Find the area of the surface:
Part of the sphere x^2+y^2+z^2=4z that lies inside the paraboloid z=x^2+y^2

Question

Find the area of the surface:
Part of the sphere x^2+y^2+z^2=4z that lies inside the paraboloid z=x^2+y^2

anonymous · Answer

@Zarkon @phi @amistre64

anonymous · Answer

Well your first equation can be rewritten as:
$$x^2+y^2+z^2-4z=0 \implies x^2+y^2+(z-2)^2=4$$
So a sphere with radius 2 centered at (0,0,2). 
So you need to find the equation that defined the boundary of the surface. Doing that doesn't seem easy however.

anonymous · Answer

I'm a bit new to this, so I'm not totally sure how to find the equation of the boundary of the surface. How would I start? Thank you

anonymous · Answer

Notice that if z=3 for both equations they reduce to:
$$x^2+y^2+1=4 \implies x^2+y^2=3; x^2+y^2=z \implies x^2+y^2=3$$
This means they meet at:
z=3,x^2+y^2=3. So we have something that looks like this:
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