Volume of a Solid, $x^2 - y^2 = a^2$ The question is

Find the volume of a solid rotated around the y axis, bounded by the given curves: $$x^2 - y^2 = a^2$$
$$x = a + h$$

I am lost by the number of variables in this question and the question does not tell me what kind of variables they are, only that they are both greater than 0.
 A: Begin by writing both functions as functions of $y$:
\begin{align}
f(y)&=\sqrt{a^2-y^2} \\
g(y)&=a+h.
\end{align}
In the upper plane, the curves intersect in the point $(a+h,\sqrt{2ah+h^2})$, so $y$ ranges from $0$ to $\sqrt{2ah+h^2}$. Using the formula of the volume of a solid of revolution, we see that
$$
V(h) = \pi\int_{-\sqrt{2ah+h^2}}^{\sqrt{2ah+h^2}} \left[g(y)^2-f(y)^2\right]dy.
$$
Can you finish from here?
Edit: I mistakenly wrote the difference squared instead of the difference of the squares. Can you see the difference?
A: This is known as a hyperboloid of revolution of one sheet. 
Forget the second equation, it only means that to get both x and y real, x should be equal or grater than a by a small amount h. $y > x$ always in the first quadrant for a quarter branch of this (rectangular) hyperbola.The limits of y are $0$ and $y_1$ in:
$$ x^2 = y^2 +1 $$
$$ V = \pi \int x^2 dy = \pi \int_0^{y1} (1+y^2) dy,   $$ where I  have chosen $y $ as the independent variable. And you can take it from there.
A: 
The region that is to be revolved about the $ \ y-$axis [marked in gold in the graph above] is bounded by the "vertical" hyperbola $ \ x^2 - y^2 \ = \ a^2 \ $ and the vertical line $ \ x \ = \ a + h \ \ , \ $ with $ \ a \ $ and $ \ h \ $ being unspecified constants.  What is asked for in the problem then is a formula for the volume of the solid of revolution expressed in terms of those constants.
The solid "generated" will be within a cylinder of radius $ \ a + h \ $ and a height $ \ \mathcal{H} \ $ which we have yet to determine from the intersections of the bounding curves and outside of the hyperboloid of one sheet identified by Narasimham. These intersections are found from
$$ x^2 \ - \  y^2 \ \ = \ \  a^2 \ \ \rightarrow \ \ y^2 \ \ = \ \ x^2 \ - \ a^2 $$ $$ \Rightarrow \ \ \mathcal{H}^2 \ \ = \ \ (a + h)^2 \ - \ a^2 \ \ = \ \ 2ah \ + \ h^2 \ \ ,   $$
so the top and bottom of the solid are found at $ \ y \ = \ \pm \mathcal{H} \ = \ \pm \sqrt{2ah \ + \ h^2} \ \ , \ $ respectively.
This suggests one relatively quick method to obtain the volume (which might be called the "Cavalieri approach"):  use a familiar formula to calculate the volume of the cylinder, then calculate the volume "inside" the hyperboloid by integrating the "vertical stack of disks" it is comprised of, and "subtract away" that volume.  The volume of the cylinder is simply $ \ V_{cyl} \ = \ \pi·(a + h)^2 · (\ \mathcal{H} - [\mathcal{-H}] \ ) $ $ = \ \pi·(a + h)^2 · 2·\sqrt{2ah \ + \ h^2} \ \ . $ The disks making up the solid hyperboloid have a "radius-squared" as a function of $ \ y \ $ given by $ \ x^2  \ = \ a^2 + y^2 \ \   $ and "infinitesimal thicknesses" of $ \ dy \ \ . \ $ Because the "stack" is symmetrical about the plane $ \ y \ = \ 0 \ \ , \ $ its volume is given by
$$ V_{hyp} \ \ = \ \ 2 \ \int_0^{\mathcal{H}} \ \pi · x^2 \ \ dy \ \ = \ \ 2 \pi \ \int_0^{\mathcal{H}} \   (a^2 + y^2) \ \ dy \ \ = \ \ 2 \pi \ · \    \left( \ a^2·y \ + \ \frac13 y^3 \ \right)|_0^\mathcal{H}$$
$$ = \ \ 2 \pi \ · \    ( \ a^2·\mathcal{H} \ + \ \frac13 \mathcal{H}^3 \ ) \ \ = \ \ 2 \pi  \mathcal{H} \ · \    ( \ a^2 \ + \ \frac13 \mathcal{H}^2 \ ) \ \ . $$
We can now determine the volume of the specified solid of revolution as
$$ \mathcal{V} \ \ = \ \ V_{cyl} \ - \ V_{hyp} \ \ = \ \ 2 \pi  \mathcal{H} \ · \  \left[ \ (a + h)^2 \ - \  \left( \ a^2 \ + \ \frac13 \mathcal{H}^2 \ \right) \ \right] \ \ $$
$$ = \ \ 2 \pi  \mathcal{H} \ · \  \left[ \ (a^2 + 2ah + h^2)  \ - \  \left( \ a^2 \ + \ \frac13 · (2ah \ + \ h^2) \ \right) \ \right] \ \ $$
$$ = \ \ 2 \pi  \mathcal{H} \ · \  \frac23  ·  (2ah \ + \ h^2)  \ \ = \ \  \frac{4 \pi}{3} · \sqrt{2ah \ + \ h^2} ·  (2ah \ + \ h^2) \ \ = \ \ \frac{4 \pi}{3} · (2ah \ + \ h^2)^{3/2}  \ \ . $$
$ \rule{15 cm}{0.5 pt} $
This problem is also amenable to either of the volume integration methods taught for dealing with solids of revolution.
If we apply the "cylindrical shells" method, the radii of the shells "run" from $ \ x \ = \ a \ $   to    $ \ x \ = \ a + h \ \ , \ $ with "heights" given by $ \ y \ - \ (-y) \ = \ 2·\sqrt{x^2 - a^2} \ \ . $  The volume integral is then
$$ \mathcal{V} \ \ = \ \   \int_a^{a+h} \ 2 \pi \ · \ x \ · \  2·\sqrt{x^2 - a^2} \ \ dx  $$
[making the substitution $ \ u \ = \ x^2 - a^2 \ \Rightarrow \ du \ = \ 2x \ dx \ \ ] $
$$ \rightarrow \ 4 \pi \ \int \ u^{1/2} \ \ \left(\frac12 \ du\right) \ \ = \ \ 2 \pi \ · \ \frac{u^{3/2}}{3/2} \ \ = \ \ \frac{4 \pi}{3} \ · \ u^{3/2} $$
[we'll evaluate the definite integral after "back-substitution"]
$$ \rightarrow \ \frac{4 \pi}{3} \ · \ (x^2 - a^2)^{3/2}|_a^{a+h} \ \ = \ \ \frac{4 \pi}{3} \ · \ [ \ ( \ [a+h]^2 - a^2 \ )^{3/2} \ - \ ( \ a^2 - a^2 \ )^{3/2} \ ] $$
$$ = \ \ \frac{4 \pi}{3} \ · \ [ \ ( \ [a^2 + 2ah + h^2] - a^2 \ )^{3/2} \ - \ 0 \ ] \ \ = \ \ \frac{4 \pi}{3}  ·    (  2ah + h^2  )^{3/2} \ \ . $$
The "disk/washer" method calculation resembles the "Cavalieri approach" we used above, except that the washers lie between the hyperboloid and the cylinder:  the "outer radii" are the cylinder's radius  $ \ a + h \ \ , \ $ and the "inner radii-squared" are $ \ x^2  \ = \ a^2 + y^2 \ \ . \  $  The volume integral for this "stack of washers" is
$$ \mathcal{V} \ \ = \ \ \int_{-\mathcal{H}}^{\mathcal{H}} \ [ \ \pi · (a + h)^2 \ - \ \pi · (a^2 + y^2) \ ] \ \ dy \ \ = \ \ 2 \pi \ \int_0^{\mathcal{H}} \ ( \  a^2 + 2ah + h^2  -  a^2 - y^2 \ ) \ \ dy $$
[again exploiting the symmetry of the "stack" about $ \ y \ = \ 0 \ \ ] $
$$ = \ \ 2 \pi  ·  \left[ \ ( 2ah + h^2 )·y \  - \ \frac13·y^3 \ \right]|_0^{\mathcal{H}} \ \ = \ \  2 \pi  ·  \left[ \ ( 2ah + h^2 )·\mathcal{H} \  - \ \frac13·\mathcal{H}^3 \ \right]   $$
$$ = \ \  2 \pi  ·  \left[ \ ( 2ah + h^2 )·(  2ah + h^2  )^{1/2} \  - \ \frac13·(  2ah + h^2  )^{3/2} \ \right] \ \ = \ \  \frac{4 \pi}{3}·(2ah + h^2  )^{3/2} \ \ . $$
