I don't know if this is going to be what you're asking for, but it's what I thought it could have been so.. take it or leave :)
It's about curvature scalar and Ricci tensor, then. We will find them for the $2-D$ surface of a sphere, that is your tennis ball (a good sphere). The Ricci tensor is calculated from the Riemann tensor, and that in turn depends on the Christoffel symbols, so we need them first. It’s easiest to find them from the geodesic equation. I hope you know a bit of General Relativity or at least Differential Geometry.
$$\displaystyle g_{aj}\ddot{x}^{\ j}+\left(\partial_{i}g_{aj}-\frac{1}{2}\partial_{a}g_{ij}\right)\dot{x}^{\ j}\dot{x}^{\ i}=0 \ \ \ \ \ (1)$$
which is formally equivalent to
$$ \displaystyle \ddot{x}^{\ m}+\Gamma_{\; ij}^{m}\dot{x}^{\ j}\dot{x}^{\ i}=0 \ \ \ \ \ (2)$$
The metric for a sphere is
$$\displaystyle ds^{2}=r^{2}d\theta^{2}+r^{2}\sin^{2}\theta d\phi^{2} \ \ \ \ \ (3)$$
where ${r}$ is the constant radius of the sphere, so
$$\displaystyle g_{\theta\theta} \displaystyle = \displaystyle r^{2}\ \ \ \ \ (4)$$
$$\displaystyle g_{\phi\phi} \displaystyle = \displaystyle r^{2}\sin^{2}\theta \ \ \ \ \ (5)$$
We have two equations arising from $1$. For ${a=\theta}$
$$\displaystyle r^{2}\ddot{\theta}-r^{2}\sin\theta\cos\theta\dot{\phi}^{2}=0 \ \ \ \ \ (6)$$
Comparing with $2$ we get, after dividing out the ${r^{2}}$:
$$\displaystyle \Gamma_{\;\phi\phi}^{\theta}=-\sin\theta\cos\theta=-\frac{1}{2}\sin2\theta \ \ \ \ \ (7)$$
For ${a=\phi}$:
$$\displaystyle r^{2}\sin^{2}\theta\ddot{\phi}^{2}+2r^{2}\sin\theta\cos\theta\dot{\theta}\dot{\phi}=0 \ \ \ \ \ (8)$$
Dividing through by ${r^{2}\sin^{2}\theta}$ and comparing with $2$ we get (remember that the second term is ${\left(\Gamma_{\;\theta\phi}^{\phi}+\Gamma_{\;\phi\theta}^{\phi}\right)\dot{\theta}\dot{\phi}}$ and that ${\Gamma_{\;\theta\phi}^{\phi}=\Gamma_{\;\phi\theta}^{\phi}})$:
$$\displaystyle \Gamma_{\;\theta\phi}^{\phi}=\Gamma_{\;\phi\theta}^{\phi}=\cot\theta \ \ \ \ \ (9)$$
All other Christoffel symbols are zero.
In $2D$, the Riemann tensor has only one independent component, which we can take to be ${R_{\theta\phi\theta\phi}}$, which can be calculated from
$$\displaystyle R_{\; j\ell m}^{i}\equiv\partial_{\ell}\Gamma_{\; mj}^{i}-\partial_{m}\Gamma_{\;\ell j}^{i}+\Gamma_{\; mj}^{k}\Gamma_{\;\ell k}^{i}-\Gamma_{\;\ell j}^{k}\Gamma_{\; km}^{i} \ \ \ \ \ (10)$$
Lowering the first index, we have
$$\displaystyle R_{\theta\phi\theta\phi} \displaystyle = \displaystyle g_{\theta i}R_{\;\phi\theta\phi}^{i}\ \ \ \ \ (11)$$
$$\displaystyle \displaystyle = \displaystyle g_{\theta\theta}R_{\;\phi\theta\phi}^{\theta}\ \ \ \ \ (12)$$
$$\displaystyle \displaystyle = \displaystyle r^{2}\left(\partial_{\theta}\Gamma_{\;\phi\phi}^{\theta}-\partial_{\phi}\Gamma_{\;\theta\phi}^{\theta}+\Gamma_{\;\phi\phi}^{k}\Gamma_{\;\theta k}^{\theta}-\Gamma_{\;\theta\phi}^{k}\Gamma_{\; k\phi}^{\theta}\right)\ \ \ \ \ (13)$$
$$\displaystyle \displaystyle = \displaystyle r^{2}\left(-\cos2\theta-0+0-\Gamma_{\;\theta\phi}^{\phi}\Gamma_{\;\phi\phi}^{\theta}\right)\ \ \ \ \ (14)$$
$$\displaystyle \displaystyle = \displaystyle r^{2}\left(\sin^{2}\theta-\cos^{2}\theta+\cos^{2}\theta\right)\ \ \ \ \ (15)$$
$$\displaystyle \displaystyle = \displaystyle r^{2}\sin^{2}\theta \ \ \ \ \ (16)$$
We can now find the Ricci tensor.
$$\displaystyle R_{ij}=g^{ab}R_{aibj} \ \ \ \ \ (17)$$
Since the metric is diagonal and ${g^{ab}}$ is the inverse of ${g_{ab}}$, we have
$$\displaystyle g^{\theta\theta} \displaystyle = \displaystyle \frac{1}{g^{\theta\theta}}=\frac{1}{r^{2}}\ \ \ \ \ (18)$$
$$\displaystyle g^{\phi\phi} \displaystyle = \displaystyle \frac{1}{g^{\phi\phi}}=\frac{1}{r^{2}\sin^{2}\theta} \ \ \ \ \ (19)$$
so, using the symmetries of the Riemann tensor,
$$\displaystyle R_{\theta\theta} \displaystyle = \displaystyle g^{ab}R_{a\theta b\theta}\ \ \ \ \ (20)$$
$$\displaystyle \displaystyle = \displaystyle g^{\phi\phi}R_{\phi\theta\phi\theta}\ \ \ \ \ (21)$$
$$\displaystyle \displaystyle = \displaystyle \frac{1}{r^{2}\sin^{2}\theta}r^{2}\sin^{2}\theta\ \ \ \ \ (22)$$
$$\displaystyle \displaystyle = \displaystyle 1\ \ \ \ \ (23)$$
$$\displaystyle R_{\phi\phi} \displaystyle = \displaystyle g^{ab}R_{a\phi b\phi}\ \ \ \ \ (24)$$
$$\displaystyle \displaystyle = \displaystyle g^{\theta\theta}R_{\theta\phi\theta\phi}\ \ \ \ \ (25)$$
$$\displaystyle \displaystyle = \displaystyle \sin^{2}\theta\ \ \ \ \ (26)$$
$$\displaystyle R_{\theta\phi} \displaystyle = \displaystyle R_{\phi\theta}=0 \ \ \ \ \ (27)$$
We can get the upstairs version of the Ricci tensor as well:
$$\displaystyle R^{\theta\theta} \displaystyle = \displaystyle g^{\theta i}g^{\theta j}R_{ij}\ \ \ \ \ (28)$$
$$\displaystyle \displaystyle = \displaystyle g^{\theta\theta}g^{\theta\theta}R_{\theta\theta}\ \ \ \ \ (29)$$
$$\displaystyle \displaystyle = \displaystyle \frac{1}{r^{4}}\ \ \ \ \ (30)$$
$$\displaystyle R^{\phi\phi} \displaystyle = \displaystyle g^{\phi\phi}g^{\phi\phi}R_{\phi\phi}\ \ \ \ \ (31)$$
$$\displaystyle \displaystyle = \displaystyle \frac{1}{r^{4}\sin^{2}\theta} \ \ \ \ \ (32)$$
The curvature scalar is
$$\displaystyle R \displaystyle = \displaystyle g_{ij}R^{ij}\ \ \ \ \ (33)$$
$$\displaystyle \displaystyle = \displaystyle r^{2}\frac{1}{r^{4}}+r^{2}\sin^{2}\theta\frac{1}{r^{4}\sin^{2}\theta}\ \ \ \ \ (34)$$
$$\displaystyle \displaystyle = \displaystyle \frac{2}{r^{2}} \ \ \ \ \ (35)$$
As we would expect, the curvature of a sphere decreases as its radius gets larger.
