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Homework Help: Volume of a Sphere Question, really bothering me!

  1. Jul 27, 2010 #1
    I am trying to find the volume of a sphere using a different method than I have before, (I can do it other ways) but it's not working and I cannot figure out why.
    [PLAIN]http://img824.imageshack.us/img824/9333/spheren.jpg [Broken]

    That's my work...what is wrong with what I did? what did I find? (where I wrote volume of disk I mean volume of sphere)
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jul 27, 2010 #2
    The way you define d theta is not correct. It should be dr, where r is the distance from the center of the sphere to the disc element of infinitesimal thickness, dr.

    So your integral will be,
    2$\int^{\pi}_0 \pi R^2 \sin^2{\theta}d r $.

    Then convert dr to d theta using the relation cos theta= r/R.
  4. Jul 27, 2010 #3
    I see that that works, thanks. But then how come when finding the surface area I can define d theta that way? (and instead of using the area of the disk I used the circumference of a ring, 2pi(Rsintheta)).
  5. Jul 27, 2010 #4
    Good question. First note that, you could use d r for the volume of sphere as well, but then your integral equation becomes:
    2$\int^{R}_0 \pi R^2 ( 1- \frac{r^2}{R^2})d r $.

    Notice the change in limits, here you follow how r changes from 0 to R.

    Then notice that the infinitesimal is defined along the direction how you imagine your quantity to change. When you are talking about the surface area, it is along the arc, (you can pretend you were rotating your head) and that's why you had everything in terms of theta and your infinitesimal quantity was d theta. When we considered the volume of the sphere and used d r, then we were looking from the center to the top of the sphere, ie. from bottom to top linearly.

    So it all depends on how you visualize your variable quantities to change. For example, the area of a triangle drawn on a plane surface can be calculated simply, but the area of a triangle drawn on the surface of the sphere will need a new form of visualization.

    But keep working at this! Your question shows that you are thinking in the right direction and after some practice you will forget the existence of such issues.

    Skip the following line if you are already confused: Differential quantities are always defined using a coordinate system as reference. Ideally, you would want to keep your coordinate system to the most basic. When you choose d r as the differential element, you essentially chose a linear coordinate system and you converted all the variable terms (theta) to r. When you choose d theta, then you chose a rotational (cylindrical) coordinate system.

  6. Jul 27, 2010 #5
    I guess I am still a little confused as to why I can't use d theta the way I did, I understand that it doesn't work, but as theta changes in my equation so does the volume of the disk, I don't intuitively see why this is wrong, why the thickness of this disk is different than the thickness of a ring, because a ring is the same thing as a disk except the disk is filled in the middle.
    Last edited: Jul 27, 2010
  7. Jul 27, 2010 #6
    Theta rotates, round and round. You have divided your sphere like an eggslicer would divide an egg. If your slices went round and round, you could have been correct, but your slices dont. They go up and down.


  8. Jul 27, 2010 #7


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    The 'thickness' (I would say 'width') of a ring is its dimension parallel to the surface of the sphere. The 'thickness' of a disk is its dimension perpendicular to the disk. They are two different things. I'm really not sure why you think they are the same.
  9. Jul 27, 2010 #8


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    As sshzp4 was saying, you need dr rather than d[itex]\theta[/itex] because intuitively, each slice will have consistent thickness if using dr. When you use the other, at [itex]\theta=\pi/2[/itex] (the right side of the sphere in the diagram), a small change of say, 1o up and down will give a slice thickness of [itex]2Rsin(1^o)\approx 0.035 R[/itex] while if we are at [itex]\theta=0[/itex] (the top of the sphere), a small change of 1o left and right will give a thickness of [itex]R-Rcos(1^o)\approx 0.00015R[/itex] from the top of the sphere. The inconsistency should be fairly intuitive.
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