z = y^2 - 3x) and what the double integral represents is the volume of the shape made between the shaded area D and the surface.
@@rowandavis2061 thank you
Can you help me with this? Integral of (secx.e^x)dx
Can you help me with this problem? I didn't get it totally. Thanks in advance @blackpenredpen! 😁
Problem:
The surface area of a rectangular box without top is said to be 108 ft². Find the greatest possible volume.
Please tell me
Why infinity subtract infinity is not equal to zero
Because it have different in size and dimensions
Is this a volume?
Where did dA=rdrdtheta came from?
Here, things get serious.
What exactly (5pi - 26) is ?
Somebody please help me with a refresher. I took calculus but it’s been a long time. What is the question here asking? I know that a integral gives us the area under a curve (and above the x-axis) but what are we finding when we find this double integral?
volume enclosed by the curve z = y^2+3x and z=0. bounded by the circles on the yx plane. aka a sort of semicircle tube ish for the base of the shape
I am trying to find out the pattern of when you use BGM of Doraemon.
But hey bprp, what is that D variable down, is it like a shorthand for the interval we are integrating
Yea, we have to integrate over that region D
@@blackpenredpen Thanks bprp!
Uhhh... why is the area negative?
YAY
Please tell surface integral, Stokes theorem
why dA = r dr d(theta)
I forgot... How is dxdy = rdrdθ ?
Use the Jacobian
solve this if you can
if log₀.₃(x-1)
Answer is a
Can we get some calc 3 hype in the chat? 🥳
Fredde ah I thought that was a good idea until cal 1 students start arguing and asking too many questions. :D
💯💯💯
Can can solve this using Green's theorem
i love you
# the graph of mod(argz)=mod(z)
Where z is the complex number
Anyone
Please ! Integrate t^n/t^2+t+1 n€R, I have absolutely no idea !!
Cptn_n3m0 bounds? If it’s from 0 to 1 the answer can be written in terms of the digamma function
One 🥧 minus 2 🥧 plus sin of 3🥧 over 2
Is It too rude to ask for greens theorem,stokes theorem,gauss divergence theorem ,line, surface and volume integrals 😂😂😂😂 ?
Jokes aside , love your videos
Check out dr. P!!
Honestly speaking I didn't think Dr.Peyam had made those ....they're gonna be helpful to clear my concepts 😍thanks a lot
I don't get it, So the area of a surface is 5*π- 26
Not an area but volume, negative means that function is more below xy plane
Cool
I think this is wrong because that area should be a positive value ?! And 5 * pi - 26 < 0 ! What am I missing ?
Sebestyén Béla The area described is below the x-axis, so the integral is negative
@@jakemoll it has nothing to with it being below the x axis
its about the function y^2 +3x that is what we are integrating
If you make a 100 double integrals you would just had to do 50 exercises. Will be easier then haha
Yeah🤘🤘, I like it .I have came second time across double integral
Bro make a video on basics of double integral plz
Read my comment plz
🙏🙏🙏🙏🙏🙏🙏🙏🙏
I would have separated the functions then wrote the integrals as products
And what is the (y^2+3x)'job? Geometrically
Its the graph of a plane in R3. The integral ur evaluating is the volume under that plane (y² + 3x) under a region D which is the region between the two circles. Its like the normal integral where you calculate the area under a function y = f(x) in an interval from a to b. Now you go up one dimension and you evaluate the volume under a graph z = f(x,y) in a region D
@@adrician oh thanks.
@@adrician While your explanation is correct, I feel it is important to mention that that is definitely not a plane.
@@andrewhaar2815 if you plot it you will see its a probolic cilinder
Or am i wrobg about the defenition of a plane? In my languag, the trabslation for it is a plane, maybe its different in the english language
But why dxdy=rdrdtheta
Hi BbRb I need you to integrate (xln(x))/(x_1) please 😍😍😍😍😍
integral-calculator(dot)com
@@Mot-dh5sx what did you mean?
Use that website
@@Mot-dh5sx ماشي شكرا الك
I never noticed before, but it really irritates me when you don't finish the theta symbol haha
Sir when 1^m=1^n (m
Hey man, great problem! With it we can “prove” all positive integers are equals, which is absurd. But, I think the solution is quite simple: we cannot conlude that m = n from 1^m = 1^n; here is why.
1^m = e^log(1^m) = e^(m*log1)
1^n = e^log(1^n) = e^(n*log1)
1^m = 1^n if and only if e^(m*log1) = e^(n*log1). To conclude that e^m = e^n (which is m = n), we should raise every member to the power of 1/log1, which is 1/0, which is impossible.
I hope I am right :)
Thanks for the solution . I was being annoyed by this problem for a long time .
Hey, I’m the first here! Can’t wait to see this pi day vid
If DumbGuy = me, I don't understand. The way I understand circles, it's A=pi*r^2.
Let r=1 and R=3
A1=pi*1^2 => 3.1415...
A2=pi*3^2 => 28.2743...
Shaded area = (28.2743 - 3.1416) / 4 = 6.2832...
His answer => (5*pi-26) = -10.2920... whaaa...?
Well, u're not at all a dumb...
See, that shaded region you see in the xy-plane is only the domain we are concerned to integrate our 2-variable function on. This 2-variable function represents a 2d curved surface in 3d space. So, we are asked to perform an integration on the curved surface OVER THAT SHADED REGION and not on the shaded region itself.
What you did, for 2d integrals it's just like saying the answer is 2 when we are asked to integrate x^5 wrt x from x=6 to 8(since you just found the region of the domain), which does not make sense. Hope this helped. Have a great day.
😘🤩😍🤩
5π - 26 < 0 ----> ERRROR!!!
How can i find a single integral in a 6 hours video????😨😨😨😨
Where my fellow high school juniors
Use the chen lu 🤣🤣🤣🤣🤣
Steve you remembered the pi day, but what about professor Hawking's death and Albert Einsteins birthday
Special video for pi day. Happy pi(π) day. #YAY
Calculus III? Well, I had this in my first semester... But nevertheless great video!
Hello,
Why is it from 1 to 3? Isnt it from -1 to -3 ??
It's the radius.
when you put x=r*cos(f) and y=r*sin(f) in x^2+y^2=1 you get r^2=1 so its r=1 and r=-1, and you know that radius cant be negative so only solution is r=1
Plugin plugin 😄😄