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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 $x = 1$.

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For finding the surface area, you're perfectly correct that the separate areas are added. (If instead you were finding the volume swept out when the stated region is revolved, you'd subtract the "inner" volume from the "outer".)

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 $x = 1$).

A region bounded by two parabolas

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