Triple integral in spherical coordinate, where am I wrong $$ \iiint_{D} z\left(x^{2}+y^{2}+z^{2}\right) \mathrm{d} x \mathrm{~d} y \mathrm{~d} z $$
D is given by $x^{2}+y^{2}+z^{2}\leq 2z$
I try to use  $ \left\{\begin{matrix} 
  x=r\sin \phi \cos \theta \\  
  y=r\sin \phi \sin \theta \\ 
z=r\cos\phi
\end{matrix}\right. $ while the Jacobian is $r^2 \sin \phi$ and $ \left\{\begin{matrix}  0\leq \theta \leq 2\pi\\
0\leq \phi \leq \frac{\pi}{2} \\0\leq r \leq 1\end{matrix}\right. $
$\Rightarrow$
$$\int^{2\pi}_0 \mathrm{d}\theta\int^{\pi/2}_0\mathrm{d}\phi\int_0^{2\cos\phi}r^4\sin^2\phi \cos \phi\mathrm{d}r  $$
$\Rightarrow$
$$2\pi\int^{\pi/2}_0\cos^6\phi-\cos^8\phi \ \mathrm{d}\phi$$
$$I_{n}=\int_{0}^{\frac{\pi}{2}} \cos ^{n} x d x\Rightarrow I_{n}=\frac{n-1}{n} I_{n-2}\text{  integral by part }$$
$$I_{2 m}=\frac{2 m-1}{2 m} \cdot \frac{2 m-3}{2 m-2} \cdot \cdots \cdot \frac{3}{4} \cdot \frac{1}{2} I_{0}\\
I_{2 m+1}=\frac{2 m}{2 m+1} \cdot \frac{2 m-2}{2 m-1} \cdot \cdots \cdot \frac{4}{5} \cdot \frac{2}{3} I_{1}\\
 I_{0}=\int_{0}^{\frac{\pi}{2}} d x=\frac{\pi}{2}, I_{1}=\int_{0}^{\frac{\pi}{2}} \cos x d x=1\\$$
$\Rightarrow$
$$2\pi\int^{\pi/2}_0\cos^6\phi-\cos^8\phi \ \mathrm{d}\phi=\frac{\pi^2}{4}$$
but the answer is $\frac{8\pi}{3}$, where am I wrong.
I used wolframalpha to calculate $\int^{\pi/2}_0\mathrm{d}\phi\int_0^{2\cos\phi}r^4\sin^2\phi \cos \phi\mathrm{d}r  $ the answer is $\frac{\pi}{8}$, am I wrong at first?

Firstly, I have a typo it is $r\leq2\cos \phi$.
Secondly, it is
$$\int^{2\pi}_0 \mathrm{d}\theta\int^{\pi/2}_0\mathrm{d}\phi\int_0^{2\cos\phi}r^5\sin\phi \cos \phi\mathrm{d}r  $$
and it is easy to integrate
$\Rightarrow$
$$-2\pi \int_0^{\pi/2}\frac{2^6}{6}\sin \phi \cos^7\phi d\phi\Rightarrow-\frac{2^7\pi}{6}\int_0^{\pi/2} \cos^7\phi d\cos\phi\\ \Rightarrow-\frac{2^7\pi}{6\times2^3}\cos^8 \phi \bigg|^{\pi/2}_0=\frac{8\pi}{3}$$
