Ytukyg

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:

$$P'_u = (cos v, sin v, 2u)$$

$$P'_v = (-usin v, ucos v, 0)$$

$bullet$ cross product between $P'_u$ and $P'_v$

$$P'_u wedge P'_v = (-2u^2cos v, 2u^2sin v, u)$$

$bullet$ cross product module

$$N(u,v) = usqrt{4u^2 + 1} $$

$bullet$ double integral

$$displaystyle iint_P usqrt{4u^2 + 1} ; du ; dv = int^{beta(v)}_{alpha(v)} usqrt{4u^2 + 1} ; du$$

where $alpha(v) = 0$ and $beta(v) = dfrac{1}{2} tan dfrac{pi}{4} = dfrac{1}{2}$

$$displaystyle int^{beta(v)}_{alpha(v)} usqrt{4u^2 + 1} ; du = displaystyle dfrac{1}{12} (4u^2 + 1)^{3/2} Big|^{u=1/2}_{u=0} = dfrac{1}{12}(2sqrt{2} - 1)$$

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} usqrt{4u^2+1} ,mathrm du , mathrm dv\
&= frac{1}{12}int_{v=0}^{pi/4}left((tan^2v+1)^{3/2}- 1right) , mathrm dv \
&= frac{1}{12}int_{v=0}^{pi/4} sec^3 v , mathrm dv - frac{pi}{48},
end{align}
etc.