Surface Area Formula: (Circumference X Arch Length)
For revolution about $x$-axis: $S = \int_a^b 2\pi f(x)\sqrt{1+[f'(x)]^2}dx$ For revolution about $y$-axis: $S = \int_c^d 2\pi h(y)\sqrt{1+[h'(y)]^2}dy$
In the surface area formulas above, I understand that the $f(x)$ will just be what $f(x)$ (aka $y$) equal (e.g. for $y=x+4$, $f(x)$ in that formula will be $x+4$) but I do not understand what $h(y)$ will be. Will it just be $f(x)$ in terms of $y$? (e.g. for $y = x+4$, $h(y) = y-4$)