I want to find the area of the portion of the sphere $x^2+y^2+z^2=4z$ inside the function $x^2+y^2=z$ using double integrals. The graph would be something like this:
Because of the nature of this functions, I assume the best way would be to find the area projected on the plane yOz in the first octant and multiply it by 4. Hence, the area would be: $$4\int \:\int \:\sqrt{1+\left(\frac{\partial x}{\partial y}\right)^2+\left(\frac{\partial x}{\partial z}\right)^2}dA$$ However, I failed every time I tried to solve this, so I would appreciate any help. I know that he correct answer is $4\pi $.
Let $x=f(x,y)=2+\sqrt{4-x^2-y^2}$, then $f_x=-\dfrac{x}{\sqrt{4-x^2-y^2}}$ and $f_y=-\dfrac{y}{4-x^2-y^2}$.
When we solve for $r^2+z^2=4z$ for $r>0$ and $z$ to get $r=\sqrt3$ and $z=3$
The surface area is $$\int\int_D\sqrt{1+f^2_x+f^2_y}\ dA$$ $$\int\int_D\dfrac{2}{\sqrt{4-x^2-y^2}}\ dA=\int_0^{2\pi}\int_0^{\sqrt3}\dfrac{2r}{\sqrt{4-t^2}}\ drd\theta=-2(2\pi)\sqrt{4-r^2}\bigg|_0^{\sqrt3}=4\pi$$