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**lilaziz1** Hey I'm stuck on this problem. I don't know if I'm doing something wrong with the surface integral or not because after simplification, I got to an integral that I can't do. Here is the question:

Evaluate the surface integral:

$\displaystyle \int\int_S yz dS$

S is the surface with parametric equations x = u^2, y = usinv, z = ucosv 0 <= u <= 1, 0 <= v <= $\displaystyle \pi/2$

This is attempt:

$\displaystyle \vec{r}_u = <2u, sinv, cosv>$

$\displaystyle \vec{r}_v = <0, ucosv, -usinv>$

$\displaystyle \vec{r}_u \times \vec{r}_v = <-u, 2u^2sinv, 2u^2cosv>$

$\displaystyle |\vec{r}_u \times \vec{r}_v| = u\sqrt{4u^2 + 1}$

$\displaystyle \int \int_D (usinv)(ucosv)|\vec{r}_u \times \vec{r}_v| dA$

$\displaystyle \int_0^1 u^3\sqrt{4u^2+1}du \int_0^{\pi/2} sinv cosvdv$

I don't know how to evaluate the integral $\displaystyle \int_0^1 u^3\sqrt{4u^2+1}du$ :\. if my steps are incorrect, I would be obliged if someone can tell me what I'm doing wrong *and* how to evaluate that integral.

Thanks in advance!