We were working through a practice problem where we were computing an integral of the hemisphere $2\leq z \leq 2+\sqrt{4-x^2-y^2}$ via spherical coordinates. We came to the realization you can shift down 2 units and it's much easier, however we were trying to compute what the bounds on $\rho$ would be if we left it elevated at $z=2$. The answer is:
$$\int_{0}^{2\pi}\int_{0}^{\frac{\pi}{4}}\int_{2\sec\phi }^{4\cos\phi }\rho ^{2}sin\phi d\rho d\phi d\theta $$
I understand how geometrically, we can compute the lower bound, but is there a way to use core concepts from trig to arrive at the upper bound, $4\cos \phi$? In other words, I'm trying to see if we can get that answer using projections onto the axes rather than substiution into the equation for $z$.
The issue is that I can't find any projection to any axis that stays fixed unlike the lower bound where the projection on the z-axis is constant at $2$.