1. ## Cylindrical Coordinates

Describe the solid by inequalities in cylindrical coordinates which is enclosed by $\displaystyle z=0, z=x+y+1$ and the two cylinders $\displaystyle x^2+y^2=4, x^2+y^2=9$

Just need someone to quickly verify this,
I have
$\displaystyle 2 \leq r \leq 3$,
$\displaystyle 0 \leq \theta \leq 2\pi$ and
$\displaystyle 0 \leq z \leq x+y+1$ (not sure how else to state that one)
is this correct?

2. I would draw it first and then work backwards. Now cylindrical coordinates are $\displaystyle {r,t,z}$ so you have $\displaystyle r$ right but from the plot below, z will depend on $\displaystyle x=r\cos(t)$ and $\displaystyle y=r\sin(t)$ and note as I increase z, the angle that t goes through changes as a function of z. But we can find that angle from solving the simultaneous equation:

$\displaystyle x^2+y^2=4$

$\displaystyle x+y+1=z$

for constant z then obtain the two angles as $\displaystyle (t_1(z),t_2(z))$ then I think it would go something like:

$\displaystyle 2\leq r \leq 3$

$\displaystyle t_1(z)\leq t \leq t_2(z)$

$\displaystyle 0\leq z\leq r\cos(t)+r\sin(t)+1$

3. Thank you very much for providing a picture, I made a careless mistake of assuming that the plane did not cut the object through the xy-plane. However, it isn't clear to me that theta is changing with respect to z from examining the plot at different angles.

I'm still a little bit confused on how to go about solving that. Would you mind explaining to me the choice of equating $\displaystyle x+y+1=z$ with $\displaystyle x^2+y^2=4$ rather than $\displaystyle x^2+y^2=9$?