I think of triple integrals as the mass of an object where the function you're integrating is the density at different points inside the object instead of as a hyper volume. It just makes more sense to when visualizing what you're doing.
...Cher Dr. Peyam, Malgré le fait que je n'ai pas encore beaucoup d'expérience avec les intégrales triples, j'ai pus suivre votre excellente présentation sans aucun problème. Je tiens à vous remercier beaucoup pour cela... Cordialement, Jan-W
I would interpret this as follows: With double Integral, the input or domain of a function f(x,y) ist the 2D-Plane x,y. With triple Integral, the domain (input) of the function, is the entire three-dimensional object 'E' --> Result is four-dimensional. But to calculate n-dimensional volumes, you can just write ∫ √g dx^n where √g = √det(g_ij) = det(A_ij) is the determinant of the matrix 'A', generated by the local base-vectors A = (R_x, R_y, R_z, …, R_n)
I've never think about triple integral is a hypervolume, cuz in my university at math classes we've learned just how to calculate them - get 3d picture, get integrating ranges, and then integrate, without knowing what we wanna to find :D But with double ints we was knowing that we finding a volume :D Thank you :D
You don't have to think of an integral as a volume or a hyper volume, etc. In this case the region (or "limits") of integration has a volume that can be calculated easily. Then we have some function which takes on a value at each point in that region. Well, you can just think of the integral as the average of that function over the entire region, multiplied by the volume of the region. Suppose again that the function represents density. Then the average of the density within the region, multiplied by the volume of the region, is pretty obviously the total mass within the region. Conversely, if we have calculated the integral, then the average density equals the total mass in the region divided by the volume.
I already knew what triple integral was but I watched it anyway. Wasn't disappointed :) P.S : I thought about hyper volume before but I prefer to think about mass. It's easier to visualize. Hyper volume isn't wrong though, the mathematics job is to make abstractions in the most general way possible.
Hi Dr. Peyam, awesome like always, I would like you answer me a question, is it possible to apply the residue theorem to non definite integrals?, I have read books about Complex analysis but I never saw it, do you know where I can get info about it just in case it’s possible.
I don't think that it's possible, because the residue theorem says that the line integral over a curve it's equal to the sum of the residues*the winding number of each point you're calculating the residue*2πi, so you have to define the integral over a closed curve in order to use the staments
This is excellent! Thank you. (C'est tout à fait apparent) And Mathematica also gets 4/3, using the following two lines of code: region = ImplicitRegion[(0 = 0), {x, y, z}]; Integrate[z, Element[{x, y, z}, region]] ----> 4/3
I'm sure there is some way to do integrals with non-orthogonal coordinates, but why would you want to? But then again, I'm sure there is a reason why you might want to.
@@RalphDratman Depending on the function you're integrating, integrating over non-cartesian coordinates might preferable (e.g. polar or cylindrical coordinates). But Dr. Peyam doesn't even mention that he assumes x, y and z together form a cartesian coordinate system, yet I think he should, because if they don't, dx dy and dz need to be variably rescaled due to location depending compression and stretching of the coordinates. I admit one might say that the one who formulated the function to be integrated already (should have?) encapsulated that rescale factor into the function.
How could you define bounds to an integral across a fractal? Would it necessarily require the use of a fractal with a dimensionality of one less then that of the original fractal? Could i define the bounds of an integral across something with another thing that has a dimensionality greater then one less then that of the original thing? (For example bounding an integral along a 2d surface with a box that's 3d)
@@drpeyam Yes, but how?... (I have never tried anything like this before) How can you, for example, define an inequality that involves "five and a half" variables? That's why i asked the second and third questions. To try to go around this issue.
Meanwhile you could have just done 1/6 det of the tetrahedron using the intercept coordinates to make a 4x4 matrix in less than 2 minutes.... math is fun and crazy at the same time
@@drpeyam oh no i just meant its very interesting to see different methods to solve the same thing lol im currently in multivarable calculus and a seperate linear algebra class and i love to see the overlap and how useful Linear algebra is thats all. Grateful for your videos and rescources
Right, that description as nested loops makes perfect sense to a programmer like me. Then you just have to understand how to multiply z by dx dy dz. Which I still find difficult to visualize since it is an infinitesimal volume. It is easy enough to compute the approximate result numerically: just set delta x, delta y and delta z to some smallish number delta, then add up the value of z in each cubical volume element with side delta. Then set delta to a smaller number and calculate again. Keep decreasing delta until the sum seems to be approaching a limit. So it's a cinch to get an approximate answer, but to solve the problem as stated we need the exact answer. So formally I guess that would be the triple limit as the deltas all go to zero.
@@Gamma_Digamma For me there is nothing "initially" about my difficulty. I've been thinking about these things for decades! I think my major problem is that I don't like the dx dy dz "infinitesimal" notation and the ill-defined methods of manipulating them. They obviously work very well for people who enjoy calculus, but I have never been able to get comfortable with them.
@@drpeyam Using your interpretation as a volume of a 4D simplex with the base being this 3D tetrahedron of volume 4/3 the only thing we have to compute is the height. It will be the biggest distance from the base i.e. the biggest possible value of z which is the same as the z-intercept which is 4 (the foot of the altitude from the 5th vertex of this simpex will be at (0,0,4)). So we just have to apply a formula for a volume of 4D simplex which is 1/4*base*altitude which in this case is the same as the volume of the base because the altitude and 1/4 cancel each other.
I think of triple integrals as the mass of an object where the function you're integrating is the density at different points inside the object instead of as a hyper volume. It just makes more sense to when visualizing what you're doing.
The Double Helix That’s actually really practical in physics too, very interesting
@@tomatrix7525 I got the idea from physics class lol
Trivial, just count infinitesimal polyhedrons until you get the answer
...Cher Dr. Peyam, Malgré le fait que je n'ai pas encore beaucoup d'expérience avec les intégrales triples, j'ai pus suivre votre excellente présentation sans aucun problème. Je tiens à vous remercier beaucoup pour cela... Cordialement, Jan-W
Merci ☺️
Someday I'll be this awesome. That was great work!
I love maths as much as you do.
Once again you left me speechless. Amazing video
Next level: nth integral
I would interpret this as follows:
With double Integral, the input or domain of a function f(x,y) ist the 2D-Plane x,y.
With triple Integral, the domain (input) of the function, is the entire three-dimensional object 'E' --> Result is four-dimensional.
But to calculate n-dimensional volumes, you can just write ∫ √g dx^n
where √g = √det(g_ij) = det(A_ij) is the determinant of the matrix 'A', generated by the local base-vectors A = (R_x, R_y, R_z, …, R_n)
This make so much sense and can relate this video alot will wish 2 understand more and more
Great !!!! (but im not capable of thinking of the fourth dimension😭
Loads of thanks sir..
You did save me indeed...
Well explained. ...
I've never think about triple integral is a hypervolume, cuz in my university at math classes we've learned just how to calculate them - get 3d picture, get integrating ranges, and then integrate, without knowing what we wanna to find :D But with double ints we was knowing that we finding a volume :D Thank you :D
You don't have to think of an integral as a volume or a hyper volume, etc. In this case the region (or "limits") of integration has a volume that can be calculated easily. Then we have some function which takes on a value at each point in that region. Well, you can just think of the integral as the average of that function over the entire region, multiplied by the volume of the region. Suppose again that the function represents density. Then the average of the density within the region, multiplied by the volume of the region, is pretty obviously the total mass within the region. Conversely, if we have calculated the integral, then the average density equals the total mass in the region divided by the volume.
I already knew what triple integral was but I watched it anyway. Wasn't disappointed :)
P.S : I thought about hyper volume before but I prefer to think about mass. It's easier to visualize. Hyper volume isn't wrong though, the mathematics job is to make abstractions in the most general way possible.
Thank you very much for this video! It is a nice computation and basic conceptual practice
And the numbers work out nicely
Hi Dr. Peyam, awesome like always, I would like you answer me a question, is it possible to apply the residue theorem to non definite integrals?, I have read books about Complex analysis but I never saw it, do you know where I can get info about it just in case it’s possible.
I don't think that it's possible, because the residue theorem says that the line integral over a curve it's equal to the sum of the residues*the winding number of each point you're calculating the residue*2πi, so you have to define the integral over a closed curve in order to use the staments
Piscopopasco junior yes, I think so, but see the next video and you are going to see what I am talking about
This is where I could find it: @ (Spanish video)
@@piscopopasco @
ua-cam.com/video/TSjXWGpsedY/v-deo.html
another way by directly determining the area of the
tetrahedron section as a function of z: I = integral from 0 to 4 of 0,5 z(1-z/4)(2-z/2)dz = 4/3
This is excellent! Thank you. (C'est tout à fait apparent)
And Mathematica also gets 4/3, using the following two lines of code:
region = ImplicitRegion[(0 = 0), {x, y, z}];
Integrate[z, Element[{x, y, z}, region]]
----> 4/3
I ones used to calculate Flux
Nicely done
Thanks.
If x, y and z were *not* orthonormal coordinates, wouldn't this triple integral have been worked out wrong?
I'm sure there is some way to do integrals with non-orthogonal coordinates, but why would you want to? But then again, I'm sure there is a reason why you might want to.
@@RalphDratman Depending on the function you're integrating, integrating over non-cartesian coordinates might preferable (e.g. polar or cylindrical coordinates). But Dr. Peyam doesn't even mention that he assumes x, y and z together form a cartesian coordinate system, yet I think he should, because if they don't, dx dy and dz need to be variably rescaled due to location depending compression and stretching of the coordinates. I admit one might say that the one who formulated the function to be integrated already (should have?) encapsulated that rescale factor into the function.
Thank you sir
Thx,dr
How could you define bounds to an integral across a fractal?
Would it necessarily require the use of a fractal with a dimensionality of one less then that of the original fractal?
Could i define the bounds of an integral across something with another thing that has a dimensionality greater then one less then that of the original thing?
(For example bounding an integral along a 2d surface with a box that's 3d)
You would use Hausdorff measure
@@drpeyam
Yes, but how?...
(I have never tried anything like this before)
How can you, for example, define an inequality that involves "five and a half" variables?
That's why i asked the second and third questions. To try to go around this issue.
Meanwhile you could have just done 1/6 det of the tetrahedron using the intercept coordinates to make a 4x4 matrix in less than 2 minutes.... math is fun and crazy at the same time
Right but how would that illustrate how to calculate triple integrals?
@@drpeyam oh no i just meant its very interesting to see different methods to solve the same thing lol im currently in multivarable calculus and a seperate linear algebra class and i love to see the overlap and how useful Linear algebra is thats all. Grateful for your videos and rescources
You keep saying x-y-z-plane. I don't think we have the same definition of "plane", there.
It's just nested loops, these multiple integrals
Right, that description as nested loops makes perfect sense to a programmer like me. Then you just have to understand how to multiply z by dx dy dz. Which I still find difficult to visualize since it is an infinitesimal volume. It is easy enough to compute the approximate result numerically: just set delta x, delta y and delta z to some smallish number delta, then add up the value of z in each cubical volume element with side delta. Then set delta to a smaller number and calculate again. Keep decreasing delta until the sum seems to be approaching a limit. So it's a cinch to get an approximate answer, but to solve the problem as stated we need the exact answer. So formally I guess that would be the triple limit as the deltas all go to zero.
@@RalphDratman and congratulations, you rediscovered the Riemann sum definition of the integral
@@Gamma_Digamma I know the definition. It's just not very intuitive for me. I'm not sure why.
@@RalphDratman sorry I was kidding back there.
I know, it is very non intuitive initially but I'm sure you get the hang of it gradually
@@Gamma_Digamma For me there is nothing "initially" about my difficulty. I've been thinking about these things for decades!
I think my major problem is that I don't like the dx dy dz "infinitesimal" notation and the ill-defined methods of manipulating them. They obviously work very well for people who enjoy calculus, but I have never been able to get comfortable with them.
Coincidence ? The result is the same as the volume of the tetrahedron.
Yeah, that’s a coincidence!
@@drpeyam Using your interpretation as a volume of a 4D simplex with the base being this 3D tetrahedron of volume 4/3 the only thing we have to compute is the height. It will be the biggest distance from the base i.e. the biggest possible value of z which is the same as the z-intercept which is 4 (the foot of the altitude from the 5th vertex of this simpex will be at (0,0,4)). So we just have to apply a formula for a volume of 4D simplex which is 1/4*base*altitude which in this case is the same as the volume of the base because the altitude and 1/4 cancel each other.
I would never say that the results (triple integral from z and triple integral from 1) will be the same.But it is so.
it's sad when you only know how to solve simple integrals :
Solving multiple integrals is just solving multiple simple integrals, one after the other.
@@nullplan01 precisely, the only difference is you can have functions in the boundaries.