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Find the surface area of the part of the cone $x^2+y^2=z^2$ with $z \geq0$ under the plane $2z=x+1$

I've sketched the figure but I'm not sure how to parameterize this one

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If you use $x=z\cos\phi$, $y=z\sin\phi$, so that on the border you have $2z=z\cos\phi+1$, i.e. $z=1/(2-\cos\phi)$, you get $$A=\int_0^{2\pi}\frac{\sqrt{2}z^2}{2}d\phi=\int_0^{2\pi}\frac{\sqrt{2}d\phi}{2(2-\cos\phi)^2}=\frac{2\sqrt{2}}{3\sqrt{3}}\pi$$ (with a high probability of a mistake).

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Looks fine to me (somewhat lowering the probability of a mistake). – joriki Oct 1 '12 at 20:25

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