For the people who are asking about the dTheta and dz part: In a solid sphere, we could think about the slices as "small cylinders". When taking slices, you have to have some ignorance about the smaller parts of the volume. In the case of small cylinders, most of the volume would be in the center and the small inconsistencies would be on the ends. So we could ignore the ends and focus on the inner part. This is done by thinking about the cylinders by their centers, that is, the points on the z-axis. In a hollow sphere, however, we cannot think about the slices as small cylinders. Because now, most of the volume is actually on the ends. So a cylinder would not be able to approximate the volume enough. So now we can look at the slices as angle differences. This time, the inconsistency will be at the angle curvature. But you can prove, by the squeeze theorem, that it is ignorable. Hope this helps.
I would be called a NERD at 18 years of age if I said that this Video is so Enriching and Insightful. BUT, at 66 years of age.. I find your videos to be an Adventure and Exploration in the search for Truth and the Laws of Nature... as if it was Scaling Mount Everest and reaching the Top!!.. very well done Michel!!.. The ease with which you present these solutions is nothing less than PURE TALENT... thank you! (ps, I am not saying that 18 year olds are nerds for appreciating science and proofs.. but you know how kids can be... :) )
Of course my husband would probably have been called a nerd at 8. One of our son's friend's gave him a t-shrit that said, "I make nerd look good!" (I guess it hereditary.) It is people like you and my husband that make "nerd" look good!
wow, I just truly enjoy going back over these Videos from years past.. :) ... this video so clearly explains the logic from beginning to end... What a great catalog of Lectures Michel !... I wonder how many DVDs would be required to record all your videos.. wow..... Thanks for your talent at teaching ...
@@MichelvanBiezen LOL... Wife.... so true !!.. Occasionally I will post a link to one of your videos on my FB page. Just to serve as a Bookmark so I can get back to that video.. IT's interesting to hear some of the comments I get from my Friends as they wonder what's WRONG with me !!.. lol. because they think Science and Math are so Long ago.... Some people do Crossword Puzzles, some do Jigsaw puzzles... my Puzzles are Math and Physics problems.. :) .. and now I get to ponder the physics of HURRICANES as the season is upon us....
To everyone asking why it is Rd(\theta), remember that it is a hollow sphere. Plus arc length formula says s = r(\theta), so in this case, s = height, r*(\theta) = R*(d\theta)
Why isn't the height dz for starters? How do you know to set the height to dTheta? So frustrated as I tried this out before watching the video and my integral ends up being cos^4(theta) not cos^3(theta). The math all makes perfect sense when I watch you write it all out, I'm just lacking the insight in the setup.
It is important to note that dz and Rd{\theta} are not interchangeable since they do not represent the same values. Observe that dz is parallel to the z-axis whereas Rd{\theta} is tangent to the curvature of the sphere. Utilizing trigonometry, you can easily prove that Rd{\theta} is, in fact, equal to dz/cos{\theta}. You can also show this by taking the derivative of z=Rsin{\theta} with respect to theta. Therefore, dz=Rcos{\theta}d{\theta}. Also, realize that cos{\theta}=x/R by definition, so Rd{\theta}=R*dz/x. When converting from polar form to cartesian form with the equation derived above, there should be no issue with the cos^4{\theta}. Originally I arrived at something like (3{\pi}mR^2)/16 with the same error... Now to address the original question, "Why isn't the height dz for starters?" Besides my explanation above that dz doesn't equal Rd{\theta}, dz simply doesn't apply in the first place due to the definition of finding arc length. When finding the difference in the volume of the hoops of the hollow sphere, we're essentially finding the surface area of the sphere with an infinitesimal thickness {\delta}r. To find the surface area, we're summing all the cross-sectional hoops' dA, all with a 'thickness' ds (ds=difference in arc length). The integral, in this case, would basically involve the sum of all the tiny arc lengths from each hoop rather than the height dz as the sum of all dz does not equal the total arc length of the hoops. The only reason why dz was used in the previous video is that a solid sphere is comprised of disks rather than hoops. When finding the dV of each disk, the thickness of the disk is now the height dz since we're basically solving an integral for finding an area under the curve. When finding the area under the curve, we can now see that it is just a summation of values of the radius of each disk times an infinitesimal width of dz. If you're still asking why dz can't be used, it's basically a question of why do we define the integral of the area under the curve with rectangular width while defining the integral of an arc length with the actual segment length of the curve. I'll leave that up to intuition.
because if it's a solid sphere you can calculate the volume of the infinitely small cylinder meaning dm=Pdv=P*pi*x^2*dz(height of the infinitely small cylinder) giving the dI=1/2*x^2*dm=1/2*x^2*Pdv=1/2*x^2*P*pi*x^2*dz which then would give the integral (1/2*pi*P times this integral)(upper limit R, lowe limit -R representing top and bottom of the sphere) ∫x^2*x^2 dz=∫x^4dz the position x then equals the distance from the top of the sphere R minus the distance from the Z axis which gives x^4=(R^2-Z^2)^2 which makes the integral easy, now in the case of a hollow sphere the volume does not equal the whole sphere but only the surface area which for cylnders are 2*pi*r*h which using the same method as with the solid sphere would give the integral ∫x^3dz which doesn't make x easily translatable into something depending of Z. So basically it has with the calculation of the volume of the slice, if the slice is hollow then you wouldn't include the "middle part" only the thickness of the shell, and calculating the thickness of that is most easily done using R.dthetha since that integral is alot easier to solve
Sir I am indian student sir you explane this topic very easy with concept । Thaks sir for this vedio because it's clear my 🧐 doubt . Thaks sir .........👍👍👍👍🙏🙏🙏🙏🙏
Sir, I have a question. Concerning about the video about the M.O.I of a hollow sphere and the solid sphere. Why did you consider for the solid sphere, the dv, to get the surface area which is (pi)x^2dx then for hollow sphere, the circumference. 2(pi)xR(d theta)(delta R)
Denver Daño I hope sir you understand my question. I'm just confused about the idea of the thing you said, the surface area for the solid sphere. then the circumference for hollow sphere. both question about the dv.
Denver Daño Hello, in the case of a solid sphere, the volume of the section should be the one of a very slim cylinder of base AREA : pi*x^2 and of height dz so the volume is dv=pi*x^2*dz For a hollow sphere, the section taken looks a lot like a hoop, a ring or an empty cylinder, but because the "walls" of the cylinder are very narrow, you can assimilate the area of the base to the circumference of the base times its width which is deltaR, so base AREA=2*pi*x^2*delta R and the height of the cylinder is R*delta (theta), so dv=2*pi*x^2*deltaR*R*delta(theta). Hope I have answered your question.
Sir, when calculating dV, are you taking in consideration the "wedge" of the outer "ring?" What term within the product creates that "wedge" of the "ring?"
No, having the limits as shown gives you the moment of inertia for the top half of the sphere. Since the moment of inertia for the bottom half will be the same, it is also the moment of inertia for the whole sphere.
For solid sphere, we integrate it along height (dz). For hollow sphere, we integrate it along arc length (r.dtheta). Why the difference? In both, the sphere is divided into same number of discs and the discs are all summed up right? The same discrepancy i see when we calculate volume & surface area of a sphere as well. It's very puzzling
That is done to make it easier to integrate. You may remember all the different techniques we learned to integrate over a volume, using cylinders, washers, etc. for the dV
I tried integrating both ways. But i get an extra cos(theta) in one of the ways. So, the answer doesn't come same. I have not seen your other videos - I will check them out now. Thanks 😊👍
@@DucNguyen-rr2ug Volume of Sphere needs to be calculated using dz instead of r.dtheta (arc length) ! To understand why? Try calculating using r.dtheta ! Essentially, instead of summing up many cylinders, you will be now summing up Trapeziums ! But a Trapezium = Cylinder + 2 Triangles on the side ! So, basically people are ignoring these triangles & only considering the cylinder to calculate the volume. I wondered why that is. So, i tried calculating the volume of the triangle, which is surface area of triangle x circumference of circle ! But surface area of triangle = 1/2 . dx . dy ! So, when dx terms are around, we usually IGNORE dx-square terms & dx-cube terms right ? And dx-square is nothing but "dx . dx" ! Similarly following the same logic, we can now ignore "dx . dy" term too ! That means we can ignore the Trapeziums & reason only with cylinders for Volume of Spheres ! But when it comes to surface area, we can continue reasoning with Trapeziums, because thats where our area element lies. Hope i made sense
@@MichelvanBiezen I tried deriving this independently but I kept hitting the same obstacle. I was just wondering if you chose to use R*dtheta rather than dz, in order to avoid multivariable calculus. I know some of the comments here have addressed this question, but I don't understand what they are saying!
@@MichelvanBiezen It makes more sense now. The moment of inertia of a hollow cone takes a similar step in the sense that it totally neglects the volume of the cone and only focuses on surface area.
Since m = density x volume and density is constant, volume is proportional to mass and thus we can use volume to relate the mass of the shell to the geometric shape.
@@MichelvanBiezen thanks for the reply! so its basically just a length x width x height calculation, which gives the volume? My problem is that since were modelling a 'washer' shouldn't you also have to take away the 'empty space' in the middle of the washer/hoop/ring?
Hello. Would it be accurate to say that the ΔR was unnecessary, as you could have simply used the Surface Area density of the sphere, given that the ratio of mass to surface area is also constant? From your calculation, it simply removes the ΔR from the top and bottom of the fraction, leaving the integration untouched.
@@MichelvanBiezen It works, simply because all that happens is you get the same equation, but the ΔRs disappear. Interestingly, when I attempted to do it another, unrelated way, I got the correct answer, but I believe it was sheer coincidence, because the reasoning is unsound. I thought you could just integrate by 'adding up' the moments of all of the hoops ( each being MR^2 = λR^3 for linear density λ) from r=0 to r=R. Although some ad hoc adjustments got me the right answer, I think it fails because there is no interpretation of dr, since each hoop MR^2) has negligible width. I don't know if this makes any sense when communicated in words.
@@MichelvanBiezen I tried integrating using dz, and the issue came with integrating (x^2 + z^2)^(3/2) dz, which leaves a convoluted solution equation + C. However, what is the reason this is not solvable? Is it because of the unknown constant? It would be nice to understand the reason it doesn't work.
Jet Erickson Tirona Also notice that I multiplied the integral by two. That way you get the result for the whole sphere, but it makes the integral easier.
Jet Erickson Tirona Hey! When I saw him integrating from 0 to pi/2 I thought he was only doing the work for one quarter of the sphere, and this baffled me. I don't know if this was also the case for you, but if so, I finally noticed that integrating from 0 to pi/2 was actually enough to solve for the top half of the sphere. Notice that going through the angle from 0 to pi/2 actually covers the entire height from 0 to R (so it's like integrating from 0 to R), while integrating from 0 to pi is equivalent to finding twice the height 2R. Hope it made some sense.
The way I think about it makes sense if you've had experience with something called volumes of revolution (usually covered in calc II, or in calc BC if you're doing AP). It's like taking just the first quadrant of a circle and rotating it about the y-axis (in this case, the z-axis). Except in this case, you just have to remember that the result should be hollow, so it resembles a typical washer method problem. In this type of problem we'd integrate from 0 to the height (in this case, pi/2 since we're using radians). Then just multiply by 2 to get the whole hollow sphere.
@@MichelvanBiezen volume of a sphere is volume = (4/3) · π · r3 and thats the dv we use for a solid sphere. But the dv for this sphere is 2pixrdodeltar. So im just curious why we would use that and not 4/3pi r3. maybe its something i missed
doesn't R*dtheta give you a small arc length? Can you still say the arc length is approx the heigth of the little segment of hollow disc? I guess you assume dtheta goes to zero? :)
@@MichelvanBiezen sorry to bother you with this stupid question..but I really don't get how the volume of the sphere is its surface area times thickness
