# Area sphere integral trapezoid elements

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.

• Why $dr$? At first, you wrote $r'(z)$. You should stick to this. – Amitai Yuval Apr 3 '18 at 16:01
• 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. – Amitai Yuval Apr 3 '18 at 16:03
• ok for the $o(dz)$ term. If I let $r'(z)dz$, I still have to deal with $dz\,dz$ (see my edit). – Smilia Apr 3 '18 at 16:39