Surface integral of $f(\vec x) = z^2$

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Let $\vec x = (x,y,z)^T$ and $f(\vec x) = z^2$. Calculate the surface integral $$ \int_{|\vec x| = \frac{1}{2}} f \, \, dO $$

my solution :

$dO = r^2 sin(u) \, \, du \, dv$ and $f(\vec x) = z^2 = r^2 cos^2(u)$.

$$ \Rightarrow \int_{|\vec x| = \frac{1}{2}} f \, \, dO = \frac{1}{16}\int_{0}^{2 \pi} \int_{0}^{\pi} \sin(u) \cos^2(u) \, \, du \, dv = \cdots $$ is the solution correct?

Thank you :)