Calculate the area of the part of the cone $\displaystyle \ z^2 = x^2 + y^2$ that is inside the cylinder $\displaystyle \ x^2 + y^2 \leq 2x$, outside the cylinder $\displaystyle \ x^2 + y^2 \leq 1$ and above the xy plane.

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Cylinder: (ucosv, usinv, u)

Attempt:

2x = 1 ---> x = 1/2 ---> y = $\displaystyle \sqrt(3)/2$

ucos(v) = 1/2

usen(v) = $\displaystyle \sqrt(3)/2$

Solving I get: u = $\displaystyle \sqrt(3)/2$

and

$\displaystyle \int_0^{2\pi}$$\displaystyle \int_0^{\sqrt{3}/3} u\sqrt{2}dudv$