Suppose we had a vector field $F = (axy^2,ayx^2,x^2\cos(\pi z))$ and wanted to calculate the surface integral of $$\int\int_SF\cdot n \ dS$$
where $S = \{(x,y,z): x^2 +y^2 = 1, 0 \leq z \leq 1/2 \}$
The question does this using cylindrical coordinates,
so we have: $F\cdot n = 2ax^2y^2 = 2ar^4\cos^2(\phi)\sin^2(\phi)$ but $r^2 = x^2 + y^2 = 1$ so $F\cdot n = 2ax^2y^2 = 2a\cos^2(\phi)\sin^2(\phi)$
now we have the Jacobian as $r$, and the question states that "using cylindrical integration we get:
$$2a \int_0^{1/2}\int_0^{2\pi} \cos^2(\phi)\sin^2(\phi) \ d\phi d z$$ but usually we have $drd\phi d z$ why only $\ d\phi d z$ in this case?
Because you're doing a surface integral (and there are precisely two degrees of freedom, i.e., two parameters, on a surface), not an integral over the solid cylinder.