mass of a body with a density function Let $V$ be the body confined by the following quadratic surfaces:
$$
x^{2} + y^{2} = 12z,\quad x^{2} = 3z,\quad z = 3
$$

*

*Compute the mass of $V$ with the density function $\rho = 1$.

*My problem is I can't figure the integration boundaries$\ldots$

*I know that substituting to spherical or cylindrical coordinates won't help because it says so in the question.

 A: $x^2+y^2 = 12$ describes a paraboloid
$x^2 = 3z$ is a parabolic cylinder
$z = 3$ is a plane.
How do the 3 surfaces intersect?  Here is my picture.

plug $z = 3$ into the other two surfaces to get
$x^2+y^2 = 36, z=3$
$x = \pm 3, z=3$
And $x^2 = 3z$ into the equation of the paraboloid.
$3z+y^2 = 12z$
$y^2 = 9z$
I would use this integral.
$M = \rho V = 4\int_0^3 \int_0^{3\sqrt z} \int_{\sqrt{3z}}^{\sqrt {12z-y^2}} \ dx\ dy\ dz$
or
$4\int_0^3 \int_{\sqrt{3z}}^{\sqrt{12z}} \int_{0}^{\sqrt {12z-x^2}} \ dy\ dx\ dz$
I am getting  $108( \frac {\pi}{3} - \frac{\sqrt{3}}{4})$
In cylindrical coordinates, I have
$4\int_0^3\int_{0}^{\frac {\pi}{3}}\int_{\sqrt{3z}\sec{\theta}}^{\sqrt{12 z}} r\ dr\ d\theta\ dz$
A: lower- and upper-bounds for the mass of the body
$M \in (M_{l},M_{u})=(93.53,108)$.

numerical verification
From numerical calculations, I obtained $M \approx 101.17 \in \left( 100.77, 108 \right)$, the improved lower- and upper-bounds mentioned in comments.
Plot of the surface the region:


calculation
Starting from $x^2 + y^2 = 12z$ and $x^2 = 3z$, we obtain $y^2 \in (9z,12z)$. This simplifies the problem a bit. To visualize the problem, take different values of $z$ and solve for $x$ and $y$,
$$
\begin{align}
& z=1 \implies |x| < \sqrt{3}, 3<|y|<2\sqrt{3} \\
& z=2 \implies |x| < \sqrt{6},3\sqrt{2}<|y|<2\sqrt{6} \\
& z=3 \implies |x| < 3, 3\sqrt{3}<|y|<6
\end{align}
$$
calculating a lower-bound for the mass of the body
Going along the $z$-axis, the area of the largest rectangle inside the truncated circle (for obtaining a lower-bound to the area $A_{l}$) is
$$
\begin{align}
A_{l}(z) &= \left( 2\times 3\sqrt{z} \right) \times \left( 2\times \sqrt{3z} \right) \\
&= 12\sqrt{3}z
\end{align}
$$
What remains is to add the height element to this rectangle which is $dz$:
$$
\begin{align}
dV_{l} &= A_{l}(z)dz \\
&= 12\sqrt{3}zdz
\end{align}
$$
Since mass is volume times density, which is just $1$ here, numerically, $dM_{l}=dV_{l}$.
Integrate along $z$ from $0$ (since $z$ is a sum of non-negative quantities, it is also non-negative) to $3$ to obtain the volume:
$$
\begin{align}
M_{l} &= \int_{z=0}^{3} dM_{l} = \int_{z=0}^{3} dV_{l} = \int_{z=0}^{3} A_{l}(z)dz = \int_{z=0}^{3}12\sqrt{3}zdz \\
&= 12\sqrt{3}\int_{z=0}^{3}zdz \\
&= 12\sqrt{3}\left[ \dfrac{9}{2} \right] \\
&= 54\sqrt{3} \approx 93.53
\end{align}
$$
calculating an upper-bound for the mass of the body
Having found a lower-bound to the mass, let's find an upper-bound. Along the $y$-axis, the larger smallest rectangle containing the horizontal sections has boundaries $y = \pm 2\sqrt{3z}$. We may calculate an upper-bound to the area as well.
Going along the $z$-axis, the area of the smallest rectangle containing the truncated circle (for obtaining an upper-bound to the area $A_{u}$) is
$$
\begin{align}
A_{u}(z) &= \left( 4\times \sqrt{3z} \right) \times \left( 2\times \sqrt{3z} \right) \\
&= 24z
\end{align}
$$
$$
\begin{align}
dV_{u} &= A_{u}(z)dz \\
&= 24zdz
\end{align}
$$
$$
dM_{u}=dV_{u}
$$
$$
\begin{align}
M_{u} &= \int_{z=0}^{3} dM_{u} = \int_{z=0}^{3} dV_{u} = \int_{z=0}^{3} A_{u}(z)dz = \int_{z=0}^{3}24zdz \\
&= 24\int_{z=0}^{3}zdz \\
&= 24 \left[ \dfrac{9}{2} \right] \\
&= 108
\end{align}
$$
$M \in (M_l,M_u)=(54\sqrt{3},108)$. That's not bad.
A: The formula for volume computations with $xy$-projection is $$V=\sum_{i=1}^n\iint_{R_i}({z_{i,max}}-{z_{i,min}})dA.$$
In our example, we have three surfaces: $z=3,z=\frac{r^2}{12}$ and $z=\frac{r^2\cos^2\theta}{3}$ in cylindrical coordinates and $dA=rdrd\theta$.
Let me determine $R_i$s if I can, as I can only imagine the solid: I can not sketch the solid but I can work algebraically. First of all, we find the intersection of surfaces: $1)$ $z=3$ and $z=\frac{r^2}{12}$ gives $r=6$. $2)$ $z=\frac{r^2}{12}$ and $z=\frac{r^2\cos^2\theta}{3}$ gives $\theta=\frac{\pi}{3}$ $3)$ $z=3$ and $z=\frac{r^2\cos^2\theta}{3}$ gives $r\cos\theta=3$, in $[0,\frac{\pi}{2}]$.
Hence, I think, a correct subdivision of quarter of the solid is the following:
$1)$ $R_1: 0\leq\theta\leq\frac{\pi}{3}\;,0\leq r\leq \frac{3}{\cos\theta}\;$, $\frac{r^2\cos^2\theta}{3}\leq z\leq 3$.
$2)$ $R_2: \frac{\pi}{3}\leq\theta\leq\frac{\pi}{2}\;,0\leq r\leq 6\;$, $\frac{r^2}{12}\leq z\leq 3$.
Therefore, due to symmetry w.r.t. $x$ and $y$ axes,
$$V=4\left(\int_{0}^{\frac{\pi}{3}}\int_0^{\frac{3}{\cos\theta}}(3-\frac{r^2\cos^2\theta}{3})rdrd\theta+\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\int_0^6(3-\frac{r^2}{12})rdrd\theta\right)=27\sqrt{3}+18\pi.$$
