Making a coordinate changes we set $u = x, v = y, w = 2z$, then the Jacobian is $\left|\dfrac{\partial(x,y,z)}{\partial(u,v,w)}\right|= \dfrac{1}{2}$, and the equation in the new coordinates is: $u+v = 4, v^2+w^2 = 16$. Observe that the second equation is now a circle centered at the origin in the $vw$ plane having a radius of $4$ instead of previously an ellipse which causes integration a bit more complicated. Thus, using cylindrical coordinates in this new coordinates, $V = \displaystyle \int \int \int_{\text{Region}(u,v,w)} \left|\dfrac{\partial(x,y,z)}{\partial(u,v,w)}\right|dudvdw = \displaystyle \int_{0}^{\frac{\pi}{2}} \int_{0}^4 \int_{0}^{4-r\sin \theta} \dfrac{1}{2}rdudrd\theta=...= 8\pi-\dfrac{32}{3}$.
In order to better understand the solid region $B$ enclosed by both surfaces in order to better evaluate our limits of integration, we can visualize the surfaces as follows:
In cylindrical coordinates, we can express the cylinder as:
$$ r = a \cos \theta \qquad (0 \leq \theta \leq \pi)$$
and the sphere as:
$$ r^2 + z^2 = a^2$$
We see that, as we traverse all the $(r,\theta)$ or $(x,y)$ points within the cylinder's projection on the $x$-$y$ plane (a circle), the values of $z$ we want are between $\pm\sqrt{a^2-r^2}$
Finally, we recall that:
$$ dV = dx \, dy \, dz = r \, dr \, d\theta \, dz $$
So:
\begin{align*}
\iiint_B \, dV &= \int_{0}^{\pi} \int_0^{a \cos \theta}\int_{-\sqrt{a^2-r^2}}^{\sqrt{a^2-r^2}} r \, dz \, dr \, d\theta \\
&= \int_{0}^{\pi} \int_0^{a \cos \theta} 2\int_{0}^{\sqrt{a^2-r^2}} r \, dz \, dr \, d\theta \\
&= \int_{0}^{\pi} \int_0^{a \cos \theta} 2r\sqrt{a^2-r^2} \, dr \, d\theta \\
&= \int_{0}^{\pi} \left[ -\frac{2}{3}(a^2-r^2)^{3/2} \right]_0^{a \cos \theta} \, d\theta \\
&= \frac{2}{3}\int_{0}^{\pi} (a^3 - a^3 \sin^3 \theta) \, d\theta \\
&= \frac{2a^3}{3}\int_{0}^{\pi} (1 - \sin^3 \theta) \, d\theta \\
&= \frac{2a^3}{12}\int_{0}^{\pi} (4 - 3\sin \theta + \sin 3\theta) \, d\theta \\
&= \frac{a^3}{6} \left[4\theta + 3 \cos \theta - \frac{1}{3}\cos 3\theta \right]_0^{\pi} \\
&= \frac{a^3}{6} \left(4\pi - 3 + \frac{1}{3} - 3 + \frac{1}{3} \right) \\
&= \frac{a^3}{6} \left(4\pi - \frac{16}{3} \right) \\
&= \frac{2a^3}{9} \left(3\pi - 4 \right)
\end{align*}
The problem that you probably ran into is parametrizing from $\theta = -\pi/2$ to $\theta = \pi/2$. This is a valid parametrization, but when you reach this point:
$$ L = \frac{2}{3}\int_{-\pi/2}^{\pi/2} \left[a^3 - (\sin^2 \theta)^{3/2} \right] \, d\theta $$
We have that:
$$ (\sin^2 \theta)^{3/2} \neq \sin^3 \theta $$
for $\sin \theta < 0$, which occurs for $-\pi/2 \leq \theta < 0$.
So we choose a parametrization where we keep $\sin \theta \geq 0$, which is why we use $0 \leq \theta \leq \pi$.
Best Answer
We have to figure out what the boundaries were in Cartesian in the first place. The innermost bounds will always be the most helpful because they give the direct equations for some of the surfaces involved. In this case we have
$$x^2+y^2 = z \hspace{24 pt} x \geq 0$$
$$x+y=0$$
And since $y \leq 0$ we have that $x \geq 0$ for the bottom surface as well. Moving on to the next bounds, $z=2y^2$ is simply the projection of the intersection of the above plane and paraboloid on to the $yz$ plane ($x = -y \implies z = 2y^2$). This means that $z=-4y$ is a completely independent surface boundary from the previous ones.
Lastly as a sanity check, $-4y = 2y^2 \implies y = 0,2$, so we do not have to cut up the region any further. This gives us enough information to figure out the bounds in polar coordinates.
$x+y=0$ is a straight vertical plane, which means it represents a $\theta = k$ surface. Given the context of the quadrant in the $xy$ plane we are in (quadrant IV), we get that $\theta = -\frac{\pi}{4},0$. The other surfaces are standard cylindrical coordinates transitions.
Setting up the integral with $z$ first we have that
$$I = \int_{-\frac{\pi}{4}}^0 \int_0^{-4\sin\theta} \int_{r^2}^{-4r\sin\theta} \frac{1}{r}\:dz\;dr\:d\theta = \pi-2$$
which is almost the same as the expression you originally obtained.