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Find the surface area of the shape bounded by paraboloid $x^2+y^2=z$ and cone $z=2-\sqrt{x^2+y^2}$.

$\cfrac{\pi(6\sqrt{2}+5\sqrt{5}-1)}{6}$

$\cfrac{\pi(\sqrt{2}+\sqrt{5}-1)}{6}$

$\cfrac{\pi(\sqrt{5}-1)}{3}$

$\cfrac{\pi(\sqrt{2}-1)}{3}$

None of the above