I would like to compute the area of the sphere of radius $R$ with the following method:

I consider section of the sphere in an orthonormal frame with Cartesian coordinates $(x,y,z)$ delimited by the two horizontal planes of elevation $z$ and $z+dz$ so that the area of this section (which is a trapezoid) is $dS = dz/2\; (2\pi r(z)+2\pi r(z+dz))$ where $r(z)$ is the radius of the circle defined by the intersection of the Sphere and the plane of elevation $z$ given by $$ r(z)=\sqrt{R^2-z^2}. $$

So the integral to compute is $$ A = \pi\int_{-R}^R (r(z)+r(z+dz)) dz $$

I can write $r(z+dz)=r(z)+dz\,r'(z) + o(dz)$ and then

$$ A = \pi\int_{-R}^R (2r(z)+ dr +o(dz))dz $$

Now how to handle for instance $dr\,dz$ ? $dz\, dz$ ?

I don't master exterior derivative of differential forms, but how can we introduce them here ?

I can also do the following computation $$ A = \pi\int_{-R}^R (2r(z)+ r'(z)dz +o(dz))dz \approx \pi\int_{-R}^R (2r(z)+ r'(z)dz)dz $$ but here I have also to deal with $dz\, dz$. If it is equal to $dz\wedge dz=0$, I would like to see why geometrically.

  • $\begingroup$ Why $dr$? At first, you wrote $r'(z)$. You should stick to this. $\endgroup$ – Amitai Yuval Apr 3 '18 at 16:01
  • $\begingroup$ As for the $o(dz)$ term, you should show that its contribution to the total integral is $<\epsilon$ for every $\epsilon>0$ and thus vanishes. $\endgroup$ – Amitai Yuval Apr 3 '18 at 16:03
  • $\begingroup$ ok for the $o(dz)$ term. If I let $r'(z)dz$, I still have to deal with $dz\,dz$ (see my edit). $\endgroup$ – Smilia Apr 3 '18 at 16:39

The error in the analysis is in your approximation by the area of a trapezoid. The area of a frustum of a cone is given by the product of the average circumference with the slant height (as opposed to vertical height). [You can see the error in your approach if you try to do arclength ignoring the hypotenuse of the right triangle.]

You will need to do a surface integral to bring in double integration, which is not what you're trying to do.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.