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!
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)
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....
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 .........👍👍👍👍🙏🙏🙏🙏🙏
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
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
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 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.
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.
@@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.
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?"
@@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
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 had a doubt @ 8:07 volume of sphere is 4/3 πR^3 but he used as 4 πR^3 why??
@@jeeva3922 it's not 4piR^3, it's 4piR^2.delta(R). The formula is V = A.delta(R) and A is the area of the sphere = 4piR^2.
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!
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)
That give you the top half of the sphere. (that is why the integral was multiplied by 2).
You can do it both ways. Try it using dz and see what happens. (It may be a more difficult integration).
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 ...
We buy external memory by the TB. -Wife
@@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....
simplifying is da best part...very true :)
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 .........👍👍👍👍🙏🙏🙏🙏🙏
You are welcome. Glad you find this helpful. 🙂
Best tutorial on youtube for this... I thought it would be easy to derive this one... Turns out its one of the harder ones lol
Thanks sir you are the best
Love from india😍
Yo I'm also from india
Why haven't you used dz for the "height" of the ring like in the solid sphere instead of Rd(theta)
then partial derivatives would have been involved and then it would have gotten more complicated
Your videos are incredibly helpful!
Thank you. We are glad you found them! 🙂
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.
Great video. Thank you!
Glad you liked it!
Why is the height of a slice of a solid sphere only dz, but here it's R.dtheta? What's the difference?
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
great I like it
It's so easy to understand
*Fascinating*
Glad you liked it. 🙂
Why we made our limits from 0 to pi /2 that means that we got 1/4 of our sphere then we should multply it by 4
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.
10/10 Would watch again
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 😊👍
@@VickysTuition i have the same problem having extra cos , can u explain it for me
@@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
you saved me in my EXCERCISE
Thank you so much sir
This video is perfect 👌. And very nice video Thank-you so much for cool video
Why dV is used? For hollow sphere we use sigma(area density or dA) because in C.O.M derivation for hollow sphere we used dA.
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.
1:10 Why take the volume (dV) of that strip and not the area (dA) since the sphere has no thickness?
Something with zero thickness does not exist. There must be some thickness
Does the radius include the thickness in your calculation?
Essentially it makes no difference since the thickness goes to zero in the limit (which is the foundation of calculus).
Thank you, you have helped me so much.
Happy to help!
@@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!
Yes, I look for the easiest way to solve the problem.
@@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.
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 instead of using twice integration from 0 to π/2 can we just straight away integrate from 0 to 2π or not?
Try it and see if you get the same result
Absolutely Lovely
Thanks!
The videos you post just perfect.
Would you please explain to me what the meaning of - dv (02:00) in a few words?
+Maor Kadosh I didn't see a "dv" but I did use "dV" dV is a infinitesimal (very small) volume. In this case it is the volume of the very small ring.
why in the calculation for dv, do you multiply by the radius of the hoop?
Circumference = 2 pi x height = R d theta thickness = dR
@@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?
Indeed. That is why the equation given only accounts for the washer part. Not the empty space.
8:07 actually the vol. Of the sphere is 4/3π R^2∆R but you used as 4 π R^2∆R??
No, the volume of the volume element is 4 pi R^2 delta R
@@MichelvanBiezen sir can u explain me further pls
Think of it as a shell (that is what it is). You multiply the surface area x thickness = 4 pi R^2 x dr
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.
There are often multiple ways to solve a problem. Try it and see if you get the same answer.
@@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.
Yes, it made sense. Going through all that effort will help you understand this much better. Way to go.
Why are we taking the horizontal discs? Shouldn't it be vertical disks?
Try it with vertical disks and see if you can get the answer (best way to learn).
Why bothered with orientation it is a sphere 😂😂
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?"
Sorry, I see that it is R(d-theta.)
Thank you sir for your lectures.
Why can't you use the height of the initial disc to be dz, wouldn't that make the intergration easier?
It is not a disk but a ring. (Try it and see if you get the same answer).
@@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.
You use the perimeter instead of the area. You can integrate dz's of areas, why can you not integrate dz's of perimeters?
great
Thank you. Glad you liked it.
Nice
Thanks
Muito boa essa explicação
Why are the integral limits up to pi/2 and not pi?
Because of the symmetry. You can pick the limit from 0 to pi/2 and then multiply the result by 2
I see it now :P Off course... thanks a lot! Your videos have helped me a lot during my exam period :)
thanks
I am confused, why I(x) = dm(x^2) instead of I = 1/2(dm*x^2) since is a disk????
It is a hollow sphere, so dm is a hollow washer (not a disk).
Thank you very much Sir, your the best. Then it is true that when you use disk then Area= pie* r^2 for shell(ring) Area = 2pie* r * h?????
Bro it has a thickness delta r
lol
thank u very much ~~
You're welcome 😊
Sir, I just wanted to know why your integral bounds were from 0 to pi/2 if you were trying to solve for the top half of the sphere?
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.
you could havev assumed delta r as dr then the problem will become difficult I don't know how did you do that
+AVDESH KUMAR
dr is delta r as the delta approaches zero.
Why is volume 4pi and not 4/3 pi?
The volume of what object?
@@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
wait nvm figured it out lol
Why is the height Rdtheta
R d(theta) is the arc length
Thank u
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? :)
Дејан Гујић There are different ways to solve this problem.
Go ahead and try it your way and see if you get the same answer.
I didn't get the volume part while calculating density of the sphere..anybody wanna help?
density = mass / volume therefore mass = density x volume
@@MichelvanBiezen yes, but how is the volume 4piR^2 times delR
@@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
It's a sphere so..shouldnt it be 4/3pi(R+ del R)^3 - 4/3piR^3 or something like that?
Because we are calculating the volume of a "shell". So we take the surface area and multiply it with the thickness.