surface area of a sphere using cartesian coordinates

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November 4th 2007, 08:06 PM
stug

surface area of a sphere using cartesian coordinates

I'm trying to derive the equation for the surface area of a sphere by using a one-dimensional integral in cartesian coordinates, but I keep getting the wrong answer. Here is my reasoning:
The circumference of a circle is 2*pi*r
For a circle in the x-y plane, the equation is x^2 + y^2 = r^2, or x = sqrt(r^2 - y^2).
If we let x, as defined above, be the radius of a circle (in the z-x plane), we can calculate the area of an element of the surface area as a ring, with circumference 2*pi*sqrt(r^2 - y^2)*dy . We should then be able to integrate with y going from -r to r and get the surface area of the sphere. However, evaluating this integral only yields pi^2 * r^2, not the right formula.
What is wrong with my method here. Am I making a wrong assumption?
November 4th 2007, 08:44 PM
kalagota

Quote:

Originally Posted by stug

I'm trying to derive the equation for the surface area of a sphere by using a one-dimensional integral in cartesian coordinates, but I keep getting the wrong answer. Here is my reasoning:
The circumference of a circle is 2*pi*r
For a circle in the x-y plane, the equation is x^2 + y^2 = r^2, or x = sqrt(r^2 - y^2).
If we let x, as defined above, be the radius of a circle (in the z-x plane), we can calculate the area of an element of the surface area as a ring, with circumference 2*pi*sqrt(r^2 - y^2)*dy . We should then be able to integrate with y going from -r to r and get the surface area of the sphere. However, evaluating this integral only yields pi^2 * r^2, not the right formula.
What is wrong with my method here. Am I making a wrong assumption?

let $f(x) = \sqrt{r^2 - x^2}$
and revolve it about x-axis..

use this formula:

$S= \int_a^b 2\pi f(x) \sqrt{1 + (f'(x))^2}dx$
November 4th 2007, 08:51 PM
stug

I've seen that formula before while looking for an answer to this problem. My question, however, is more like why doesn't my approach work? What does it leave out? Or perhaps, what does that extra sqrt(1 + f'(x)^2) term take into account?
November 4th 2007, 09:00 PM
kalagota

Quote:

Originally Posted by stug

I've seen that formula before while looking for an answer to this problem. My question, however, is more like why doesn't my approach work? What does it leave out? Or perhaps, what does that extra sqrt(1 + f'(x)^2) term take into account?

as i've understood my notes (a year and a half ago), the $\sqrt((1 + (f'(x))^2)) dx$ is the height of the curve at a point x.
November 4th 2007, 09:21 PM
stug

Now that I think about it, I believe that sqrt(1 + f'(x)^2) is the arc-length, since we can change sqrt(1 + (dy/dx)^2)dx into sqrt(dx^2 + dy^2).

However, I still don't really see why this is necessary. I thought that the point of dealing with infinitesimals was that changes in the function shape were small enough to be negligible over a dx or dy. In other words, can't we just view this as summing the surface area of a bunch of cylinders with infinitesimal height, rather than taking into account the slope of their sides (as the arc-length term seems to be doing)?
November 4th 2007, 09:43 PM
kalagota

Quote:

Originally Posted by stug

..
However, I still don't really see why this is necessary. I thought that the point of dealing with infinitesimals was that changes in the function shape were small enough to be negligible over a dx or dy. In other words, can't we just view this as summing the surface area of a bunch of cylinders with infinitesimal height, rather than taking into account the slope of their sides (as the arc-length term seems to be doing)?

hehe, that thing i cant answer well.. maybe others can explain it..
November 6th 2007, 02:12 PM
stug

anyone?
July 30th 2012, 02:20 PM
Skip

Re: surface area of a sphere using cartesian coordinates

First, you're right, that square root thing is the arc length, not just the height. Second, you're sort of correct in that, as you let dx get real small, dy also gets real small, so that square root LOOKS LIKE it is resolving to just the square root of one, which gets you to your result. But while dy is getting small, dy RELATIVE TO dx may not be getting small at all. That's what dy/dx is. At the equator, dy is actually changing quite rapidly wrt dx, so dy/dx is rather large, and if you don't have the full square root thing in there, you'll miss a lot of the surface area.