# Surface area of a sphere within a paraboloid

Calculate the surface area of the following: 1) the portion of the sphere $x^2 + y^2 + z^2 = 16z$ that lies within the paraboloid $z = x^2 + y^2$

Attempt:so by rearranging the eqn given we have a sphere with centre coordinates $(0,0,8)$ and radius 8. Setting the eqns equal to each other ( the eqn of the paraboloid and sphere) I get $z + (z-8)^2 = 64$ which $=> z=15, z = 0$. I am sort of stuck from here. I tried saying that in spherical polars, $z=r\cos\phi$ so by putting this into the eqn $(z) + z^2 = 16z,$ I get $r\cos\phi + r^2\cos^2\phi = 16r\cos\phi$ but this gives an undefined angle. What did I do wrong which led to an undefined angle - any hints on how to proceed with the question?

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Hint: the question is rotationally symmetric, so you are better off using cylindrical instead of spherical coordinates. Drawing a picture (by symmetry your picture can be just inside the $y$-$z$ plane) you see that the portion of the sphere inside the paraboloid is the portion where the $z$ coordinate is at least 15.

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Does there not exist a part of the sphere enclosed within the paraboloid beyond z= 15? I drew the picture and it seems to be the case? Should we use cylindrical because there is symmetry with respect to one of the axes? – CAF Nov 7 '12 at 16:20
I meant to say there exists a portion of the sphere below z=15? – CAF Nov 7 '12 at 16:45
@CAF: if you solved the quadratic equation correctly, there shouldn't be a portion of the sphere below $z = 15$. The intersection points of the sphere and the paraboloid are the boundaries of those regions. The other region is bounded above by $z = 0$, and hence is a single point. – Willie Wong Nov 8 '12 at 8:39
Here is the graph. And yes, we use cylindrical symmetry because there is symmetry with respect to the $z$ axis. – Willie Wong Nov 8 '12 at 8:43