$$f= (x+y) $$ $$S = r(u,v) = \langle 4\cos(u),4\sin(u),v\rangle$$ $u$ bounds are $[0,\pi/2]$, $v$ bounds are $[0,9]$.
My work:
$$r_u \times r_v = \langle 4\cos(u), -4\sin(u)\rangle$$ $$\Vert r_u \times r_v\Vert = 4$$
$$\int_0^9\int_0^{\pi/2}(4\cos(u)+4\sin(u))\,4\,du\,dv= 72?$$
First of all u need to tell us what u are integrating. Here i am assuming that u want to calculate
$\int\int_S (x + y) dS$
U are correct in that $dS=\| n(u,v) \|$ where n is the normal on surface S
$r_u'(u,v)=(-4\sin{u},4\cos{u},0)$
$r_v'(u,v)=(0,0,1)$
So in that we get $$ \|n(u,v)\|=\|r_u'(u,v)\times r_v'(u,v)\|=\|(4\cos{u},4\sin{u},0)\|=\sqrt{16}=4 $$
With that we have
$$ 16\int_{0}^{9}\int_{0}^{\pi/2} \cos{u}+\sin{u} \space dudv=16*9*\int_{0}^{\pi/2} \cos{u}+\sin{u} \space du $$
$$ =144(\sin{u}-\cos{u})|_{0}^{\frac{\pi}{2}}=144(1-0-(0-1))=2*144=288 $$