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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

enter image description here

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 $r(1-x/h)$?

Quanto
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Keith
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  • I would suggest integrating the area of the rectangles with base parallel to the cut diameter. While I don't particularly understand your solution, you cannot integrate volume to get volume, you can sum volumes to get volumes, or you can integrate area. – Gabe Aug 29 '19 at 01:58
  • Ok, my explanation may cause confusion. But is the result $\pi r^2 h/6$ a right one? – Keith Aug 29 '19 at 02:04
  • I meant to sum the volume elements $r^2(1-x/h)(\pi/2)dx$ to get the total volume. – Keith Aug 29 '19 at 02:10
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    The integral would be $\int_0^r 2\sqrt{r^2-x^2}\frac{xh}{r}$ which does not give what you got. This uses vertical rectangles parallel to the diameter of the base semicircle. I understand what you are trying to do now, and unfortunately the shapes formed by your slices are not semicircles. Notice how the curvature remains one of radius r, but the straight length gets shorter. You could instead integrate the areas of the sectors*, and you should receive the same answer as my integral. – Gabe Aug 29 '19 at 02:25
  • Oh they are not semicircles... – Keith Aug 29 '19 at 02:38
  • That was what I was anxious about...and now that I am wrong...Thank you so much for your explanation. – Keith Aug 29 '19 at 02:43
  • And does your integral have to include $dx$? You simply missed it? – Keith Aug 29 '19 at 02:46
  • Yes of course there should be a dx, sorry about that – Gabe Aug 29 '19 at 18:32

1 Answers1

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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 $h$.

enter image description here

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

$$V = r^2 h\int_0^{\pi/2} (\theta-\sin\theta\cos\theta)\sin\theta d\theta = \frac{2}{3}r^2h$$

Quanto
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