I'm having a lot of trouble finding from where to where I have to integrate when splitting up a triple integral into 3 integrals.
I've already posted a question regarding this but while that helps for that specific problem I'd like to know what technique I generally need to apply to solve it.
Here's an example of the type of problem I'm talking about:
I need to calculate the following in cylindrical coordinates:
$$\iiint_K \sqrt{x^2+y^2+z^2}\,dx\,dy\,dz$$ $K$ is bounded by the plane $z=3$ and by the cone $x^2+y^2=z^2$.
That question in particular can be found here Calculating $\iiint_K \sqrt{x^2+y^2+z^2}\,dx\,dy\,dz$., but again, I'm not looking for an answer to that particular integral in this question, I'm merely asking for a good way to solve most of these types of problems.
The key is to somehow translate the equations defining the bounding surfaces of the region of integration into a system of inequalities that explicit define the region through bounds on the coordinates.
Consider the example problem you provided. The region $K$ is defined implicitly to be the region between two bounding surfaces, the plane $z=3$ and the (double-)cone $x^2+y^2=z^2$. Let's convert to cylindrical coordinates as the problem suggests. Substituting $x=\rho\cos{\phi}$ and $y=\rho\sin{\phi}$, the equation for the plane is unchanged, but the equation for the cone becomes:
$$z^2=x^2+y^2=\rho^2\cos^2{\phi}+\rho^2\sin^2{\phi}=\rho^2\\ \implies z=\pm\rho.$$
We're only interested in the cone corresponding to the solution $z=+\rho$ since this is the cone intersected by the plane $z=3$. Now, any point in this region $K$ between the cone $z=\rho$ and plane $z=3$ will have $z$ coordinates satisfying the inequality,
$$\rho\le z\le 3.$$
The lower bound on $\rho=\sqrt{x^2+y^2}$ is of course $\rho=0$, so we actually have:
$$0\le\rho\le z\le 3.$$
This last inequality above allows you to immediately read off the limits of integration. If you want to integrate with respect $\rho$ first, then you would integrate $\rho$ from $0$ to $z$, and next you would integrate $z$ from $0$ to $3$. Similarly, if you wanted to integrate with respect to $z$ first, then you would integrate $z$ from $\rho$ to $3$, and next you would integrate $\rho$ from $0$ to $3$.