Flux through a side of a cylinder

My troubles come with calculating the flux perpendicular to the cylinder's axis (ie, radial direction; $$S_3$$) through the surface. What I'd do is: $$\iint_{R} v \cdot n \frac{dxdz}{|n \cdot j|} = \int_{0}^{3} \int_{0}^{2} (\frac{4x^2}{y} - 2y^2) dxdz$$

But it doesn't yield $$48\pi$$.

The book provides another method which indeed yields the expected solution: Why am I wrong?

I don't really understand the book's method; so if you want to provide an explanation on that as well I'd be grateful for it.

Thanks

You posed well the integral, but some things have to be fixed: the range for $$x$$ is $$-2\leq x\leq 2$$; the integral has to be done for $$y=\sqrt{4-x^2}$$, one half of the cylinder, and for $$y=-\sqrt{4-x^2}$$, the other half and, further, we are dealing with the absolute value of $$y$$ in $$|n \cdot j|$$, so we have to be careful with the signs in some expressions: $$y^3/|y|=y^2$$ if $$y\geq0$$ but $$y^3/|y|=-y^2$$ if $$y\lt0$$
$$\iint_{R} v \cdot n \frac{dxdz}{|n \cdot j|} = \int_{0}^{3} \int_{-2}^{2} \left(\frac{4x^2}{y} - 2y^2\right) dxdz+\int_{0}^{3} \int_{-2}^{2} \left(\frac{4x^2}{-y} + 2y^2\right) dxdz=$$
$$= \int_{0}^{3} \int_{-2}^{2} \left(\frac{4x^2}{\sqrt{4-x^2}} - 2(4-x^2)\right) dxdz+\int_{0}^{3} \int_{-2}^{2} \left(\frac{4x^2}{\sqrt{4-x^2}} + 2(4-x^2)\right) dxdz=$$
$$=2\int_{0}^{3}dz \int_{-2}^{2} \left(\frac{4x^2}{\sqrt{4-x^2}}\right) dx=48\pi$$