Suppose the volume of a bowl is defined by the following inequalities: $$z\geq x^2+y^2-1 {}, \\z\leq \sqrt{x^2+y^2}+1,\\z\geq 0$$ What is the value of its volume?
So I reasoned that the sought after volume is the volume of an elliptical paraboloid minus its bottom, (where $z<0$ ), and minus an inscribed cone. I think that the volume can be calculated with a triple integral. First with respect to $z$:
$$\int_{x^2+y^2-1}^{\sqrt{x^2+y^2}+1}dz=(\sqrt{x^2+y^2}+1)-(x^2+y^2-1)$$
Now we can integrate by x and y by first transform into polar coordinates: $$x^2+y^2=r^2\\{Jacobian}:r$$
So $$2\pi\int_1^2 (\sqrt{r^2}+2-r^2)r dr=2{\pi}[\frac{r^3}{3}+r^2-\frac{r^4}{4}]_1^2=\frac{19\pi}{6}$$
Nevertheless this is the wrong answer and I'm not quite sure where I've made a mistake. The answer is supposed to be $\frac{29\pi}{6}$.
Vertex of the cone is at $z = 1$ and it forms for $z \geq 1$. The circular paraboloid has vertex at $z = -1$ and it intersects the cone at $z = 3$. We are interested in the volume of the paraboloid above $z = 0$ outside the cone.
Intersection of cone and paraboloid using cylindrical coordinates,
$r^2 - 1 = r + 1 \implies r^2 - r - 2 = 0$.
That gives us $r = 2$ and $z = 3$.
So one way to integrate is paraboloid volume between $0$ and $1$ (as there is no cone) and then between $z = 1$ and $3$.
$V = \displaystyle \int_0^{2\pi} \int_1^3 \int_{z-1}^{\sqrt{z+1}} r \ dr \ dz \ d\theta \ + \ \int_0^{2\pi} \int_0^1 \int_{0}^{\sqrt{z+1}} \ r \ dr \ dz \ d\theta$
As far as your method, that works too but you also need to add the volume for $0 \leq r \leq 1$ where $z$ is between the plane $z = 0$ and the cone.
So add the component $\displaystyle \int_0^{2\pi} \int_0^1 \int_{0}^{1+r} r \ dz \ dr \ d\theta$.