Question

Find the surface area of x^2+y^2+z^2=9 that lies above the cone z= sqrt(x^@+y^2)

Answer 1

The cone equation gives

$z=\sqrt{x^2+y^2}\implies z^2=x^2+y^2$

which means that the intersection of the cone and sphere occurs at

$x^2+y^2+(x^2+y^2)=9\implies x^2+y^2=\dfrac92$

i.e. along the vertical cylinder of radius $\dfrac3{\sqrt2}$ when $z=\dfrac3{\sqrt2}$.

We can parameterize the spherical cap in spherical coordinates by

$\mathbf r(\theta,\varphi)=\langle3\cos\theta\sin\varphi,3\sin\theta\sin\varphi,3\cos\varphi\right\rangle$

where $0\le\theta\le2\pi$ and $0\le\varphi\le\dfrac\pi4$, which follows from the fact that the radius of the sphere is 3 and the height at which the sphere and cone intersect is $\dfrac3{\sqrt2}$. So the angle between the vertical line through the origin and any line through the origin normal to the sphere along the cone's surface is

$\varphi=\cos^{-1}\left(\dfrac{\frac3{\sqrt2}}3\right)=\cos^{-1}\left(\dfrac1{\sqrt2}\right)=\dfrac\pi4$

Now the surface area of the cap is given by the surface integral,

$\displaystyle\iint_{\text{cap}}\mathrm dS=\int_{\theta=0}^{\theta=2\pi}\int_{\varphi=0}^{\varphi=\pi/4}\|\mathbf r_u\times\mathbf r_v\|\,\mathrm dv\,\mathrm du$
$=\displaystyle\int_{u=0}^{u=2\pi}\int_{\varphi=0}^{\varphi=\pi/4}9\sin v\,\mathrm dv\,\mathrm du$
$=-18\pi\cos v\bigg|_{v=0}^{v=\pi/4}$
$=18\pi\left(1-\dfrac1{\sqrt2}\right)$
$=9(2-\sqrt2)\pi$