Volume of water in tilted cylinder - Why does answer not depend on $\pi$?

Question

A question is as follows: Consider a open top cylinder with radius $R$ and height $H$ full of water and tilt the cylinder to pour the water until the water surface at the base of the cylinder intersects the diameter of the base.

Find the volume of the water remaining.

My approach is to use multivariable integration, and I considered freezing the water and tilting the cylinder back to upright position, then setting the coordinate axis on the center of base of the cylinder. Then I found the plane containing the surface of water by finding 3 points:

(0,0,0), (R,0,0), (0,R,H), on the surface of water in the cylinder, which gave the plane $z=\frac{H\cdot y}{R}$

Finally, I integrated this using cylindrical coordinates, with z from 0 to the plane, and $D$ the region that is a semicircle $D = \{(x,y):x^2+y^2\leq R^2, y\geq0\}$

$$\iint_D\int_{z=0}^{\frac{H\cdot r\cdot \sin \theta }{R}}r dzdA,$$ where $D$ is just $\theta$ from $0$ to $\pi$ and $r$ from $0$ to $R$.

This yields the answer $\frac{2HR^2}{3}$. I find it suspicious the answer does not depend on $\pi$, so what did I do wrong? Or does the answer really not depend on $\pi$?

Answer 1

It is not surprising that integration involving circular region may result in a numerical outcome that no longer involves $\pi$.

It may be illustrative to consider integrating the mass of a unit disk $x^2+y^2\le 1$. If the density is uniformly 1, then its mass is just $\pi$. On the other hands, if the density varies as $|y|$, then its mass is integrated as,

$$\int_{x^2+y^2\le1} |y|dxdy=\int_0^{2\pi}\int_0^1 r^2|\sin\theta|dr d\theta = \frac43$$

which on longer depends on $\pi$. This should not be surprising because the variable density $|y|$ introduces compensating circular features and in the end 'nuetralizes' $\pi$.

Your volume integration

$$\int_{x^2+y^2\le R^2, \ y\ge 0}\frac{H }{R}y \ dxdy =\frac HR\int_0^{\pi}\int_0^1r^2\sin \theta dr d\theta=\frac23HR^2$$

is correct. The analogy here is that the height $\frac HR y$ is the equivalent of the variable density that provides extra circular features and eventually 'nuetralizes' $\pi$.

Answer 2

I think your answer is correct. Basically, after freezing the water and standing it upright, you are integrating a bunch of vertical rectangles with width $2\sqrt{R^2 - y^2}$ and height $Hy / R$, between $y = 0$ and $y = R$, so you have

$$ \int_0^R 2 \frac{y H}{R} \sqrt{R^2 - y^2} ~\mathrm{d}y $$

which you can evaluate using $u$ substitution and has no $\pi$.

You can alternatively slice it using horizontal slices, each area will be a piece of the circle and I believe the integral to look something like (up to possibly some numeric factors)

$$ \int_0^H R^2 \cos^{-1}(z / H) - \frac{z R^2}{H} \sqrt{1 - \frac{z^2}{H^2}} ~\mathrm{d}z $$

both portion of the integrals can be evaluated directly and be seen to not involve $\pi$.