Oceans and volume What would be the math formulas to calculate the increase in the height of the ocean if one drop of water were released into it.  Assuming that everything is static.  How would you solve that question. What would be the numbers?
 A: It's the volume of the drop divided by the surface area of the ocean.
A: To add some context to Michael Hardy's answer:
Volume is a quantity proportional to the product of three length measurements:
$$V = k_V L_1 L_2 L_3.$$
Area is a quantity proportional to the product of two length measurements, which need not be the same measurements we use to measure volume:
$$ A = k_A \hat{L_1} \hat{L_2}.$$
Depth is just a length measurement.
Dividing volume by area yields something that is proportional to a length measurement:
$$ \frac{V}{A} = \frac{k_V L_1 L_2 L_3}{k_A \hat{L_1} \hat{L_2}} = kL_1 = D.$$
So the change in depth is some constant multiplied by the ratio of drop volume to ocean surface area. A rough estimate is to assume the drop is a sphere, and the ocean is circular. This is accurate up to a constant factor, but the estimate is going to be completely dominated by the orders-of-magnitude difference between the drop's volume and the ocean's area.
In other words, if you mis-estimate the surface area of the ocean by a factor of 2, it doesn't matter, because the droplet is so small, comparatively.
A: The volume of the ocean can be approximated by $V(h) = \int_{r_0}^{r_0+h} A(r) dr$, where $h$ is the height of the ocean (above the Earth's surface), and $A(r)$ is the surface area of the ocean if it was at height $r$ (measured from the Earth's center).
We want to estimate $\delta$, where $V(h_0+\delta)-V(h_0) = V_{\text{drop}}$, and $h_0$ is the initial height of the ocean. We approximate $V(h_0+\delta)-V(h_0) \approx V'(h_0) \delta$, which gives $\delta = \frac{V_{\text{drop}}}{V'(h_0)}$.
From the above formula, we have $V'(h_0) = A(r_0+h_0)$, and since $h_0$ is small compared to $r_0$, we can approximate $V'(h_0) \approx A(r_0)$.
The area can be approximated by $A(r) = \alpha 4 \pi r^2$, where $\alpha$ is the fraction of the Earth's surface covered by the oceans, so we end up with 
$$\delta \approx \frac{V_{\text{drop}}}{\alpha 4 \pi r_0^2}$$
Plugging in some numbers, we have $\alpha \approx 0.7$, $r_0 \approx 4,000 \text{ miles}$, and taking a drop to have radius $\frac{1}{8}''$, we have $V_{\text{drop}} \approx 1.6 \times 10^{-11}\text{ miles}$, which gives $\delta \approx 10^{-19}\text{ miles}$ or $(7 \times 10^{-15}) {''}$. (We have $h_0 \approx 2.5 \text{ miles}$, so the area approximation is reasonable.)
