<s>[INST] Find the surface area of part of cone $x^{2} = y^{2} + z^{2}$ which lies inside the sphere $x^{2} + y^{2} + z^{2} = 2z$. Find the surface area of part of cone $x^{2} = y^{2} + z^{2}$ which lies inside the sphere $x^{2} + y^{2} + z^{2} = 2z$
Intersection of both surfaces on $yz$ plane will be $ y^{2} + (z-1/2)^{2} = 1/4$ 
Surface area $= \int \sqrt{1 + (y/x)^{2} + (z/x)^{2}} dydz$
$= \int (\sqrt{ (x^{2} + y^{2} + z^{2}})/x) dydz$
$= \int ( \sqrt{ 2(y^{2} + z^{2})})/(\sqrt{y^{2} + z^{2}}) dydz$
$= \int \sqrt{2} dydz$
Since area of the circle $ y^{2} + (z-1/2)^{2} = 1/4$ is $ \pi /4$ so the surface area should be $( \sqrt{2} \pi)/4$
But answer is given $ \sqrt{2} \pi$
So where did I go wrong$?$
 [/INST] The condition $ x^2 = y^2 + z^2 $ implies the existence of two cones
$x = \sqrt {y^2 + z^2}$ and $x = -\sqrt {y^2 + z^2}$ so the result should be doubled, as you only used the first cone. 
However, the expected answer still remains different.
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