Mass and center of mass of solid 
Let $S$ be the part of the solid $z = 4 - x^2 - y^2$ where $z$ $\geq 3$, and let $T$ be the solid enclosed by $S$ and the plane $z = 3$.
  Assume $T$ has constant density equal to $1$. Calculate the mass and the coordinates of the center of mass of $T$.

So to find the mass I solved $$\int_0^{2\pi} \int_0^1 \int_3^4rdzdrd\theta = \pi$$ and then found the $z$ coordinate for the center of mass to be $$ \frac1\pi\int_0^{2\pi} \int_0^1 \int_3^4rzdzdrd\theta = \frac72$$ Am I correct so far? Would it also be correct to asume that the senter of mass must lie on the $z$-axis so that the center of mass has the coordinates $(0,0,\frac72)$?
By the way, is it possible to solve this using only a double integral, or must a triple integral be used?

The triple integral is the easiest I think. You have the condition
$$4-r^2=z\ge3$$
So that $r\le1$ is your upper bound for $r$. The lower bound is clearly $0$. Now for $z$, you were given that $3\le z\le4-r^2$, so the mass integral becomes
$$m=\int_0^{2\pi}\int_0^1\int_3^{4-r^2}dz\,r\,dr\,d\theta=2\pi\int_0^1(1-r^2)r\,dr=\left.2\pi\left(-\frac14\right)(1-r^2)^2\right|_0^1=\frac{\pi}2$$
Then
$$m\bar z=\int_0^{2\pi}\int_0^1\int_3^{4-r^2}z\,dz\,r\,dr\,d\theta=2\pi\int_0^1\frac12\left[(4-r^2)^2-9\right]r\,dr=\pi\left[\left(-\frac16\right)(4-r^2)^3-\frac92r^2\right]_0^1=\frac{5\pi}3$$
So
$$\bar z=\frac{m\bar z}m=\frac{10}3$$
And this is between $3$ and $4$ as it should be. Your mistake lay in not having the correct upper bound for $z$. $\bar x=\bar y=0$ because when you do the $\theta$ integrals ou will find that
$$\int_0^{2\pi}\cos\theta\,d\theta=\int_0^{2\pi}\sin\theta\,d\theta=0$$
