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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}
I will only give an answer for the Cartesian method, to help you on your way.
For both Cartesian and Cylindrical Polar methods, you need to integrate between $z = 0$ to $z = 6$ the cross section $x^2 + y^2 = 1, y≥\frac{1}{2}$
In Cartesian, we clearly have $\frac{1}{2} ≤ y ≤1$ and then $-\sqrt{1-y^2}≤x≤\sqrt{1-y^2}$ from our equation of the circle. So our integral is thus: $$\int_{z=0}^6\int_{y=\frac{1}{2}}^1\int_{x=-\sqrt{1-y^2}}^{x=\sqrt{1-y^2}}\mathrm{d}x\mathrm{d}y\mathrm{d}z$$ I recommend making the substitution $y=\sin\theta$ to solve the integrand with the square root.
The Cylindrical Polar method is very similar: just remember your Jacobian.
Best Answer
The projection of the solid on the $xy$ plane is the region enclosed by the parabolas $y = x^2$ and $x = y^2$. These intersect at $(0,0)$ and $(1,1)$ and form the shape of a leaf. Then:
$0 \le x \le 1$ and $x^2 \le y \le \sqrt{x}$.
$z$ moves from $0$ towards the plane $x + y + 21 = z$. Therefore:
$$V = \int_0^1 \int_{x^2}^{\sqrt{x}} \int_0^{x + y + 21} dz dy dx$$