the volume of the solid limited by the surfaces $x^2+z^2=4$ and $y^2+z^2=4$. I want to find the volume of the solid limited by the surfaces $x^2+z^2=4$ and $y^2+z^2=4$.
I have broken the volume into sixteen pieces, one of which can be represented by the following section:
$0\leq x \leq 2 $
$0\leq y \leq x $
$0\leq z \leq \sqrt{4-y^2} $
Whenever I integrate over this section, I get a volume with $\pi$, whilst the answer I want is $\dfrac{128}{3}$.
I used the integral:
$$\int_{0}^{2}\int_{0}^{x} \sqrt(4-y^2) dydx$$
I solved this with trig substitution, which left me with an inverse sin which went to $\dfrac{\pi}{2}$.

The region of integration is even with respect all variables, thus
$$V=8\int_{0}^{2}\int_{0}^{\sqrt{4-z^2}}\int_{0}^{\sqrt{4-z^2}}dxdydz$$
$$V=8\int_{0}^{2}\int_{0}^{\sqrt{4-z^2}}\sqrt{4-z^2}dydz$$
$$V=8\int_{0}^{2}(4-z^2)dz$$
$$V=8(4z-\frac{1}{3}z^3)|_{0}^{2}=\frac {128}{3} $$
