# Thread: Surface area of a sphere using integration.

1. ## Surface area of a sphere using integration.

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 $\displaystyle 4\pi r^2$. However from the figure:

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

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

$\displaystyle SA=2\pi \int _{-r}^r \sqrt{r^2-z^2}dz=\pi ^2r^2$

2. ## 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 $\displaystyle z$, 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).

3. ## Re: Surface area of a sphere using integration.

EDIT - see next post

4. ## Re: Surface area of a sphere using integration.

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

$\displaystyle SA = 2 \pi \int_{-R} ^R \sqrt{r^2-z^2} ds = 2 \pi \int_{-r} ^r \frac {r \sqrt{r^2-z^2}} {\sqrt{r^2-z^2}} dz = 2 \pi r \int _{-r} ^r dz = 4 \pi r^2$ .

5. ## Re: Surface area of a sphere using integration.

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

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

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

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

So, why can you do it for the volume but cannot use the same idea for surface area??

6. ## Re: Surface area of a sphere using integration.

Note that dz = dS at theta = 0. So it can't be that $\displaystyle sin \theta = \frac {fz}{ds}$. In additiobn, if yuo try the integral using that you get an incorrect result.

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.

7. ## 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.