Finding the limits of integration in triple integrals without using the figure. As an example I was trying to integrate the function $f(x,y,z) = x$ on the set $B = \{(x,y,z) \in R^3 \mid 4y^2 + z^2 \le 16, x \le \sqrt{4y^2 + z^2}, x+y \ge -1  \}$.
I notice that by swapping to cylindrical coordinates I can obtain an easier region of integration so I put $y = \frac{r \cos \theta}{2}, z = r \sin \theta, x =t$.
To obtain $$B = \{ (r,\theta, t) \in [0, + \infty[ \times [0 , 2 \pi] \times R \mid  r^2 \le 16, t \le r, t + \frac{r \cos \theta}{2} \ge -1\}$$ 
Usually here I was taught to draw a figure and find the projections graphically, I would like someone to teach me how to do this without utilizing the figure. Like a step by step analytical approach.
As a start I see that the $t \in B \iff -\frac{r \cos \theta}{2}  -1 \le t \le r$ so now when I go to analyse the equations of the other variables should I treat $t$ as fixed?
Usually my questions are more clear and direct, I apologize. Since it could be a long and tedious answer a link to a source that explains a non graphical method to these kinds of problems will suffice.
 A: 
Cylindrical polar coordinates 

For convenience let our order of integration be
$$
\mathrm{\iiint_V g(r,\theta,t)\,dt\,dr\,d\theta}
$$
where $\mathrm{g}$ is the appropriate function. 
As we're integrating w.r.t $\;\mathrm{t}\;$ first, we fix $\mathrm{r}$ and $\theta$ and then figure out the limits. From the description of $B$ it is clear that
$$
\mathrm{-1-{r\cos\theta\over2}\le t\le r}
$$
Now that we're done integrating w.r.t $\;\mathrm{t}\;$, our integral looks like
$$\mathrm{\iint_S h(r,\theta)\,dr\,d\theta}$$
Notice how the integrand now only depends on $\mathrm{r,\theta}$ and not on $\mathrm{t}$. If $\mathrm{t}$ remains even after integration w.r.t $\;\mathrm{t}$ has been performed you're doing something wrong. This happens because definite integrals involving single-variable functions result in constant values, not functions e.g. $$
\mathrm{\int_0^1t\,dt=0.5\ne\text{a function of t (assuming we don't regard constants as functions)}}
$$
So to figure out the limits for $\mathrm{r}$ we have to fix only $\mathrm{\theta}$. Your change to cylindrical coordinates assumes $\mathrm{r\ge0}$. So we have
$$
\mathrm{0\le r\le4}
$$
Although we don't have any $\mathrm{\theta}$ dependence in the limits, we easily could have it with a set representing a different volume than $B$. Finally, your integral becomes
$$
\mathrm{\int_0^{2\pi}\int_0^4\int_{-(1+{r\over2}\cos\theta)}^r\color{blue}{t}\color{red}{r\over2}\,dt\,dr\,d\theta}=24\pi
$$
where $\mathrm{t}$ is the function we were originally integrating and $\mathrm{r\over2}$ is the absolute value of the Jacobian determinant of the coordinate transformation or you could look at it as $\mathrm{{r\over2}dtdrd\theta}$ is the infinitesimal volume element in your cylindrical coordinate system.

Cartesian coordinates

For convenience we integrate w.r.t $\;\mathrm{x}$, then $\mathrm{z}$ and then $\mathrm{y}$. According to the first representation of $B$ we have
$$\mathrm{
-1-y\le x\le\sqrt{4y^2+z^2}
}$$
We also have
$$\mathrm{
z^2\le16-4y^2\implies -\sqrt{16-4y^2}\le z\le\sqrt{16-4y^2}
}$$
and finally for $\mathrm z$ to have real limits we must have
$$\mathrm{
16-4y^2\ge 0\implies y^2\le 4\implies -2\le y\le 2
}$$
so your integral becomes
$$\mathrm{
\int_{-2}^2\int_{-\sqrt{16-4y^2}}^{\sqrt{16-4y^2}}\int_{-1-y}^{\sqrt{4y^2+z^2}}x\,dx\,dz\,dy=24\pi
}$$
