Calculate the integral $\iiint\limits_{D}dxdydz$ over domain D Let $a\in (-1,1)$ and domain $D=\{(x,y,z)\in \Bbb R:x^2+y^2+z^2<1,z>a\}$.
I try to calculate the integral $$\iiint\limits_{D}dxdydz$$ with the use of cylindrical coordinates, but I couldn't find the limits of the integration. Do you have any idea about the limits of the integral ?

As in our region $D$ we have
$$ x^2+y^2+z^2 < 1 $$
and
$$
z > a,
$$
so we can take $r$, $\theta$, and $z$ as follows:
$$
0 \leq r <  1,
$$
$$
0 \leq \theta \leq 2 \pi,
$$
and
$$
a < z < 1.
$$
Moreover, we note that, since
\begin{align}
x &= r \cos \theta, \\
y &= r \sin \theta, \\
z &= z, 
\end{align}
therefore we find that
\begin{align}
\frac{\partial(x, y, z) }{\partial(r, \theta, z)} &= \left| \begin{matrix} \frac{\partial x}{\partial r } & \frac{\partial x}{\partial \theta } & \frac{\partial x}{\partial z } \\
\frac{\partial y}{\partial r } & \frac{\partial y}{\partial \theta } & \frac{\partial y}{\partial z } \\
 \frac{\partial z}{\partial r } & \frac{\partial z}{\partial \theta } & \frac{\partial z}{\partial z } 
  \end{matrix} \right| \\
&= \left| \begin{matrix} \cos \theta & - r\sin \theta & 0 \\
\sin \theta & r \cos \theta & 0 \\ 
0 & 0 & 1  \end{matrix} \right| \\
&= \left| \begin{matrix} \cos \theta & -r \sin \theta \\ \sin \theta & r \cos \theta  \end{matrix} \right| \\
&= r \cos^2 \theta - \left(-r \sin^2 \theta \right) \\
&= r,
\end{align}
and so
\begin{align}
\int\int\int_{D}\, dx \, dy\, dz &= \int_{z = a}^{z = 1} \int_{\theta = 0}^{\theta = 2 \pi} \int_{r = 0 }^{r = 1}   r d r\, d \theta \,  dz \\
&= \int_{z = a}^{z = 1} \int_{\theta = 0}^{\theta = 2 \pi} \frac{1}{2} d \theta \,  dz \\
&= \int_{z = a}^{z = 1} \frac{1}{2} (2 \pi ) dz \\
&= \int_{z = a}^{z = 1} \pi dz \\
&= \pi (1 - a).
\end{align}
