Explain how a multiple integral may be expressed in a different coordinate system

I can't see much relation between the first and second parts of this question (but I haven't actually tried doing the question). As for the second part, you seem to be along the right lines. The region $R$ doesn't change when you change coordinates; all that changes is the way you express it. If $u,v,w$ is a change of coordinates, then this means that $u,v,w$ are all bijective functions of $x,y,z$. In particular, they have inverses, so $x,y,z$ are bijective functions of $u,v,w$, and so the cone is parametrised by $z(u,v,w)^2 = x(u,v,w)^2 + y(u,v,w)^2$. Then the limits of the integral will be precisely the set of $(u,v,w)$ for which this equation is satisfied.

I think that makes sense to me, but I don't understand how it's possible to get the new limits in a useable form to finish the question without knowing what the functions relating u, v and w to x, y and z are - because $z(u,v,w)^2 = x(u,v,w)^2 + y(u,v,w)^2$ doesn't seem to have a solution.

How could you possibly solve the equation without knowing what $u,v,w$ are? If you know what they are then you might have a hope of solving it. So by saying "solve $z^2=x^2+y^2$ for $u,v,w$", you've said as much as you can until you know what $u,v,w$ are. But the question is asking a general question, so you're not expected to give an explicit form of an answer (not least because such an expression doesn't exist); the specifics come later on.

For instance, we could use cylindrical polar coordinates $(u,v,w)=(r, \varphi, z)$, so that we have $x = r \cos \varphi$, $y = r \sin \varphi$ and $z = z$. Then the equation of the surface, $z^2 = x^2 + y^2$, becomes $z^2 = r^2$. Note that this doesn't depend on $\varphi$, so $\varphi$ can take any value and as long as $r=\left|z\right|$ the equation is still satisfied. (This makes sense since all the equation above tells you is that for a given z, the cross-section of the graph through that point on the z-axis is just a circle of radius |z|, which is why it looks like a cone [or rather two cones stuck together at their tips].) So we can integrate over $\theta, z$ by taking $1 \le z \le 2$ (which is the given limit) and $0 \le \theta < 2\pi$, and setting $r=z$ in the integral and multiplying by the Jacobian.

This is just one example of a parametrization. (It just happens to be probably the easiest parametrization to use in this case.)

I think I understand what you're saying. Thank you very much for your help.

The answer I got for the final part seems a bit funny, PiLn2, but I think the method was right, and that's what matters.

That's the right answer, so you probably did something right