Triple integrals (Find volume): The solid bounded by the sphere $r = 2 cos$ $ \phi $ and the hemisphere $r = 1$, $z$ $\ge$ $0$

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Here is the exact question: https://imgur.com/a/cBQC8su!

My particular question regards the range of $\phi$; $\phi$ certainly lives between $0$ $\le$ $\phi$ $\le$ $\frac {\pi}{2}$. $\rho = 1$ intersects with $ \rho = 2cos\phi$ at $\frac {\pi}{3}$. I thought that the range of $\phi$ would be $0$ $\le$ $\phi$ $\le$ $\frac {\pi}{2}$, where $0$ $\le$ $\phi$ $\le$ $\frac {\pi}{3}$ to $\frac {\pi}{3}$ $\le$ $\phi$ $\le$ $\frac {\pi}{2}$. Any helpful tips? The book lists the answer as $\frac {5\pi}{12}$.

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The intersection $S$ of the two unit balls is a lens shaped object whose volume can easily be computed by the "washer method". One obtains $${\rm vol}(S)=2\cdot\int_{1/2}^1\pi(1-r^2)\>dr={5\pi\over12}\ .$$