Find the area of the cap cut from the sphere $x^2 + y^2 + z^2 = 2$ by
the cone $z = \sqrt{x^2 + y^2}$. Doing this implicitly is straightforward, but I'm wondering what I'm doing wrong when I try to do this explicitly.
Using the parametrization $r(u,v)=(u cos(v),u sin(v), u)$ and taking the cross product I get $|r_u \, x \, r_v| = \sqrt{2}r$
which makes the surface area integral
$A = \sqrt{2} \int_{0}^{2\pi}{rdrd\theta} = \pi\sqrt{2}$
The book I'm self-studying from (Thomas) doesn't say when I can use explicit vs. implicit. Is this one that I just need to use implicit for? I assume there's something funky with my parametrization but I can't figure it out. Any help would be greatly appreciated!
Best Answer
The parametrization does not correspond to the cap. Try this parametrization. $$r(u, v) = (u\cos v, u\sin v, \sqrt{2-u^2})$$
Then $$r_u = \left(\cos v, \sin v, \frac{-u}{\sqrt{2-u^2}}\right)$$ and $$r_v = (-u\sin v, u\cos v, 0)$$ One can see geometrically and algebraically that $r_u$ is perpendicular to $r_v$. Thus, the surface element is $$\left\lVert \frac{dr}{du} \times \frac{dr}{dv}\right\rVert = \left\lVert \frac{dr}{du} \right\rVert \left\lVert \frac{dr}{dv}\right\rVert = u\sqrt{1 + \frac{u^2}{2-u^2}} = \frac{\sqrt{2}u}{\sqrt{2-u^2}}$$ Thus, the surface area is $$\iint_{[0,1] \times [0, 2\pi]} \frac{\sqrt{2} u}{\sqrt{2-u^2}} \: dudv = \sqrt{2} \int_{0}^{2\pi} \Big[\sqrt{2-u^2}\Big]_1^0 \: dv = \sqrt{2}(\sqrt{2}-1)2\pi = \boxed{(4 - 2\sqrt{2})\pi}$$