Triple Integral - My answer seems too large Consider $R=[(x,y,z)\in \mathbb{R}^3 | 1 \leq x^2 + y^2 \leq 16, 0 \leq z \leq y+4]$. Then calculate the integral
$$I=\int_R(x-y)dV$$
I thought about the region and found that as $€ \leq x^2 + y^2 \leq 16,$ we have $x \in [1,4], y \in [1,4]$. Then we can have $z \in [0,8]$
I then changed the integral to $\int_1^4 \int_1^{16-x^2}\int_0^{y+4}(x-y)dzdydx$, which I found to be $\frac{93461}{70} \approx 1335.16$
However, if I was to make the region larger, say taking the  circle $x^2+y^2=16 \ (a=4)$ and making it into a spheroid with $z=8 \ (c=4)$, then I would get volume $\frac{4}{3} \pi 4^24=\frac{256}{3}\pi \approx 268.08$
Hence I know that my answer is far too large, so I have probably found the incorrect limits of integration?
 A: You need to split your integral if you are setting it up in cartesian coordinates. Given the region is between two cylinders of radius $1$ and $4$ bound by two planes $z= 0$ and $z = y + 4$, I would recommend cylindrical coordinates.
In cylindrical coordinates, $x = r \cos \theta, y = r\sin\theta, z= z$.
So we have,
$0 \leq z \leq 4 + r \sin\theta$
$1 \leq r \leq 4$
$0 \leq \theta \leq 2\pi$
So the integral becomes,
$ \displaystyle \int_0^{2\pi} \int_1^4 \int_0^{4 + r\sin\theta} r^2 (\cos\theta - \sin\theta) ~ dz ~ dr ~ d\theta$
But you must note that $x$ being an odd function and the symmetry of the region about YZ plane, its integral over the region would be zero. So you should just evaluate integral of $ - y$.

If you are setting it up in cartesian coordinates, see the projection of the region in xy-plane.

The bounds of $z$ is straightforward as mentioned by you. But bounds of $x$ and $y$ are different for $|x| \leq 1$ and for $1 \leq |x| \leq 4$. A more straightforward way would be to evaluate over both cylinders separately and then subtract. Again, you can avoid integrating $x$ as mentioned above.
$ \displaystyle I_1 = \int_{-1}^1 \int_{ - \sqrt{1-x^2}}^{ - \sqrt{1-x^2}} \int_0^{y+4} (x-y) ~ dz ~ dy ~ dx$
$ \displaystyle I_2 = \int_{-4}^4 \int_{ - \sqrt{16-x^2}}^{ \sqrt{16-x^2}} \int_0^{y+4} (x-y) ~ dz ~ dy ~ dx$
and finally, $I = I_2 - I_1$
A: Use cylindrical coordinates. Your integral becomes$$\int_0^{2\pi}\int_1^4\int_0^{r\sin(\theta)+4}r^2\bigl(\cos(\theta)-\sin(\theta)\bigr)\,\mathrm dz\,\mathrm dr\,\mathrm d\theta,$$which is equal to:\begin{multline}\int_0^{2\pi}\int_1^4r^2 (\cos (\theta )-\sin (\theta )) (r \sin (\theta )+4)\,\mathrm dr\,\mathrm d\theta=\\=\int_0^{2\pi}\left(-\frac{255}{4} \sin ^2(\theta )-84 \sin (\theta )+84 \cos
   (\theta )+\frac{255}{4} \sin (\theta ) \cos (\theta )\right)\,\mathrm d\theta=-\frac{255}4\pi.\end{multline}
