Why do disks on planes grow more quickly with radius than disks on spheres? 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?
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 $0 &lt; r &lt; \pi R$) 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 $0 &lt; \sin u &lt; u$, for $u &gt; 0$, we have for $0 &lt; r &lt; \pi R$ that
$$\frac{d}{dr} A_{\text{sphere}} (r) = 2 \pi R \sin \frac{r}{R} &lt; 2 \pi R \left(\frac{r}{R}\right) = 2 \pi r = \frac{d}{dr} A_{\text{plane}} (r)$$
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 $r &gt; \pi R$: 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.
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.