Find the volume of the vase I am told that the shape of a vase can be modelled by rotating the curve with equation $16x^2-(y-8)^2=32$ between $y=0$ and $y=16$ completely about the $y$-axis. I'm asked to find the volume of the vase. 
I would like to solve this using triple integrals. I can see that I need to compute something of the form $$\iiint_D r\,dr\,d\theta \,dz$$
I'm not sure where I need to go from here though. 
I appreciate that this may not be the simplest way to do this but this is the way I've been asked to solve the problem.
 A: $r = x\\
r^2 - (y-8)^2 = 32\\
r = \sqrt {\frac{32 + (y-8)^2}{16}}$
$\int_0^{2\pi}\int_0^{16}\int_0^{\sqrt{\frac{32 + (y-8)^2}{16}}}r\, dr\, dy\, dt$
Before we integrate lets make the substitution $y' = y-8$
$\int_0^{2\pi}\int_{-8}^{8}\int_0^{\sqrt{\frac{32 + y'^2}{16}}} r\,dr \,dy'\, dt\\
\int_0^{2\pi}\int_{-8}^{8}\frac 12 r^2 |_0^{\sqrt{\frac{32 + y'^2}{16}}} \,dy'\, dt\\
\int_0^{2\pi}\int_{-8}^{8}\frac 12 (\frac{32 + y'^2}{16}) \,dy'\, dt\\
\int_0^{2\pi} y' + \frac 1{96} y'^3 |_{-8}^{8} \, dt\\
\int_0^{2\pi} \frac {80}{3}\, dt\\
\frac {160\pi}{3}$
A: I am studying it as well so I won't presume to be official and ultimately correct. I would note that for every $0 \le y\le 16$, the equation gives you two symmetric $x$ values, $x=\pm {\sqrt{(y-8)^2+32}\over 4}$. Rotating the curve about the $y$-axis, you get that for every $0 \le y_0\le 16$, there is a flat area about the $y$-axis confined by the circle $x^2+z^2=x_0^2=({\sqrt{(y_0-8)^2+32}\over 4})^2={{(y_0-8)^2+32}\over 16}$. 
Now, set $y=h$ (the height of the figure). $x=r\cos \theta, z=r\sin \theta$, where $r\le{\sqrt{(h-8)^2+32}\over 4}$. i.e. Cylindrical coordinates. I will add my calculations as soon as I am done...
I get a volume of ${160\pi\over 3} $ which is also arrived at using wolframalpha.
I then let $0\le \theta\le 2\pi$, and get: 
$\int_{0}^{16}\int_{0}^{2\pi}\int_{0}^{{\sqrt{(h-8)^2+32}\over 4}}rdrd\theta dh=2\pi\int_{0}^{16}{1\over 2}{(h-8)^2+32 \over 16}dh={\pi\over 16}\int_{0}^{16}((h-8)^2+32)dh={\pi\over 16}\int_{0}^{16}(h^2-16h+96)dh$.
