Surface area of part of sphere inside paraboloid

Question

Question was to find the surface area of part of sphere $x^2+y^2+z^2=4z$ that lies inside paraboloid $z=x^2+y^2$. I solve problem by taking range of azimuthal angle from 0 to 30 and taking parameters that we take for sphere centered at origin. I am doing mistake somewhere because actual answer is different. What mistake am I making?

Answer 1

The paraboloid intersects the sphere along the circle $$ \cases{ x^2+y^2=3,\\ z=3. } $$ The part of the sphere inside the paraboloid is then a spherical cap, with base radius $a=\sqrt3$ and height $h=1$.

Answer 2

I think your azimuthal range is incorrect. Take cross-sections in the $xz$-plane for simplicity. The curves $x^2 + z^2 =4z$ and $z=x^2$ intersect when $$ x^2 + (x^2)^2 = 4(x^2) \implies x^4 -3x^2 = 0 $$ So $x=0$ or $x=\pm\sqrt{3}$.

If you translate the sphere with radius $2$ back down to the origin, a point with $x$-coordinate $\sqrt{3}$ has $\sin\phi = \frac{\sqrt{3}}{2}$. So $\phi = \frac{\pi}{3}$, or $60^\circ$, not $30^\circ$ as you were calculating.