Let $C$ be a solid cylinder with the height $h$ and whose base is a circle of radius $r$. And let $D$ be the result of slicing $C$ by a plane whose intersection with on base circle is a point and whose intersection with the other base circle is a diameter. Then I have to calculate the volume of $D$. A crude picture is as below
I think the voluem of $D$ is gained by accumulating the volume of each left half-disk of radius $r(1-x/h)$ with the infinitesimal height $dx$, where $x$ ranges from $0$ to $h$. As a result, the answer is
$$\int_0^hr^2(1-x/h)^2(\pi/2)dx=\pi r^2h/6.$$
Is my argument correct? Could anyone please explain more persuasively that the sliced bases at height $x$ forms a half-disk of radius $r(1-x/h)$?
Best Answer
The integral setup is incorrect. The sliced disk at height $x$ is not half circle. Rather, it is a cut of the circle by a chord starting $2r$ at the base and diminishing to $0$ at the height $h$.
The cut can be viewed as a circle sector, though, minus the overlapping triangle. Assume the angle of the sector at $x$ is $2\theta$, The area of the cut is given by
$$r^2\theta-r^2\sin\theta\cos\theta$$
The volume integral is, then,
$$V = r^2 \int_0^h (\theta-\sin\theta\cos\theta)dx$$
Furthermore, it can be shown that
$$\cos\theta = \frac{x}{h}$$
After the change of variable from $x$ to $\theta$, the volume integral becomes
$$V = r^2 h\int_0^{\pi/2} (\theta-\sin\theta\cos\theta)\sin\theta d\theta = \frac{2}{3}r^2h$$