Evaluate the surface integral
$$\int_{S} (z + x^{2}y)\,dS.$$
$S$ is the part of the cylinder $y^{2} + z^{2} = 4$ that lies between the planes $x = 0$ and $x = 3$ in the first octant.
I did the calculation
Integrate[(Sin[v] + u^2 Cos[v]) 2, {v, 0, Pi/2}, {u, 0, 3}]
and got the answer $24$.
But it showed it is not correct, why?
The given region can be parameterized by
$$\mathbf r(u,v)=\langle x(u,v),y(u,v),z(u,v)\rangle=\langle u,2\cos v,2\sin v\rangle$$
where $0\le u\le3$ and $0\le v\le\dfrac\pi2$, as you suggest in your integration setup. However, just replacing $x,y,z$ with their parametric forms isn't enough. You also have to account for the change in coordinates, similar to the way you need to compute the Jacobian in an area/volume integral upon changing coordinates.
This is accounted for by taking the norm of the cross product of the partial derivatives of $\mathbf r(u,v)$ with respect to its independent variables.
$$\mathbf r_u=\langle 1,0,0\rangle$$ $$\mathbf r_v=\langle 0,-2\sin v,2\cos v\rangle$$ $$\implies\left\|\mathbf r_u\times\mathbf r_v\right\|=2\sqrt{\cos^2v+\sin^2v}=2$$
So the integral is given by
$$\iint_S(z+x^2y)\,\mathrm dS=\color{red}2\int_{u=0}^{u=3}\int_{v=0}^{v=\pi/2}(2\sin v+2u^2\cos v)\,\mathrm dv\,\mathrm du=\color{red}{48}$$
so that, in fact, the value of the integral should be twice what you found.