Results 1 to 7 of 7
Like Tree3Thanks
  • 1 Post By SlipEternal
  • 1 Post By ebaines
  • 1 Post By ebaines

Math Help - Surface area of a sphere using integration.

  1. #1
    Junior Member Dark Sun's Avatar
    Joined
    Apr 2009
    From
    San Francisco, California
    Posts
    36
    Thanks
    1

    Cool Surface area of a sphere using integration.

    Surface area of a sphere using integration.-volumesphere.jpg
    I know that there is a mistake here, but I need help finding what my error actually is! I know that surface area of a sphere of radius r is 4\pi r^2. However from the figure:

    radius of a cross-section: x=\sqrt{r^2-z^2}
    Circumference of a cross-section: C=2\pi \sqrt{r^2-z^2}

    So, one should be able to integrate to get the surface-area, but the answer does not agree:

    SA=2\pi \int _{-r}^r \sqrt{r^2-z^2}dz=\pi ^2r^2
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,895
    Thanks
    759

    Re: Surface area of a sphere using integration.

    The error is you are not integrating in the correct direction. You are integrating with respect to z, but you should be integrating with respect to the tangent vector to the sphere at each point, or more specifically, the arc length. I am not good a drawing pictures, but this website should help you understand what I am talking about: (link).
    Thanks from Dark Sun
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor ebaines's Avatar
    Joined
    Jun 2008
    From
    Illinois
    Posts
    1,103
    Thanks
    321

    Re: Surface area of a sphere using integration.

    EDIT - see next post
    Last edited by ebaines; July 1st 2014 at 09:18 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor ebaines's Avatar
    Joined
    Jun 2008
    From
    Illinois
    Posts
    1,103
    Thanks
    321

    Re: Surface area of a sphere using integration.

    Quote Originally Posted by Dark Sun View Post
    radius of a cross-section: x=\sqrt{r^2-z^2}
    Circumference of a cross-section: C=2\pi \sqrt{r^2-z^2}
    In doing the integration the width of each strip of surface area is not dz, but rather ds = dz/\cos \theta, where  \theta is the angle from the center orf the sphere to the strip. But  \cos \theta = \frac {\sqrt {r^2 - z^2}} r, and so your integral becomes:

     SA = 2 \pi \int_{-R} ^R \sqrt{r^2-z^2} ds = 2 \pi \int_{-r} ^r \frac {r \sqrt{r^2-z^2}} {\sqrt{r^2-z^2}} dz = 2 \pi r \int _{-r} ^r  dz = 4 \pi r^2 .
    Last edited by ebaines; July 1st 2014 at 09:16 AM.
    Thanks from Dark Sun
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member Dark Sun's Avatar
    Joined
    Apr 2009
    From
    San Francisco, California
    Posts
    36
    Thanks
    1

    Re: Surface area of a sphere using integration.

    I see what you are saying, except I think it should be : \sin \theta =\frac{dz}{ds}=\frac{\sqrt{r^2-z^2}}{r} .

    I am still having trouble finding the conceptual error. Because why can't we approximate using cylinders? It seems to be legitimate in the case of Volume of a Sphere:

    radius of cross-section: x=\sqrt{r^2-z^2}
    Area of cross-section: A=\pi (r^2-z^2)

    Then the volume of a differential would be a cylinder: area times height \pi (r^2-z^2)dz.

    Integrate: V=\pi \int _{-r}^r r^2-z^2 dz=\frac{4}{3}\pi r^3

    So, why can you do it for the volume but cannot use the same idea for surface area??
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor ebaines's Avatar
    Joined
    Jun 2008
    From
    Illinois
    Posts
    1,103
    Thanks
    321

    Re: Surface area of a sphere using integration.

    Note that dz = dS at theta = 0. So it can't be that  sin \theta = \frac {fz}{ds}. In additiobn, if yuo try the integral using that you get an incorrect result.
    Surface area of a sphere using integration.-spherearea.jpg

    As for why you must include this term - it's because the error between ds and dz does not become vanishingly small as dz approaches zero for values of z near r. For the volume integral on the other hand the small error yo have by ignoring the angles at the end of teh cylinder gets vanishingly small as dz approaches zero for all values of z.
    Thanks from Dark Sun
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member Dark Sun's Avatar
    Joined
    Apr 2009
    From
    San Francisco, California
    Posts
    36
    Thanks
    1

    Re: Surface area of a sphere using integration.

    I see what you mean about the sine. I was using the wrong angle :/

    Also, I see that the error does not vanish when doing the surface-area with cylindrical cross-sections. Clearly when looking at the hemisphere, the frustum is a closer estimate than the cylinder. And, since they do not converge to the same number, the cylinder cannot be a correct differential.

    I am still having trouble showing that the curved bubble over the straight edge of the frustum vanishes when you take infinitely many frustum cross-sections.

    I also want to show that the error vanishes when using cylindrical cross-sections for the volume. I am trying to bound it above by something, and then show that it converges to the straight-line-base. Not sure exactly how to go about it.

    Thank you for your help, I truly appreciate it.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Surface area of a sphere
    Posted in the Calculus Forum
    Replies: 1
    Last Post: June 11th 2010, 12:51 AM
  2. Sphere surface area
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 14th 2010, 04:01 AM
  3. surface area of a sphere
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: April 11th 2010, 12:44 AM
  4. Surface area of a sphere
    Posted in the Calculus Forum
    Replies: 6
    Last Post: August 2nd 2009, 03:02 PM
  5. Need help with surface area of this sphere
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 26th 2008, 02:47 PM

Search Tags


/mathhelpforum @mathhelpforum