Is my procedure correct?
Calculate paraboloid area portion of equations
$$P \equiv (u \cos v, u \sin v, u^2) $$
with $0 \leq v \leq \dfrac{\pi}{4}$ and $0 \leq u \leq \dfrac{1}{2}\tan v$
This is my attempt:
$\bullet$ vector first derivatives:
$$P'_u = (\cos v, \sin v, 2u)$$
$$P'_v = (-u\sin v, u\cos v, 0)$$
$\bullet$ cross product between $P'_u$ and $P'_v$
$$P'_u \wedge P'_v = (-2u^2\cos v, 2u^2\sin v, u)$$
$\bullet$ cross product module
$$N(u,v) = u\sqrt{4u^2 + 1} $$
$\bullet$ double integral
$$\displaystyle \iint_P u\sqrt{4u^2 + 1} \; du \; dv = \int^{\beta(v)}_{\alpha(v)} u\sqrt{4u^2 + 1} \; du$$
where $\alpha(v) = 0$ and $\beta(v) = \dfrac{1}{2} \tan \dfrac{\pi}{4} = \dfrac{1}{2}$
$$\displaystyle \int^{\beta(v)}_{\alpha(v)} u\sqrt{4u^2 + 1} \; du = \displaystyle \dfrac{1}{12} (4u^2 + 1)^{3/2} \Big|^{u=1/2}_{u=0} = \dfrac{1}{12}(2\sqrt{2} - 1)$$
Thank you in advance
The partial derivative $P_u'$ shouldn't have a minus sign in the second component. Otherwise, everything was fine until you inserted everything into the integral. Notice that your bound for $u$ depends on $v$, so you should have \begin{align} \int_P \, \mathrm dA &= \int_{v=0}^{\pi/4} \int_{u=0}^{(\tan v)/2} u\sqrt{4u^2+1} \,\mathrm du \, \mathrm dv\\ &= \frac{1}{12}\int_{v=0}^{\pi/4}\left((\tan^2v+1)^{3/2}- 1\right) \, \mathrm dv \\ &= \frac{1}{12}\int_{v=0}^{\pi/4} \sec^3 v \, \mathrm dv - \frac{\pi}{48}, \end{align} etc.