Volume of domain bounded by $f(x, y) = \ln(x^2 + y^2)$ The full question is as follow:

Given the function
$$
f(x, y) = \log(x^2 + y^2) 
$$
for $x^2 + y^2 > 0$, using cylindrical coordinates $r, \theta$ and $z$, find the volume of the (closed) domain, $D$, which is bounded by: the graph of the function $f(x, y)$, the xy plane, and the plane given by $z = 10$. Moreover, $(0, 0, 0) \in D$.

My attempt
First intuition tells me that the domain is an "expanding circle" of sorts as $z$ changes.
Since $D$ is bounded by the xy plane and the plane given by $z = 10$, the range of $z$ is $0 \lt z \le 10$.
Finding the bounds of $r$, given $0 \lt z \le 10$,
\begin{align*}
\log(x^2 + y^2) &= 0 \\
x^2 + y^2 &= 1
\end{align*}
and
\begin{align*}
\log(x^2 + y^2) &= 10 \\
x^2 + y^2 &= e^{10}
\end{align*}
which are circles with radius $1$ and $e^5$ respectively.
Hence, the volume of D is
\begin{align*}
D &= \iiint_V dz \ dy \ dx \\
&= \int_0^{2\pi}\int_1^{e^5}r \int_{\log r^2}^{10} dz \ dr \ d\theta \\
&= \int_0^{2\pi}\int_1^{e^5} 10r - r \log r^2 \ dr \ d\theta \\
&= \int_0^{2\pi}\left(5r^2 - \frac 1 4 (\log r^2)^2\right)\Big|_1^{e^5} d\theta \\
&= \int_0^{2\pi} 5e^{10} - 30 \ d\theta \\
&= 2\pi(5e^{10} - 30)
\end{align*}
I am not sure if the final answer is correct and whether my reasoning for the integration bounds is right as well. Any advice would be greatly appreciated!
 A: Two mistakes -
i) you are leaving the volume of the cylinder in the middle with $0 \leq r \leq 1$. It has volume of $10 \pi$.
ii) The integral of $~ r \ln r^2$ is not correct.
I would rather use the disc method so it is just one integral and the integral is more straightforward too.
$z = \ln(x^2+y^2) = \ln r^2 \implies r = e^{(z/2)}$
So, $ V = \displaystyle \int_0^{2\pi}\int_0^{10} \int_0^{e^{(z/2)}} r ~dr ~ dz ~ d\theta = (e^{10} - 1) \pi$

Edit: Adding diagrams which should help you visualize your mistake that I mentioned in point $(i)$.
Here is a $3D$ diagram and as the lower bound of $r$ you have is $1$, your integral set up is not integrating over $0 \leq r \leq 1$ and is hence leaving out the volume in the middle. See the base of the left out cylinder volume, marked in black.

Now let me re-iterate the same by showing 2D projection in xz plane. The integral setup of yours only considers the region shaded in turquoise color but misses the region in the middle that is shaded in grey.

