Volume of the solid bounded by the sphere $x^2 + y^2 + z^2 = 9$ and paraboloid $8z = x^2 + y^2$ 
Find the volume of the solid bounded above by the sphere $x^2 + y^2 + z^2 = 9$ and below by the paraboloid $8z = x^2 + y^2$

I'm having some trouble finding the correct limits of integration in cylindrical coordinates. 
I appreciate some help. I'm prepping for my final and this is a question on an old final exam.
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\begin{align}\color{#66f}{\large V}&
=\left.\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}
\dd x\,\dd y\,\dd z\right\vert_{\large{x^{2}\ +\ y^{2}\ +\ z^{2}\ <\ 9\atop
z\ >\ \pars{x^{2}\ +\ y^{2}}/8}}
=\left.\int_{0}^{2\pi}\int_{-\infty}^{\infty}\int_{0}^{\infty}r\,\dd r\,\dd z\,\dd\phi\right\vert
_{{\large r^{2}\ +\ z^{2}\ <\ 9\atop z\ >\ r^{2}/8}}
\\[5mm]&=2\pi\left.\int_{0}^{\infty}\int_{0}^{\infty}r\,\dd r\,\dd z\right\vert
_{{\large r\ <\ \root{9 - z^{2}}\atop {r\ <\ \sqrt{8z}\atop 0\ <\ z\ <\ 3}}}
=\color{#66f}{\large%
2\pi\int_{0}^{3}\int_{0}^{\min\pars{\root{9 - z^{2}},\root{8z}}}r\,\dd r\,\dd z}
\end{align}

However,
  $$
\root{9 - z^{2}}<\root{8z}\ \imp\
\pars{z - 1}\pars{z + 9}>0\ \imp\ \pars{~z<-9\ \mbox{or}\ z>1~}
$$

\begin{align}
\color{#66f}{\large V}&=2\pi\int_{0}^{1}\int_{0}^{\root{8z}}r\,\dd r\,\dd z
+\int_{1}^{3}\int_{0}^{\root{9 - z^{2}}}r\,\dd r\,\dd z
\\[5mm]&=2\pi\bracks{\int_{0}^{1}\half\,\pars{8z}\,\dd z
+\int_{1}^{3}\half\,\pars{9 - z^{2}}\,\dd z}
=2\pi\braces{\left. 2z^{2}\right\vert_{0}^{1}
+\left. {9 \over 2}\,z - {1 \over 6}\,z^{3}\right\vert_{1}^{3}}
\\[5mm]&=2\pi\bracks{2 + {27 \over 2} - {27 \over 6} - {9 \over 2} + {1 \over 6}}
=\color{#66f}{\large{40 \over 3}\,\pi}
\end{align}
A: Sphere:
$$r^2+z^2=9\qquad z=\sqrt{9-r^2}.$$
Paraboloid:
$$r^2=8z,\qquad z=\frac{r^2}8.$$
Intersection:
$$\sqrt{9-r^2}=z=\frac{r^2}8,$$
$$9-r^2=\frac{r^4}{64},$$
Can you continue?
A: Integrate from $z=r^2/8$ to $z = \sqrt{(9-r^2)}$(positive because the volume lies above the z axis)
Determine r as the radius for which the two objects intersect and integrate from 0 to the radius of intersection and from 0 to $2\pi$.
I had trouble doing these myself but the i came across this shadow method. In your head you imagine a light source on the object above the z axis. To find the limits of integration you just integrate z between the two given surfaces, that is write z as a function of x and y or r and theta. Finally you integrate over the shadow that object casts on the planar axis (in this case the circle with radius = r of intersection) and $2\pi$.
Sorry for my bad English and my inability to write the equation properly, hope this helps.
