Tilting a cylindrical drinking glass until 1/2 the base is exposed [UPDATE: Just found this: https://www.youtube.com/watch?v=ikY8wyZvTMQ it's part one of a video series explaining how to do these questions.]
If you have a glass that is 10in tall and 4in in diameter and then tilt it so water pours out until 1/2 the base is exposed

how would you write an integral for the volume of the water? I know that it has something to do with making an equation for the area of a triangle and integrating it but I don't know anything past that.
[I deleted the flawed revolution method work thanks to the comment by DanielV]
 A: If you align the cylinder so that 


*

*the axis of the cylinder is the $x$ axis

*the base of the cylinder is at $x=0$

*the top of the cylinder is at $x=1$

*the point the water is pouring out of the cylinder is $[x, y, z] = [1, -1, 0]$
Then for the plane of the surface of the water you get the equation $x + y = 0$ (use the 3 known points $[1, -1, 0], [0, 0, \pm 1]$).
Then setting up the integral 
$$\int_{x = 0}^{x = 1} {\rm d}x~\int_{z = -z(x)}^{z = +z(x)} {\rm d}z~ y_\text{water surface} - y_\text{cylinder edge}$$
$$\int_{x = 0}^{x = 1} {\rm d}x~\int_{z = -\sqrt{1 - x^2}}^{z = +\sqrt{1 - x^2}} {\rm d}z~ (-x) - (-\sqrt{1 - z^2})$$
Which by symbolic integration (using a computer) I get is equal to $2/3$, which suggests there is probably an easier way to do this.
Since you chose 10 inches tall and 4 inches in diameter that results in $10 \times (4/2)^2 \times 2/3 = (26 + 2/3)$ cubic inches total volume.
A: Let $r$ be the cylinder’s radius and $h$ its height. All of the cross sections of the volume taken orthogonal to the diameter that bounds it at the cylinder’s base are similar to a right triangle with side lengths $r$ and $h$. Placing the cylinder so that its base is centered at the origin and this diameter lies on the $x$-axis, integrate these cross sections: $$\int_{-r}^r \frac12 rh \left({\sqrt{r^2-x^2}\over r}\right)^2dx = \frac h{2r} \int_{-r}^r r^2-x^2\,dx = \frac23 hr^2.$$ For your cylinder, $r=2$ and $h=10$, therefore the remaining volume of water is $80/3$ cubic inches.
A: This is a variant of volume of Archimedes hoof. The problem has been carefully laid out and solved in the published article The Method of Archimedes: Propositions 13 and 14 and there are some animations and Mathematica$^{\circledR}$ code here: The Method of Archimedes: Propositions 13 and 14. Of particular interest is that the solution can be expressed as a single integral in many ways. Moreover, Archimedes gave an exact solution without the benefit of integration at all.
