As shown on the above image, I have a cylinder with its part cut out. The cylinder is cut out by the plane including three points(A, B, and C). The line BC is the diameter of the base circle and the point A is the midpoint of the semicircle as shown above. The plane divides the original cylinder into two parts, and I want to know the volume of the smaller part.
I calculated as follows.
First, I assumed that the radius of the base circle is r, and the height of the cylinder is h.
Then I figured out the area of the cut part at height z.
$$S(z) = 2 \int_{(\frac{r}{h})z}^{r} \sqrt{r^2-x^2}\ dx $$
$$= r^{2}[cos^{-1}(\frac{z}{h}) – \frac{z}{h} \sqrt{1-(\frac{z}{h})^2}]$$
Finally, I integrated this area by the z-axis to earn the volume.
$$ V = \int_0^h S(z) dz$$
$$ = \frac{2}{3}r^2h $$
I am confused by this result because, at first(before I calculated the volume by integration) I thought the volume should be $ \frac{1}{4}\pi r^2h$, which is the quarter of the original cylinder. I thought so because the plane seemed to cut the exact half of the left side of the cylinder.
Please examine my calculation to check if I had made a mistake in the process of integration. Thank you.
Best Answer
The volume enclosed cannot be a quarter of the cylinder's total volume, because there is no symmetry that equates the desired volume with the part of the half-cylinder that lies above the planar cut. As such, there is no reason to expect that these two volumes are equal.
The volume calculation can be performed by taking cross-sectional slices perpendicular to the $y$-axis, which will be rectangles of width $$W(y) = 2 \sqrt{r^2 - y^2}$$ and height $$H(y) = h \frac{y}{r}.$$ Thus the differential volume of a representative slice for some $y \in [0,r]$ is $$dV = W(y) H(y) \, dy = \frac{2hy}{r} \sqrt{r^2 - y^2} \, dy.$$ The desired volume is therefore $$V = \int_{y=0}^r dV = \frac{h}{r} \int_{y=0}^r 2y \sqrt{r^2 - y^2} \, dy = \frac{h}{r} \left[-\frac{2}{3} (r^2 - y^2)^{3/2} \right]_{y=0}^r = \frac{2}{3} r^2 h.$$
It is an instructive exercise to compute the volume of the aforementioned region above the plane, which is given by the integral
$$V' = \int_{y=0}^r h\left( 1 - \frac{y}{r}\right) 2\sqrt{r^2 - y^2} \, dy.$$
Even without evaluating it, we can see that it cannot be equal to $V$. Indeed,
$$V' = \int_{y=0}^r 2h \sqrt{r^2 - y^2} \, dy - V,$$
and the remaining integral term is simply the volume of half of the cylinder, with value $\frac{\pi}{2} r^2 h$.
As an exercise, how would you set up the volume integral for $V$ if you were to take cross-sectional slices perpendicular to the $x$-axis instead of the $y$-axis? Instead of rectangles, what shape do the cross-sections have?