Derivative of volume is surface of sphere,coz: think u have a sphere with radius R ,and inside of that sphere there another smaller sphere with radius r volume of the big sphere is V=4/3 π R³ volume of small sphere v=4/3πr³ Volume between big and small sphere ΔV=V-v and width of that volume Δr=R-r When that width of the volume between 2 spheres becomes soo small it will be surface of sphere,mathematically lim ∆V/ Δr when Δ r->0 its surface of sphere,and this expresses derivative of V..
Be careful about the equasion you used: Y=sqrt(r^2-x^2) has two solutions it has a positive solution and a negative solution. So it would still be a circal. Also you said it was a function. But it is a one to many, so by definition it is not a function. But essentially wouldn't the derivation, if done with the correct tow solutions work out anyways, rather than doing it one way?
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.
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.
@injustice fellow Not quite sure about the mathematical explaination, but if you think of a circle and increase the circumfrence, the area increases (change in amount -> derivatives in play). Same thing goes for the cone from the video. If you increase its (average) radius, the surface area increases. This is again a situation in which a function (the area) depends on a changing variable (the radius), which indicates a differential relationship between them. If you apply this to the surface area and the volume, you find similar results. This time being that an increase in area, also increases the volume (since everything in circular/round). -> Differentiating the volume (consider the volume as the function) gives the surface area.
Imagine a sphere of radius r then if radius is increased by dr then there is a shell of extra volume dV now as dr is very small the outer surface area and the inner surface area are basically the same say S so the volume of the shell would be equal to the area of a cuboid with base area S and height dr So dV = S.dr Or dV/dr = S
@injustice fellow think u have a sphere with radius R ,and inside of that sphere there another smaller sphere with radius r volume of the big sphere is V=4/3 π R³ volume of small sphere v=4/3πr³ Volume between big and small sphere ΔV=V-v and width of that volume Δr=R-r When that width of the volume between 2 spheres becomes soo small it will be surface of sphere,mathematically lim ∆V/ Δr when Δ r->0 its surface of sphere,and this expresses derivative of V..
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
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?
I believe this follows directly from the definition of an integral. When an area is changing according to some function, the boundary which is changing is by definition the rate of change of that area, the derivative of said function. This is seen most clearly by looking at the area under a curve. It's well known that the area under a curve is given by the integral, and taking the derivative of this integral brings you back to the original function that generated the curve. But what is this function but a description of the boundary of the area which is changing? As x increases, the height of the curve at x is the right-most boundary of the area described by the integral, so the area changes according to the height of that boundary.
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).
I had my exam in volume last week! I searched high and low for any video on your channel and now you make one! BUT STILL, LOVE YOUR EXPLANATION! GUESS I'll Come back during finals.
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.
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
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)
@matrices: That will work for any shape whose volume can be built up by normal (perpendicular) integration of its surface area. EDIT: I have now verified this for the regular tetrahedron and the cube, as well as a cylinder whose height = diameter of its base, using as r, the radius of the inscribed sphere. In all these cases, dV/dr = S. I believe it will also work for the other 3 regular polyhedra; and for any solid whose shape is kept constant as its size changes, using the radius of the inscribed sphere, at least so long as that sphere is tangent to every "face" of the solid. Fred
Volume can be defined as the sum of all the areas of the spheres an the radius decreases by dr. This is the definition of intergal so intergal of A dr= 4/3 *pi r^3 so differentiate both side to get 4 pi r^2
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
The problem is finding dL... What if this is one one the wrong calculation that gives right results? In this kind of task only note the 1st order differences. What if you sum section of cones by infinite sum instead? The dL is like piece of cone, but in this calculation case its piece of cylinder. Reminds the case lim x->inf f(x)/g(x), where the largest polynomial terms determines the results.
The volume is clearly the integral of the surface area from 0 to r with respect to r. Similar to how the area of a circle is the integral of the circumference from 0 to r. So basically, one must be careful here because it's not gonna be dx but rather dr in that case.
The derivative of the volume formula is the surface area formula. Geometrically speaking this is because a really tiny change in the radius can be interpreted as wrapping another surface area around the sphere.
So, how come we can approximate the volume using 'cylindrical' discs but not for the surface area ? In both cases we are integrating about a 'curved' shape whose gradient at any point is changing.
No, it is not just a coincidence. If you imagine the sphere as a summation of thin spherical shells, each of thickness dr, each incremental shell has a surface area of 4*pi*r^2. Multiply by the thickness, and get 4*pi*r^2*dr as the volume of each thin spherical shell. Add this up, i.e. integrate, and get integral 4*pi*r^2 dr. As you can see, the integral of the surface area relative to radius, is the volume. The fundamental theorem of calculus tells us that this also implies that the surface area is the radius-derivative of the volume.
something weird is if you take the formula for the volume of a sphere (V=4/3π*r^3) and then take the partial derivative of V with respect to r (𝛿V/𝛿r) you get the surface area of the sphere (SA=4πr²) not sure if there is any connection but thought i would throw it out there
You can use this procedure to calculate the space in any sphere of N-dimensions by integrating the previous (N-1) dimension sphere and adding a small piece of sphere into the new dimension (analog to adding a Z component into the circle and making it into a disc)
I mean for the volume, even if dx tend to 0 the radius of the first base is smaller of de radius of the end base : it’s when you “slice the sphere” Can we just ignore it ?
In a math class we were told that in th xy plane if we have a parallelogram and its base has sqrt(something) then the height of that parallelogram will have the same sqrt(of the same thing) Ex: If the base is =2sqrt(13) Height must have a sqrt(13) in it Anyone knows why? Or a proof..! Thx😊
well the calculus connection between the surface and the volume kinda looks like a derivative/Integral relationship. if i differentiate the volume in respect to r, gimme a sec V=4/3pi r^3 dV/dr= 4pi r^2. but given i suck at anything that even gets near geometry i have no clue why it should be the derivative. (or the integral. i guess it´s similar to if you integrate a circles circumference over the radius you get the Area inside, which is why i´m not surprised in the slightest (but as i don´t even know why this is the case except ,well that´s what you get for integrating/differentiating with respect to r i wouldn´t dare to confidently say it is even the same connection.)
The Surface area is the derivative of the volume: dV/dr = d/dr(4/3pir^3) = 4pir^2 = S. That is also true for a circle, but is it true for other things?
Please sir upload many more videos specially shortcut on differentiation intregation trigonometry permutations combination complex number inverse function
I watched your Gabriel's Horn video yesterday and it occurred to me after watching this one: would a one dimensional hyperphere have a volume of 2r and a surface area of zero?
I was a math major from 1970 to 1974 Back when chalk was still being used! My professor loved to hammer on the blackboard Till the chalk splinters flew, Especially across the cracks between the board sections. He used a metal cylinder chalk holder And enjoyed torturing us students By the screeching it made upon the board When the chalk was worn down. These videos bring back many memories of those days But I wonder what our professor would have Thought about the blackpenredpen technique??!!
I've recently had a hard time understanding why the same logic for dV does not apply to dS. dV = pi*r*r*dy because one is making a small cross section with radius r and height dy. But to me it seems like that implies: dS = 2pi*r*dy because one is making a small cylinder with radius r and height dh, so the surface area would be 2pi*r*dy. Why is this not the case? I can see what is wrong by taking this to the 2d world, and applying this logic to finding the perimeter of a triangle by "integration": Isocoles Triangle with base D and height H. If one thought: dP(erimeter) = 2dy one would obviously get P = 2y | 0 to H, which would mean P = 2H, obviously wrong. That essentially amounts to an unfinished square. It has to be 2S where S = slant height. But I still don't grasp why it works for volume, but not surface area (I initially stumbled upon this by trying to find SA of a cone). Can you please explain or link me to a video where you have?
With regards to volume you integrate the function pi*r*r over the domain [0-r], so at each point between 0 and r you have an area which either has x or y representing r , if you choose it to be x then you integrating over y and vice versa. The reason you have to use double integration with surface area is because you don't have a function to integrate over a domain. Remember polar coordinates (rcosx,rsinx) , x in radians here. For every sinx there is a corresponding cosx albeit at different radians over an interval. So if you decide to integrate 2pi*rsinx then you must multiply rsinx by r again since for every height there is an equal horizontal, you will realize this when you look at sine and cosine graphs. Integrate 2pir*r*sinx over interval [0-pi] and you will get 4pi*r*r , then working backwards to find the derivative will make sense.
Sir for volume you take cylinders of thickness dx but for surface area you take small frustem of dl why ? I am not able understand can you please help why we not take the hollow cylinders in the surface area case plz ..
All the heights coming from the circumference must be squeezed into a length equal to radius. Note; from the interval 0-pi/2 , if r is 1 then pi/2 is larger than 1 likewise if radius is 2 then for that interval it's 0-pi and so on.
I think that, from your interesting approach that, it is a requisite knowledge for understanding; at least from my perspective, Einstein's Relativistic equation.
Actually why do we have to integrate surface of cones? In the dx world „every line is stright”, so every (truncated) cone should just be a cylinder. So it seems enough to integrate cylinder side area. dS = 2 pi sqrt(r^2-x^2) Am I wrong?
I've always wondered why the intuitive answer gives you the wrong one: rotating a semicircle about its diameter 360°. V = (1/2)(πr^2)(2π) = π^2 r^2 SA = (1/2)(2πr)(2π) = 2π^2 r EDIT: I'm *still* wondering
Pappus' Theorem says that, for any figure of revolution, be it a linear figure revolved into a surface, or a plane figure revolved into a solid, its measure (area or volume) will be the measure of the original (length or area), times the length of the arc swept out by its centroid. [And this is only for cases where the figure being revolved is all on one side of the axis of revolution.] When you start with a semicircle, the centroid of the semicircular arc is in a different place than the centroid of the semicircular disk. Also, finding those centroids is basically the same problem as finding the resulting area or volume of the sphere/ball; that is, you get the same integral to evaluate, to find the centroid as to find the area or volume. Actually, an easier case is making a torus by revolving a circle about a straight line not intersecting the circle, in order to find its surface area and its volume. For that, the centroid is just the circle's center in both cases. And as a point of interest, Pappus' Theorem also works in higher dimensions, n ≥ 2. Note that the axis in n dim. is (n-2)-dimensional. So in the plane, it's a point (the center of rotation). It works, e.g., to find the area of a circle, by revolving its radius around one endpoint. You can use it to find the volume of a cylinder, or a cone, or a frustum of a cone. It's one of the greatest, yet least-taught tools in mathematics. Fred
@@NotYourAverageNothing That's right. The figure can touch, but not cross, the axis. So it doesn't really help in finding surface & volume of a sphere/ball. It works best on figures with enough symmetry that no fancy calculation is required to find areas and centroids. Fred
I remember my high school science teacher pointing it out as well that the surface area of a sphere is the derivative of it volume. He then extrapolated it to a cube. If a cube with edge r has a volume of r^3, then following the same logic would have its surface area as 3(r^2). But obviously that'a not the case; it's 6(r^2). So does this pattern only work in a sphere?
yes , because this pattern is a consequence of how derivatives work, not any intutive application we can think about if you wanna think about it, a simplification is that when you augment the radius of a sohere by a tiny r, its volume will increase by approximattely this tiny r times its surface area simimilarly, if you increase the side of a cube by a tiny l, the volume will increase by this tiny l times 3x its side². think of it as adding tiny nudges with despiaseble thicknesses to three adjacent sides of the cube
At 1:13, 2:27, 3:41, 4:56, 6:10, 7:24, 8:38, 9:52, 11:06, 12:20, 13:34, and 14:48 (so basically every ~74 seconds), the message from the first few seconds pops up for a frame, it is unsettling and catches me off guard, trying to watch the video :/ Edit: this has been fixed since I posted my comment. Also the calculus connection between the surface area and the surface area of the sphere is the rate of change of the volume of the sphere, so the volume is a primitive function of the surface area of the sphere.
I am very sorry I actually did NOT mean to edit the video like that. I am actually not sure why the video turned out to be that way. I can reupload another one later.
@@diegomullor8605 The reason might be that either your monitor's/phone's framerate is comperatively low or your browser's video player does not support this framerate, those pictures show up for a single frame or so.
The differential element, the cone in this instance, no longer has a horizontal small thickness, dx. For the cone, the small thickness is slanted and is represented as dL, which is not as simple as the dx or dy you are used to.
@cypke: dx still works; it's what you have to put in front of it, that's different! When finding the volume, the volume element is just a "plate," whose volume is a cross-section area times the thickness; dV = A(x)dx. When finding the surface area, the area element, dS, is a sloped, or beveled surface; the lateral surface, not of a cylinder, but of a frustum of a cone. So calling the slant height of that frustum dL, as bprp does, dL = √(dx²+dy²) = √(dx² + [ d[√(r²-x²)] ]²) = etc. = r dx/√(r²-x²); and then, dS = 2π√(r²-x²) dL = 2πr dx Fred
