If my integral is $$\int_{B(0,R)}x^2+y^2 dxdydz$$
is this $$\int_0^R\Biggr(\int_0^{2\pi}\Biggr(\int_0^{\pi}(r\cos\phi\sin\theta)^2+(r\sin\phi\sin\theta)^2\vert r^2\sin\theta\vert d\theta d\phi dr$$
the proper transformation into spherical coordinates? Since we have $f=x^2 +y^2$, and $g=(r\cos\phi\sin\theta,r\sin\phi\sin\theta,r\cos\theta)$, then $f(g(r,\phi,\theta)=(r\cos\phi\sin\theta)^2+(r\sin\phi\sin\theta)^2$ and the determinant of the Jacobian of $g$ is $r^2\sin\theta$.
Yes, it is. But your integrand can be made even a bit easier, since:
$$(r\cos\phi\sin\theta)^2+(r\sin\phi\sin\theta)^2=r^2\sin^2\theta(\cos^2\phi+\sin^2\phi)=r^2\sin^2\theta$$
so your whole integrand is $\;r^4\sin^3\theta\;$ , since $\;\sin\theta\ge0\;$ for $\;0\le\theta\le\pi\;$ .
Also observe that your $\;\theta,\,\phi\;$ are interchanged with what most people (non-mathematicians, indeed) are used to, since $\;\theta\;$ is usually the azimut and $\;\phi\;$ the vertical angle, and for you and many mathematicians is exactly the other way around. No problem, but it's something that must be taken into account.