# Surface area of a sphere using polar coordinates

I am trying to find the surface area of a sphere of radius r by adding the circumferences of circular rings whose radii range from 0 to r. Because the radius ranges from 0 to r, the surface area equals

$$SA = 2\int_{0}^{r}2\pi rdr$$

$$SA = 2\pi r^2$$

What is wrong about this approach? I know it doesn't take into account the slope but as the height of each circular ring gets infinitesimally small if the rings were stacked on top of each other wouldn't they accurately approximate a sphere?

• You may find this relevant: math.stackexchange.com/questions/12906/… Commented Oct 11, 2020 at 21:06
• But the radius of the rings is $R \sin \theta$ and width of the arc is $Rd\theta$. Commented Oct 11, 2020 at 21:19
• In spherical coordinates the area element is $$\mathrm{d}A=r^2\sin\phi~\mathrm{d}\theta\mathrm{d}\phi$$ I can derive this if you wish. So we need to integrate $$\text{S.A}=\int_\limits{\partial B(0,R)}\mathrm{d}A$$ $$\text{S.A}=\int_0^\pi\int_0^{2\pi}R^2\sin\phi~\mathrm{d}\theta\mathrm{d}\phi$$ $$=R^2\int_0^\pi2\pi\sin\phi~\mathrm{d}\phi$$ $$=2\pi R^2(-\cos\pi-(-\cos0))$$ $$=4\pi r^2.$$ Commented Oct 11, 2020 at 22:00
• Nevertheless we have $2 \int_0^r 2 \pi r \, du=4 \pi r^2$ Commented Oct 14, 2020 at 13:39
• @Raffaele We would get that if anyone was trying to evaluate that integral. We actually have $2 \int_0^r 2 \pi u \, du$, if you want to distinguish the variable inside the integral from the variable at the upper bound. Commented Oct 14, 2020 at 17:12

You might compare your approach to the surface of a sphere with the approach taken in Why is area of a surface of revolution integral $$2\pi y~ds$$? not '$$dx$$'? or Areas versus volumes of revolution: why does the area require approximation by a cone?
The problem is that either the curved surface of the disk or the exposed flat area is less than the surface area of a frustum that (on the upper half of the sphere) connects the upper edge of one disk to the upper edge of the disk below. If you were measuring the area of a cone, that frustum would be the exact area; when measuring a sphere, a "slice" of the sphere looks a lot more like a conical frustum than like either the side of a cylinder or an annular ring, and its area is relatively much closer to that of the frustum than to the areas of either of the other two objects (where "relatively close" means the ratio between the two areas is close to $$1$$).
In this answer to one of the other questions I detailed how the calculations differ when computing the area of a cone, where it is easy to see how this works (because we have other ways of computing the area of a cone without calculus, and because the relative error is the same no matter what part of the cone you look at). For a sphere, the argument is a bit more subtle, and the error in your calculation varies over the surface of the sphere: at the "pole" of the sphere (near the axis) your approximation is quite good, and could be used to find the area of a small spherical cap with pretty good accuracy; but near the "equator" of the sphere (where the radius of your ring is nearly $$r$$) you are missing almost all the area of each slice of the sphere no matter how thin you slice it.