A glass in the shape of a circular cylinder of radius r

A glass in the shape of a circular cylinder of radius $r$ contains water. The axis of the glass is tilted at angle $\alpha$ to the horizontal. A segment of the base with central angle $2\varphi$ is covered by water. Find the volume of the water.

We set up coordinates so the wetted patch of the base is the circular segment nearest the point $(r,0)$, symmetric about the $x$-axis.

Step 1: Locate the chord bounding the wet segment

A circular segment of the disk $x^2+y^2\le r^2$ with central angle $2\varphi$, symmetric about the positive $x$-axis, is bounded by the arc from $(-\varphi)$ to $(\varphi)$ (in polar angle) and by the chord joining its endpoints $(r\cos\varphi,\pm r\sin\varphi)$. That chord is the vertical line $x=r\cos\varphi$, and the segment itself is the region

\[D=\{(x,y): x^2+y^2\le r^2,\ x\ge r\cos\varphi\}.\]

Step 2: Write the water depth as a linear function

Because the base plane and the (horizontal) water surface are both planes, the depth of water directly above a base point is a linear function of $x$ alone (by symmetry it does not depend on $y$). It must vanish along the chord $x=r\cos\varphi$ and increase toward the deepest point $(r,0)$, so

\[h(x)=(x-r\cos\varphi)\tan\alpha,\qquad r\cos\varphi\le x\le r.\]

This is positive exactly on $D$ and negative beyond the chord — consistent with the rest of the base lying dry, above the water line.

Step 3: Set up the volume integral

For each $x$ the segment $D$ has $y\in[-\sqrt{r^2-x^2},\sqrt{r^2-x^2}]$, so

\[V=\int_{r\cos\varphi}^{r}(x-r\cos\varphi)\tan\alpha\cdot 2\sqrt{r^2-x^2}\,dx = 2\tan\alpha\int_{r\cos\varphi}^{r}(x-r\cos\varphi)\sqrt{r^2-x^2}\,dx.\]

Step 4: Split and evaluate the two pieces

Write the integral as $J_1-r\cos\varphi\cdot J_2$, where

\[J_1=\int_{r\cos\varphi}^{r}x\sqrt{r^2-x^2}\,dx = \left[-\frac{1}{3}(r^2-x^2)^{3/2}\right]_{r\cos\varphi}^{r} = \frac{1}{3}(r^2-r^2\cos^2\varphi)^{3/2} = \frac{r^3\sin^3\varphi}{3},\]

\[J_2=\int_{r\cos\varphi}^{r}\sqrt{r^2-x^2}\,dx = \left[\frac{x}{2}\sqrt{r^2-x^2}+\frac{r^2}{2}\arcsin\frac{x}{r}\right]_{r\cos\varphi}^{r} = \frac{\pi r^2}{4}-\left(\frac{r^2}{2}\sin\varphi\cos\varphi+\frac{r^2}{2}\Big(\frac{\pi}{2}-\varphi\Big)\right),\]

using $\arcsin(\cos\varphi)=\frac{\pi}{2}-\varphi$. This simplifies to $J_2=\dfrac{r^2}{2}(\varphi-\sin\varphi\cos\varphi)$.

Step 5: Combine

\[V=2\tan\alpha\left[\frac{r^3\sin^3\varphi}{3}-r\cos\varphi\cdot\frac{r^2}{2}(\varphi-\sin\varphi\cos\varphi)\right] = r^3\tan\alpha\left[\frac{2}{3}\sin^3\varphi-\varphi\cos\varphi+\sin\varphi\cos^2\varphi\right].\]

Since $\frac{2}{3}\sin^3\varphi+\sin\varphi\cos^2\varphi=\sin\varphi\left(1-\frac{1}{3}\sin^2\varphi\right)$, this reduces to

\[\boxed{V=r^3\tan\alpha\left(\sin\varphi-\frac{1}{3}\sin^3\varphi-\varphi\cos\varphi\right)}\]

Step 6: Check limiting and special cases

As $\varphi\to0$ (an empty glass — the wet patch shrinks to nothing): $\sin\varphi,\ \sin^3\varphi,\ \varphi\to0$ and $\cos\varphi\to1$, so $V\to0$, as expected.

At $\varphi=\pi/2$ (chord through the center, exactly half the base wet): $\sin\varphi=1,\ \cos\varphi=0$, giving $V=r^3\tan\alpha\left(1-\frac{1}{3}\right)=\dfrac{2r^3\tan\alpha}{3}$.

As $\varphi\to\pi$ (the wet segment expands to the whole disk — a completely full base): $\sin\varphi\to0,\ \cos\varphi\to-1,\ -\varphi\cos\varphi\to\pi$, so $V\to\pi r^3\tan\alpha$, matching the direct computation $\iint_{\text{disk}}(x+r)\tan\alpha\,dA=\tan\alpha\left(0+r\cdot\pi r^2\right)=\pi r^3\tan\alpha$ for a completely full tilted base. All three checks confirm the formula, and it was additionally verified against direct numerical integration of the double integral for several intermediate values of $\varphi$.