I am trying to solve the following integral:
$$\int _{-2}^2\:\int _0^{\sqrt{4-y^2}}\:\int _0^{\sqrt{4-x^2-y^2}}y^2\sqrt{x^2+y^2+z^2}\,\text{d}z\text{d}x\text{d}y$$
I originally just substituted through starting from $\text{d}z$ and working out until $\text{d}y$ and got $16$ as my answer but when I checked using spherical coordinates I got a negative answer.
Edit: This was my initial attempt -
(1) $\int _{-2}^2\:\int _0^{\sqrt{4-y^2}}\:\int _0^{\sqrt{4-x^2-y^2}}y^2\sqrt{x^2+y^2+z^2}\,\text{d}z\text{d}x\text{d}y$
(2) $\int _{-2}^2\int _0^{\sqrt{4-y^2}}\left(y^2\sqrt{x^2+y^2+\left(\sqrt{4-x^2-y^2}\right)^2}\right)-y^2\sqrt{x^2+y^2+\left(0\right)^2}\,\text{d}x\text{d}y$
(3) $\int _{-2}^2\int _0^{\sqrt{4-y^2}}2y^2-y^2\sqrt{x^2+y^2}\,\text{d}x\text{d}y$
(4) $\int _{-2}^2\left(2y^2-y^2\sqrt{\left(\sqrt{4-y^2}\right)^2+y^2}-2y^2\sqrt{\left(0\right)^2+y^2}\right)\,\text{d}y$
(5) $\int _{-2}^2-2y^2+y^3\,\text{d}y$
$= 16$
In spherical coordinates, the integration becomes
$\displaystyle \int_{0}^{\pi} \int_{0}^{\frac{\pi}{2}} \int_{0}^2 r^5 \sin^3 \theta \sin^2 \phi dr d\theta d\phi $
And this evaluates to
$ \left(\dfrac{2^6}{6} \right) \left( \dfrac{\pi}{2} \right) \left( \dfrac{2}{3} \right) = \dfrac{ 32 \pi }{ 9 } $