Let $f: [-1,1] \to \mathbb{R}$ be a continuously differentiable function such that $f(-1) = f(1) = 0$ and $0<f(x)\le 1$ for all $x \in (-1,1)$. Let $S$ be the surface in $\mathbb{R}^3$ obtained by revolving the curve $y=f(x)$ around the $x$-axis and $V$ the volume inside $S$. Determine the volume of the object obtained by removing this volume $V$ from the inside of the sphere of radius $2$ centered at $0$.
Edit: I think evgeny's hint (see below) in the comments is a good one to go with, so I computed the volume V of the surface S with triple integration of the 1-function, with
$-1 \le x \le 1$
$-1 \le z \le 1$
-min{f(x), $\sqrt(1-x^2)$} $\le y \le$ +min{f(x), $\sqrt{1-x^2}$},
as my bounds of integration.
The volume of the object is then $\frac {32\pi}{3}$ - 8*min{f(x), $\sqrt{1-x^2}$}.
Feel free to comment on this answer or offer another solution. My only concern is that we don't seem to have used the $C^1$ condition on f. (Unless it was given primarily so that we know the function is smooth enough from x=-1 to x=1, e.g., no sharp corners along the curve.)
Hint.
You have $$V=\pi \int_{-1}^1 f^2(x)dx$$ This equation is coming from the computation of the volume in cylindrical coordinates $$V=\int\int\int_V dxdydz=\int_{-1}^1\int_0^{2 \pi}\int_0^{f(x)} r dx d\theta dr$$
Now, the issue is that the object limited by the surface $S$ might not be included in the sphere of radius equal to $2$ and centered at the origin.