# Surface area of solids of revolution

The question is to find the area of the surface that is generated by revolving the region bounded by $y^2 = x + 3$, $y^2 = 4x$ and $y\geqslant 0$ about the $x$-axis. We use the formula and find the areas $S_1$ (from $x=-3$ to $x=1$) and $S_2$ (from $x=0$ to $x=1$) for the curves respectively. Then we should find the total area by $S_1 - S_2$ but I don't understand why. I think that we should find it by $S_1 + S_2$ because $S_1$ generates the outer surface and $S_2$ generates the inner surface of the 3D shape between $x = -3$ and $x = 1$. I could not understand the reason when I asked it to the teacher. And I did not understand why we choose $x = 1$ as the end point because the curves reach beyond $x = 1$.

For your question about bounds of integration, the "region bounded by $y^{2} = x + 3$ and $y^{2} = 4x$" refers to the bounded (i.e., "finite") region defined by the inequalities $$y^{2} - 3 \leq x \leq \tfrac{1}{4}y^{2},$$ which lies between the lines $x = -3$ and $x = 1$ (and whose "right-hand boundary component" lies between $x = 0$ and $x = 1$).