When calculating the volume of a solid of revolution defined by curves, I must necessarily consider only the intersections of these curves as the region of integration (i.e all pairs $(x,y)$ simultaneously satisfied by the curves) or is it correct to consider the region bounded by curves?
To further clarify the question, consider the following exercise from a Calculus textbook used at my university:
- Calculate the volume of the solid generated by the revolution around the axis $y $, from the set of pairs $(x, y) $ such that $ 0≤x≤2, 0≤y≤\dfrac{x^2}{2}+1$ and $y≥x^2-1$.
The figure given in the book (as well as the answer), takes into account the region delimited by these curves:
I believe that if we consider pairs $(x, y) $ that simultaneously satisfy all the inequalities of the statement, we should exclude the interval $ 0≤x < 1 $, because if $ x = 0 $ for example, not all inequalities would be satisfied simultaneously (we would have $y≥0$ and $y≥-1$).
Also related to this question, there is this other exercise in the book:
- Calculate the volume of the solid generated by the revolution around the axis $y$, of the set of pairs $(x,y)$ such that $0≤x≤2, 0≤y≤x^2 $ and $y≥ (x-1)^{1/2}$.
My friend says that all these problems should be considered points that satisfy all the curves simultaneously. Therefore, in problem $2$, the interval $0≤x≤1$ must be excluded (he doesn't agree too with the solution of the first exercise provided by the book, as he assumes that the same interval should be excluded). However, I believe that the problem $2$, the integration region includes the interval $0≤x≤1$. I find it intuitive that the piece of the curve $y = x^2$ in this interval is part of the solid that is obtained when the set is rotated around the y axis.
I appreciate any clarification on these questions.
Problem 1: When $x=0$, the second and third inequalities reduce to $0 \le y \le 1$ and $y \ge -1$, which is the same thing as saying just $0 \le y \le 1$, since for these values of $y$ the inequality $y \ge -1$ holds automatically. There's nothing contradictory about that. So the picture is correctly drawn; the region with $0 \le x < 1$ should be included.
Problem 2: I think this problem is unclear and ought to be reformulated in an unambiguous way. If I'm mindreading, I would guess that the intention was that the region $0 \le x < 1$ should be included (why write $0 \le x \le 2$ otherwise?), but on the other hand I would not consider it true that (for example) $(x,y)=(0,0)$ satisfies $y \ge \sqrt{x-1}$, since the right-hand side isn't even defined for $x=0$. So if you're reading it just as it's written, then indeed you need to have $x \ge 1$ in order to satisfy $y \ge \sqrt{x-1}$.