>I see a disconnect between the example in the video, where he derives the area of a circle with the concentric rings, and my experiment. In that example, dr is the difference between the radii of each of the rings, and in my example it has nothing to do with the radii of my cylinders, only the thickness.
You've correctly identified the issue.
First, I'd rename dr "dx" to avoid confusion. You're integrating across the x axis and assuming that the radius of your ~~sphere~~ disk is equal to x. That assumption is wrong.
Funnily enough, it doesn't seem to make a difference in the actual answer if we put x or R there as the radius of the disk. As long as we limit our integral properly( from [-R, R] or take 2 times [0,R](due to symmetry)) . Is there some other reason for this ? Or is it also because of symmetry?
Edit: Nvm, this is absolute bs we won't get the factor of 1/3 only. Sorry for the trouble.
If we choose dx for the width of the disks and take the integral, r just becomes a constant and gets pulled out of the integral. The symbols we use don't matter in themselves, as long as we're consistent. It was inconsistent of me to have r and dr in my integral that way since the width of the disks is not related to their radii
Yes, that is indeed correct. I hadn't written down anything and saw that we would get a factor of 4 outside both ways. And forgot about the 1/3. So thought that the integral calculation turn it same, but it actually does not work out to be the same
I'll give you a hint.
Take a point along the x axis, say at half the distance. x = r/2, say. What is the radius of the concentric circle at that point?
You will. You are on the right track. Now calculate the area of that circle.
What you are doing is imagining the integral as a stack of concentric disks from left to right; at each x, you know the area of the disk as it relates to r and x. Let's call that A\_x, the area of a disk at position x. The volume of that disk is A\_x dx. Now integrate it to get the total volume.
The fundamental problem is that you imagine dr to be an infinitesimal move along the x direction, but the 'r' of the disk is along the y axis.
ma guy, you are trying to calculate the volume by looking at the sphere as a "rotation body", hence you rotate a halfcircle around the x axis, meaning that your "r", isn't "x", but instead its pythagoras -> r = (R^2 - x^2)^(1/2) while integrating to dx. there's the logic falacy. Going through the x axis changing r isn't linear like you did, but it's a curve (which is clearly bigger than a cone)
You’re deriving the volume of a cone with height of R and slapping a two on it. I see what you were trying to do but you’re assuming the radius of a cross sectional circle depends linearly on the distance to the center like a cone does.
Someone else generally pointed it out, but I wanted to put it in plainer language. The idea you have is to find volumes of infinitesimally small cylinders. So what you're integrating is πr²dh, not by dr.
Sounds like you figured out the error. I think need step is for you to really understand *why* the volume differential is what it is in spherical coordinates.
https://mathworld.wolfram.com/SphericalCoordinates.html
rsin(phi) projects a point on the surface onto the x-y plane
rd(theta) sweeps out an arc on that plane
Then the last two terms are d(phi) and dr
Integrating gives the volume of a sphere
I'm not sure whether it is rigorous enough, and for entire picture it would better be triple or double integral (revolution around axis), but this technically works - You're just integrating over a volume of a hemisphere. To account for the entire sphere Your bounds of integration should be [-R,R].
A sphere has elements varying in all three dimensions. So, you need to use triple integration to address this issue. The equation you are using reads " Volume of two identical discs of area pi\*r\^2 and thickness of dr and varies from r=0 to r=R." If you can imagine it, the object turns out to be two cones joint at their vertex (like this [https://files.askiitians.com/cdn1/images/2017318-102251603-2455-3-sections-of-a-cone.jpg](https://files.askiitians.com/cdn1/images/2017318-102251603-2455-3-sections-of-a-cone.jpg))
In order to find volume of a sphere by integration, just imagine a sphere element and calculate it's volume (dV). You should first learn about spherical co-ordinate system to do this. Integrate it over the angles and radius and then get the total volume.
instead of r, you need to write the expression for the height of a circle at a certain point on the x-axis.
Remember the formula for the circle: x\^2 + y\^2 = R\^2
From this it follows that: y = sqrt( R\^2 - x\^2 )
Plug this in instead of r and it should work
>I see a disconnect between the example in the video, where he derives the area of a circle with the concentric rings, and my experiment. In that example, dr is the difference between the radii of each of the rings, and in my example it has nothing to do with the radii of my cylinders, only the thickness. You've correctly identified the issue. First, I'd rename dr "dx" to avoid confusion. You're integrating across the x axis and assuming that the radius of your ~~sphere~~ disk is equal to x. That assumption is wrong.
Ah I see. Obvious in retrospect. Thank you!
His argument can work upon realizing this. The volume of the disc at x would be pi*(R^2 - x^2 ) dx. Now proceed as before.
Funnily enough, it doesn't seem to make a difference in the actual answer if we put x or R there as the radius of the disk. As long as we limit our integral properly( from [-R, R] or take 2 times [0,R](due to symmetry)) . Is there some other reason for this ? Or is it also because of symmetry? Edit: Nvm, this is absolute bs we won't get the factor of 1/3 only. Sorry for the trouble.
If we choose dx for the width of the disks and take the integral, r just becomes a constant and gets pulled out of the integral. The symbols we use don't matter in themselves, as long as we're consistent. It was inconsistent of me to have r and dr in my integral that way since the width of the disks is not related to their radii
Yes, that is indeed correct. I hadn't written down anything and saw that we would get a factor of 4 outside both ways. And forgot about the 1/3. So thought that the integral calculation turn it same, but it actually does not work out to be the same
Are you sure about that?
Lol thank you, it was such bs 😂. Sorry.
I'll give you a hint. Take a point along the x axis, say at half the distance. x = r/2, say. What is the radius of the concentric circle at that point?
√(3)/2*R... I'm not understanding what you're getting at haha
That's what the radius of the concentric circle is. That's not what your integral is saying the radius of the concentric circle is.
You will. You are on the right track. Now calculate the area of that circle. What you are doing is imagining the integral as a stack of concentric disks from left to right; at each x, you know the area of the disk as it relates to r and x. Let's call that A\_x, the area of a disk at position x. The volume of that disk is A\_x dx. Now integrate it to get the total volume. The fundamental problem is that you imagine dr to be an infinitesimal move along the x direction, but the 'r' of the disk is along the y axis.
ma guy, you are trying to calculate the volume by looking at the sphere as a "rotation body", hence you rotate a halfcircle around the x axis, meaning that your "r", isn't "x", but instead its pythagoras -> r = (R^2 - x^2)^(1/2) while integrating to dx. there's the logic falacy. Going through the x axis changing r isn't linear like you did, but it's a curve (which is clearly bigger than a cone)
You’re deriving the volume of a cone with height of R and slapping a two on it. I see what you were trying to do but you’re assuming the radius of a cross sectional circle depends linearly on the distance to the center like a cone does.
Someone else generally pointed it out, but I wanted to put it in plainer language. The idea you have is to find volumes of infinitesimally small cylinders. So what you're integrating is πr²dh, not by dr.
Sounds like you figured out the error. I think need step is for you to really understand *why* the volume differential is what it is in spherical coordinates. https://mathworld.wolfram.com/SphericalCoordinates.html rsin(phi) projects a point on the surface onto the x-y plane rd(theta) sweeps out an arc on that plane Then the last two terms are d(phi) and dr Integrating gives the volume of a sphere
I'm not sure whether it is rigorous enough, and for entire picture it would better be triple or double integral (revolution around axis), but this technically works - You're just integrating over a volume of a hemisphere. To account for the entire sphere Your bounds of integration should be [-R,R].
That's what the factor of 2 out the front is meant to achieve. Their problem is more fundamental than that.
This is the volume of the cone of radius R and height R. Because in that case, dr=dx, unlike here.
A sphere has elements varying in all three dimensions. So, you need to use triple integration to address this issue. The equation you are using reads " Volume of two identical discs of area pi\*r\^2 and thickness of dr and varies from r=0 to r=R." If you can imagine it, the object turns out to be two cones joint at their vertex (like this [https://files.askiitians.com/cdn1/images/2017318-102251603-2455-3-sections-of-a-cone.jpg](https://files.askiitians.com/cdn1/images/2017318-102251603-2455-3-sections-of-a-cone.jpg)) In order to find volume of a sphere by integration, just imagine a sphere element and calculate it's volume (dV). You should first learn about spherical co-ordinate system to do this. Integrate it over the angles and radius and then get the total volume.
This gave me a bit of a chuckle, lol
You just (almost) figured out the volume of a cone lol. Well done
woa you were close, off by a factor of 2 which is due to the other half
Instead of integrating using disks, consider using circular shells (hollow spheres)
Does that method integrate the surface area of the shells?
You can consider each d[shell] to be surface area yea
You are taking width of your small cylinder as dr which is wrong. It is not dr. dr is from centre to periphery
instead of r, you need to write the expression for the height of a circle at a certain point on the x-axis. Remember the formula for the circle: x\^2 + y\^2 = R\^2 From this it follows that: y = sqrt( R\^2 - x\^2 ) Plug this in instead of r and it should work