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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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    [Here is the graph](http://fooplot.com/plot/mrca8falo0). And yes, we use cylindrical symmetry because there is symmetry with respect to the $z$ axis.2012-11-08