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
Best Answer
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 $y\lt0$
$$\iint_{R} v \cdot n \frac{dxdz}{|n \cdot j|} = \int_{0}^{3} \int_{-2}^{2} \left(\frac{4x^2}{y} - 2y^2\right) dxdz+\int_{0}^{3} \int_{-2}^{2} \left(\frac{4x^2}{-y} + 2y^2\right) dxdz=$$
$$= \int_{0}^{3} \int_{-2}^{2} \left(\frac{4x^2}{\sqrt{4-x^2}} - 2(4-x^2)\right) dxdz+\int_{0}^{3} \int_{-2}^{2} \left(\frac{4x^2}{\sqrt{4-x^2}} + 2(4-x^2)\right) dxdz=$$
$$=2\int_{0}^{3}dz \int_{-2}^{2} \left(\frac{4x^2}{\sqrt{4-x^2}}\right) dx=48\pi$$
The solution you cited uses cylindrical coordinates, far more easier as they adapt to the symmtry the problem has.