We are given $y^2/a^2-x^2/b^2=1$, $y>0$ . If we rotate the hyperbola around the $y$ axis the shape is similar to a glass.
What will the volume of water inside the glass be, in order to fill the glass up to $y=A$, $A>a$?
Following the given hint I found $$x=b*(y^2/a^2-1)^{1/2}.$$
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So in your drawing you show that you are almost near to find the good start.
Take half of the trapezoidal stripe you indicated, going from the y axis to the hyperbola.
Its width is $dy$, one base is $x(y)$ and the other $x(y+dy)=x(y)+x'(y)dy$.
When you rotate it by $2\pi$ you get a conic frustum of height $dy$ and volume $\pi x(y)^2 dy$ because you know that higher order infinitesimal can be waived.
Then just integrate the above along $y$, between the bounds you need.
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I do the problem in cylindrical coordinates as follows: Converting the equation of hyperbola to hyperboloid around $z$ can be simply done by replacing $y\to z$ and $x\to r$ as: $$\frac{z^2}{a^2}-\frac{r^2}{b^2}=1$$ (I figured this by reconciling $r$ with $x$ in the $x-z$ plane)
Now the integral becomes terribly simple: $$\int_0^{2\pi}\int_a^A\int_0^{\sqrt{b^2(\frac{z^2}{a^2}-1)}} rdrdzd\theta=\pi b^2\int_a^A\Bigg(\frac{z^2}{a^2}-1\Bigg)dz=\pi b^2\Bigg(\frac{(A^3-a^3)}{3a^2}-(A-a)\Bigg)$$
The green axis in your picture is the $y$ axis. You can calculate the volume in two ways: disks or shells.
The disks are perpendicular to the $y$ axis, have a thickness $dy$ and the radius is equal to $x$ at a given $y$. In this case your integration variable is $y$. So the volume of the disk is $\pi x^2 dy=b^2(y^2/a^2-1) dy$. You know the minimum $y$ is $a$, and the maximum is $A$.
The shell method means that you need to have cylindrical shells, with the axis parallel to the $y$ axis. The radius of the cylinder is $x$, the thickness is $dx$. The volume of the shell is then $2\pi x h(x)\mathrm dx$, where $h(x)$ is the height of the shell. The "bottom" of the shell starts at $y=a\sqrt{x^2/b^2+1}$, and the "top" is at $y=A$, so $h(x)=A-a\sqrt{x^2/b^2+1}$ The integration range starts from $0$ and goes on until the height is $0$, or $y(x)=A$. $$\frac{A^2}{a^2}-\frac{x_{max}^2}{b^2}=1$$ You can now get the upper limit of $x$ from the above equation.