Why do disks on planes grow more quickly with radius than disks on spheres?

Question

In the book, Mr. Tompkins in Wonderland, there is written something like this:

On a sphere the area within a given radius grows more slowly with the radius than on a plane.

Could you explain this to me? I think that formulas shows something totally different:

The area of a sphere is $4 \pi r^2$, but the area of a disk on the plane is $\pi r^2$.

UPDATE:
Later in this book there is:

If, for example, you are on the north pole, the circle with the radius equal to a half meridian is the equator, and the area included is the northern hemisphere. Increase the radius twice and you will get in all the earth's surface; the area will increase only twice instead of four times if it were on a plane.

Could someone explain me this?

Answer 1

The claim is not that the (surface) area, $4 \pi r^2$, of a sphere of radius $r$ grows more slowly than the area, $\pi r^2$, of a disk of radius $r$ on the plane. Rather, the book is asserting that the area of a disk on a sphere grows more slowly with radius than does a disk on the plane.

If we pick a point $P$ on a sphere of radius $R$, we can ask which points on $R$ are a fixed distance $r$ from $P$ along the surface of the sphere---(for $0 < r < \pi R$) by symmetry these points comprise a circle. The area the book means is the region inside this circle (i.e., in the region bounded by the circle and containing $P$).

Now (for $0 \leq r \leq \pi R$), the central angle between $P$ and a point a distance $r$ from $P$ along the sphere is just $\frac{r}{R}$. Thus, if we declare $P$ to be the north pole of the sphere and use the usual colatitude-longitude coordinates on the sphere, we find that the area $A_{\text{sphere}}(r)$ of the region $D$ of points with a distance $r$ from $P$ is $$A_{\text{sphere}}(r) = \iint_D dA = \int_0^{2 \pi} \int_0^{\frac{r}{R}} \sin \phi \,d\phi \,d\theta = 2 \pi R^2 \left(1 - \cos \frac{r}{R}\right).$$

Now, the rate of growth (w.r.t. radius $r$) of the area $A_{\text{plane}}(r)$ of a disk on the plane is $$\frac{d}{dr} A_{\text{plane}} (r) = 2 \pi r,$$ and the rate of growth of the region $D$ with respect to the radius $r$ measured along the surface of the sphere is $$\frac{d}{dr} A_{\text{sphere}} (r) = 2 \pi R \sin \frac{r}{R}.$$ Since $0 < \sin u < u$, for $u > 0$, we have for $0 < r < \pi R$ that $$\frac{d}{dr} A_{\text{sphere}} (r) = 2 \pi R \sin \frac{r}{R} < 2 \pi R \left(\frac{r}{R}\right) = 2 \pi r = \frac{d}{dr} A_{\text{plane}} (r)$$ as claimed.

Note that no point is further than $\pi R$ from $P$ (and only the point $-P$ antipodal to $P$ is exactly that distance away). So, the claim is trivially true for $r > \pi R$: By that point, the sphere is already covered but the disk in the plane keeps growing.

Remark We can extract a little more information by inspecting the Taylor series of the ratio $$\rho(r) := \frac{A_{\text{sphere}}(r)}{A_{\text{plane}}(r)}$$ around $r = 0$. Expanding to third order gives $$\rho(r) = 1 - \frac{1}{12 R^2} r^2 + O(r^4).$$ Now, we can read off the Gaussian curvature of the sphere at the point $P$ by inspecting the quadratic term; importantly, we can do this for any surface, so comparing the rates of growth of disks on general surfaces with those on the plane is important insofar as it yields a reasonably concrete interpretation of curvature.

Answer 2

Theorem: Let $0 < r \leq 2R$ be real numbers, $S_{1}$ a Euclidean sphere of radius $R$, $S_{2}$ a Euclidean sphere of radius $r$ centered at a point $O$ of $S_{1}$, and $D_{r}$ the portion of $S_{1}$ inside $S_{2}$.

The area of $D_{r}$ is $\pi r^{2}$.

---

Now let $D'_{r}$ be the disk of radius $r$ in $S_{1}$, i.e., the portion of $S_{1}$ at distance at most $r$ from $O$, measured along the surface of $S_{1}$. Since a chord of length $r$ subtends an arc of length strictly greater than $r$, the disk $D_{r}'$ is strictly smaller than $D_{r}$, and therefore has smaller area.

---

Proof of Theorem: The region $D_{r}$ (bold arc) is a zone on a sphere of radius $R$. To calculate its width, consider a longitudinal section:

Let $\theta$ be the angle subtended at the center of $S_{1}$ by a chord of length $r$. The indicated right triangles share an interior angle, and consequently are similar, with $\frac{\theta}{2}$ as interior angle. Reading from the right triangle whose hypotenuse is horizontal, $$ R\sin\tfrac{\theta}{2} = \tfrac{r}{2}. $$ On the other hand, reading from the right triangle with the dashed vertical side shows that $D_{r}$ has width $h = r\sin\frac{\theta}{2}$. By Archimedes' theorem, the area of $D_{r}$ is $$ 2\pi Rh = 2\pi R \cdot r\sin\tfrac{\theta}{2} = 2\pi r \cdot R\sin\tfrac{\theta}{2} = \pi r^{2}. $$

Answer 3

This is effect of double curvature. When Gauss curvature $K$ is negative (hyperbolic geometry) it grows faster than in a plane ($K=0$) or on a sphere ($K >0 $).

IIRC (it is in the Geometry & Imagination book by Hilbert/Cohn Vossen?) the words elliptic, hyperbolic derive their name due to the way area develops in relation to the plane.)

It becomes clearer by Gauss-Bonnet thm considering integral curvature $ \int K dA $.

However in simpler terms consider an area A spanned by constant K:

If $ A\cdot K $ is constant then $ dA/ A = - dK/K $ so hyperbolic case grows faster.

If you google for sea coral shapes with Daina Taimina's Crochet work, you appreciate that hyperbolic patches grow faster than flat ones or spheres in that order, the plane is more hyperbolic than sphere in that sense. I had to mention hyperbolic as I believe it is relevant for an answer here.

Answer 4

If one interprets the area of a sphere as its surface area, then the formula would be $4\pi r^2$. The area of a circle is $\pi r^2$. Therefore the surface area of a sphere grows 4 units when the area of a circle grows 1 unit, which is not slower. I'm sure I'm misinterpreting this, though.