Lebesgue integral, Cavalieri's principle Merry Christmas,
Can you prove my answer:
Let $B_r^n (p):=\{x\in\mathbb R^n \mid |x-p|\le r\}$ 
be the Ball with radius $r\in\mathbb R_{+}$
and origin $p\in\mathbb R$ and dimension $n\in\mathbb N$.
Evaluate $\lambda_3(B_1^3(0)\cap(B_r^2(0)\times\mathbb R))$,
when $r\in (0,1)$
Answer:
I evalute the internal term:
$$B_1^3(0)\cap(B_r^2(0)\times\mathbb R)
=\{(x,y,z)\in\mathbb R^3\mid x^2+y^2+z^2\le 1;x,y,z\in (-1,0)\cup (0,1)\}$$
And with the Principle of Cavalieri, I get following integral:
$$\int_{-1}^1 \int_{-1}^1 \int_{-\sqrt{1-x^2-z^2}}^{\sqrt{1-x^2-z^2}} \, dy \, dx \, dz=\frac{4}{3}\pi$$
Do you agree?

The fact that the region is within $B_r^2(0)\times\mathbb R$ says $x^2+y^2\le r^2$; the fact that it is within $B_1^3(0)$ says $x^2+y^2+z^2\le 1$.  Your condition that $x,y,z\in (-1,0)\cup (0,1)$ fails to say that $x^2+y^2+z^2\le 1$.  It is also not at all clear why you excluded $0$.
Your bounds of integration, in
$$
\int_{-1}^1 \int_{-1}^1 \cdots\cdots \, dx \, dz
$$
fail to constrain the pair $(x,z)$ to the region it should be in.
You could write
$$
\iiint_{B_1^3(0)\cap(B_r^2(0)\times\mathbb R} 1 \, d(x,y,z) = \int_0^{2\pi} \int_0^r \int_{-\sqrt{1-s^2}}^{\sqrt{1-s^2}} 1 \, dz \,(s\,ds) \,d\theta.
$$
Alternatively, you could use spherical coordinates.
In Cartesian coordinates, one can write
$$
\int_{-r}^r \left( \int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}} \left( \int_{\sqrt{1-x^2-y^2}}^{\sqrt{1-x^2-y^2}} 1\,dz \right) \,dy \right) \, dx.
$$
