# Solving a double integration in parametric form

I need help to find the surface area of the part of the paraboloid that lies in the first octant, meaning $x,y,z\ge 0$, of $z=5-x^2-y^2$.

So I found that I need to parametrize the paraboloid using cylindrical coordinates: $${\bf x}(u,v) = (u \cos v, u \sin v, 5 – u^2),$$with $0 \leq u \leq \sqrt{5}$ and $0 \leq v \leq \pi/2$. Then compute: $$\int_0^{\pi/2} \int_0^\sqrt{5}\|{\bf x}_u \times {\bf x}_v\|\,{\rm d}u\,{\rm d}v.$$ But I need help doing this. Can someone please help me continue on with this?

#### Solutions Collecting From Web of "Solving a double integration in parametric form"

$\vec x_{u}=(\cos v,\sin v,-2u)$

$\vec x_{v}=(u\sin v,-u\cos v,0)$

so

$\vert \vec x_{u}\times \vec x_{u}\vert =$

$\begin{vmatrix} \vec i&\vec j&\vec k\\ \cos v&\sin v&-2u\\ u\sin v&-u\cos v&0 \end{vmatrix} =\vert (2u^{2}\cos v)\vec i+(2u^{2}\sin v)\vec j-u\vec k\vert =u\sqrt {4u^{2}+1}$

from which we obtain

$\int_0^{\pi/2} \int_0^\sqrt{5}u\sqrt {4u^{2}+1}dudv=\frac{\pi}{24}(21\sqrt{21}-1)$.