Answer
$ V=\frac{4}{3} \pi \left[ h^{2}+2ah \right] ^{\frac{3}{2}} $
Work Step by Step
{Step 1 of 5}:
First, find the points of intersection of the curves $ x^{2}-y^{2}=a^{2} $
or $ x=\sqrt[]{a^{2}+y^{2}} $ and $ x=a+h $ (where a>0, h>0)
$ \sqrt[]{a^{2}+y^{2}}= \left( a+h \right) $
$ a^{2}+y^{2}= \left( a+h \right) ^{2} $
$ y^{2}= \left( a+h \right) ^{2}-a^{2} $
$ y^{2}=a^{2}+h^{2}+2ah-a^{2} $
$ y^{2}=h^{2}+2ah $
$ y^{2}=h \left( h+2a \right) $
$ y= \pm \sqrt[]{h \left( h+2a \right) } $
{Step 2 of 5}:
As an example, take $ a=1 $ and $ h=1 $ and sketch the curve
{Step 3 of 5}:
Rotate this shaded region about the y-axis. Rotate this to produce a solid. If we cut a Horizontal slice from this solid, we get a washer.
The outer radius of the washer is $ a+h $
The inner radius of the washer is $ \sqrt[]{a^{2}-y^{2}} $
The cross-sectional area of the washer
$ A \left( y \right) = \pi \left[ \left( a+h \right) ^{2}- \left( \sqrt[]{a^{2}-y^{2}} \right) ^{2} \right] $
$ = \pi \left[ a^{2}+h^{2}+2ah-a^{2}-y^{2} \right] $
$ = \pi \left[ \left( h^{2}+2ah \right) -y^{2} \right] $
{Step 4 of 5}:
The volume of the solid
$ V= \int _{-\sqrt[]{k \left( k+2a \right) }}^{+\sqrt[]{k \left( k+2a \right) }}A \left( y \right) dy $
Let $ h \left( h+2a \right) =h^{2}+2ah=k $
{Step 5 of 5}:
Then $ V= \pi \int _{-\sqrt[]{k}}^{\sqrt[]{k}} \left( k-y^{2} \right) dy $
$ = \pi \left[ ky-\frac{y^{3}}{3} \right] _{-\sqrt[]{k}}^{\sqrt[]{k}} $
[By the fundamental theorem of calculus, part 2]
$ = \pi \left[ \left( k\sqrt[]{k}-\frac{ \left( \sqrt[]{k} \right) ^{3}}{3} \right) - \left( -k\sqrt[]{k}+\frac{ \left( \sqrt[]{k} \right) ^{3}}{3} \right) \right] $
$ = \pi \left[ 2k\sqrt[]{k}-\frac{2}{3} \left( \sqrt[]{k} \right) ^{3} \right] $
$ = \pi \left[ 2 \left( \sqrt[]{k} \right) ^{3}-\frac{2}{3} \left( \sqrt[]{k} \right) ^{3} \right] $
$ = \pi \left[ \frac{4}{3} \left( \sqrt[]{k} \right) ^{3} \right] $
$ = \pi \left[ \frac{4}{3}k^{\frac{3}{2}} \right] $
$ V=\frac{4}{3} \pi \left[ h^{2}+2ah \right] ^{\frac{3}{2}} $
