Proving the volume and the surface area of a sphere by using integrals

Proving the volume of a sphere with radius r 0:00
Proving the surface area of a sphere with radius r 6:14

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[waikeanng]

Hi, blackpenredpen.Today is my birthday.May i have this pleasure have your birthday blessing?

[blackpenredpen]

Did u click the video bc of the thumbnail? If so, give this video a like! Thanks!

[VladTepesh]

One of the greatest mathematical awakenings I received was during a lesson on area and volume. x² and x³, as x increases, volume increases faster than area, and this is why even though you may not look like you've gained or lost weight on the surface (area), your volume has. That made mathematics personally relevant for me at that moment.

[matthewstevens]

That comment at the start was pretty inspiring. Can't exactly give Such a story, but your videos have helped massively. Keep it up!

[BarbecueBigMac]

Integral of surface area in terms of r (4πr²), 4πr³/3 = 4/3πr³ which is volume

[adrianparism]

Careful with notation. You used r for two different things: the fix, constant radius of the sphere, and the variable radius of the slices. They should have different names.

[helloitsme]

This might be my favorite video of yours, I've always wondered how this would work out with integrals

[dlkl]

The derivative of the volume is the surface area formula 🤔

[coisasdemacho]

Wow, Toma Kolev's words almost made me cry

[divyanshaggarwal]

I dont think that the integration steps for the surface area will be more difficult because....

A TRUE MATHEMATICIAN NEVER CONSIDER A SOLVABLE PROBLEM A DIFFICULT PROBLEM

[myuu]

I actually realized on my own about a year ago that the derivative of a sphere's volume is equal to its surface area. What I also realized is that this is not limited to 3D spheres, it also applies to 2D circles. The derivative of a circle's area is equal to its circumference. It seems that in both 3D spheres and 2D spheres (circles) the boundary of the object is equal to the derivative of the interior that is bound within it. It makes me wonder if the same is true for spheres in higher dimensions. Is the 3D surface of a 4D sphere equal to the derivative of the sphere's hypervolume?

[hjk]

I did the same thing for a torus as well. It's amazing how these formulas can be derived w/ calc! #yay

[rahilkeshav]

We can do it by trig also. The surface area :

thickness of elemental disc = dt
From circle relation, dt = r * dg
( i am using g instead of theta )
Circumference of disc = 2 pi r
But r = R sin g from triangle formed
Surface area = integral (2 pi r dt)
= integral 2 pi R sin g R dg
= 2 pi R^2 integral sing dg from -pi/2 to pi/2
= 2 pi R^2 * 2
= 4pi R^2

[beatoriche]

I really like the following explanation for why you get the volume when integrating the surface area of a sphere with respect to the radius: imagine slicing up a sphere of radius R into many concentric spheres with radii ranging from 0 to R and labeling any specific radius of a tiny sphere r. The volume of the big sphere should be all those differences in volume between adjacent individual spheres added up, right? Two adjacent spheres with radii only a tiny distance apart will have approximately the same surface area, so the change in volume is approximately 4*pi*r^2*dr. So we have a sort of weighted sum of individual surface area values from 0 to R – i. e. an integral of 4*pi*r^2 with respect to r from 0 to R.
Of course, to make this rigorous, you’d have to formulate everything in terms of limits, but I still like this argument because it generalizes nicely into higher dimensions and cuts to the core of what’s going on.
Also, I find Toma’s story really touching. It’s really unfortunate that people have to go through this, and hearing that he managed to overcome this horrible situation and that he’s now doing better feels good. Since there’s a good chance he’s reading this, I’d like to wish him the best of luck and enjoyment in his studies.

[thommunism]

the derivative of the volume (4/3 *pi*r^3) is the surface area (4*pi*r^2), just like how the derivative of a circle's area (pi*r^2) is the circumference (2*pi*r). Interestingly, for a square the perimeter (4*length) is twice the derivative of the area (length^2) and same with the cube's volume (length^3) and surface area (6*length^2).

[micogeolo]

Wow. I really appreciate guys that knows numbers ☺️☺️

[rob]

The derivative of the volume of a sphere w.r.t. its radius is the surface area. To see why, notice that we could have derived surface area first and then derived volume from it by working in polar coordinates.
The volume of a thin shell of the sphere at a distance r from the centre is dV = S(r)dr
in other words, dV/dr = S(r)

[GameMaster-pzpw]

This is probably completley wrong, but I think there might be a pattern for the area, volume, ect. for anything in a circular shape in any dimension, which is this: where d= the dimension, V=(d+1)/3•pi•r^d. In the second dimension, the formula would be (2+1)/3•pi•r^2, which simplifies to pi•r^2. In the third dimension, the formula would be (3+1)/3•pi•r^3. Both of these are the correct formula. Assuming this is correct, the volume of a 4d sphere would be 5/3•pi•r^4

[NFSMW]

Thank you so much have developed a way to solve such a lot

[Edward-gbdx]

Your videos are awesome 🙂 thank you for making maths more understandable