This looks like a homework problem because it is. I'm stuck at the portion where I solve for fractional volumes.
Suppose you are a part of a team designing a water tank in the shape of a hyperboloid. The tank is to have a top radius $a$ of $2$ meters, a bottom radius $b$ of $4$ meters, and a height $h$ of $4$ meters. In addition, the design requires the placement of sensors at $1/8$, $1/4$, $3/8$, $1/2$, $5/8$, $3/4$ and $7/8$ of the tank’s capacity. Using a double integral in polar coordinates, what are the appropriate heights at which the sensors should be placed. The side is given by \begin{equation} \frac{r^2}{a^2} - \frac{b^2 - a^2}{a^2h^2}(h - z)^2 = 1 \end{equation} or \begin{equation} z = h\left(1 - \sqrt{\frac{r^2 - a^2}{b^2 - a^2}}\right). \end{equation}
First off, your formula as it currently stands seems to be incorrect. I guess the formula you want is
$$\frac{r^2}{a^2}-\frac{\left(b^2-a^2\right)\left(h-z\right)^2}{a^2h^2} = 1$$
since otherwise the units resp. dimensions don't work out.
Consider the tank made up of disks, or cylinders of infinitesimal height. Each such disk has volume $\pi r^2\,\mathrm dz$. To integrate over these, it would be useful to have $r^2$ in terms of $z$, instead of the other way round.
$$r^2 = a^2+\frac{\left(b^2-a^2\right)\left(h-z\right)^2}{h^2}$$
Now you can integrate over these:
$$V(z) = \int_0^z \left(a^2+\frac{\left(b^2-a^2\right)\left(h-t\right)^2}{h^2}\right) \,\mathrm dt = \tfrac14z\left(z^2-12\,z+64\right)$$
The total volume of your tank is $V(4)=32\,\mathrm m^2$. So you want to solve this for $z$ for the appropriate fractions of that volume. You get these approximate results:
$$\begin{array}{rl} V(z) & \;z \approx \\\hline 4 & 0.26265 \\ 8 & 0.55510 \\ 12 & 0.88645 \\ 16 & 1.2707 \\ 20 & 1.7305 \\ 24 & 2.3046 \\ 28 & 3.0531 \\ \end{array}$$
I performed the computation of the integral as well as the numeric results using Wolfram Alpha. Solving cubic equations manually is no fun.