The title is the statement of the problem. I want to calculate the volume between those paraboloids with a triple integral.
I started the exercise doing the next:
$z=z:\ x^2+(y-1)^2=\dfrac{5}{2} -x^2 -y^2$ Then
$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x^2+(y-\dfrac{1}{2})^2=1$ this is a circle with radius 1 and it's center is at $(0.\dfrac{1}{2})$
So making a coordenates change: $x=r \ cos \theta$ and $y=r \ sin \theta \ + \dfrac{1}{2}$ with $0≤r≤1$ and $0≤ \theta≤2 \pi$ and $r^2-2 \ r \ sin\theta+ \dfrac{9}{4}≤z≤ r \ sin\theta -r^2+\dfrac{11}{4}$
The volume can be calculated by
$$V = \int_0^{2\pi} \int_{0}^1 \int_{ r \ sin\theta -r^2+11/4}^{{r^2-2 \ r \ sin\theta+ 9/4}} r \ dz \ dr \ d \theta$$
I'm not sure if I found rigth the limits $r$ and $ \theta$ and $z$ of this triple integral correctly
The volume can be integrated in cylindrical coordinates, with the integrating region in the $xy$-plane given by the circle $x^2+(y-\dfrac{1}{2})^2=1$.
Re-center the circular region with $u=x$ and $v=y-\frac12$. Then, the region of integration is a unit circle $u^2+v^2=1$ and the enclosing surfaces $z=x^2+(y-1)^2$ and $z=\dfrac{5}{2} -x^2 -y^2$ becomes
$$z_1=u^2+(v-\frac12)^2,\>\>\>\>\>z_2=\dfrac{5}{2} -u^2 - (v+\frac12)^2$$
In polar coordinates $u = r\cos\theta$ and $v = r\sin\theta$, we have $z_2-z_1 = 2(1-r^2)$ and the volume integral,
$$V = \int_{-\frac\pi2} ^{\frac\pi2} \int_{0}^1 (z_2-z_1)\>r dr d\theta =2\int_{-\frac\pi2} ^{\frac\pi2} \int_{0}^1 (1-r^2)\>r dr d\theta=\frac\pi2$$