Evaluate $$I=\iint_S (2x^2-y^3)dS$$ where $S$ is surface of cylinder $x^2+y^2=4$ between planes $z=0$ and $z=3$.
Here $\vec {n} = (x/2,y/2,0)$. By symmetry of cylinder around $y=0$, given integral becomes $I=\iint_S (2x^2)dS $ .
Now projecting on $yz$ plane (let S projection on $yz$ plane be $R$), $$I=\iint_R (2x^2)\frac {dy dz } {(\vec n \cdot \vec i)} = \int_{z=0}^{z-3} \int_{y=-2}^{y=2} 2x dy dz $$
Here again $x=\sqrt{4-y^2}$. Solving above integral I am getting $24 \pi$
But if i use cylindrical coordinates on 1st equation (in 1st line) i am getting $48 \pi$ . $$\int_{\theta=0}^{2 \pi}\int_{z= 0}{3} 2x^2-y^3 dS$$, where $dS=2d\theta dz$ $x=2\cos \theta$, $y = 2 \sin \theta$ $$= \iint((16 \cos^2 \theta - 16 \sin^3 \theta d) \theta dz = 48 \pi.$$
Please point me where i am making error.
Your evaluation in cylindrical coordinates is correct. We have that $dS=2d\theta dz$ and by simmetry the integral of $y^3$ is zero. Hence $$I=\iint_S 2x^2dS=\int_{z=0}^3\int_0^{2\pi} (2(2\cos(\theta))^2)2d\theta dz=48\int_0^{2\pi}\cos^2(\theta)d\theta=48\pi.$$ In cartesian coordinates, your evaluation should be doubled because you are considering just the semicircle given by $x=\sqrt{4-y^2}$ where $x\geq 0$.
An "easy" evaluation can be done by using Gauss divergence theorem. Let ${\bf F}:=(4x,0,0)$ then $$\iint_S 2x^2dS=\iint_S {\bf F}\cdot{\bf n}\,dS=\int_V\text{div}({\bf F})dV-\iint_D {\bf F}\cdot{\bf n}\,dS=4\cdot(2\pi\cdot 2\cdot 3)-0=48\pi$$ where $V$ is the solid cylinder and $D$ is the union of the two disks at $z=0$ and $z=3$.