# Tilting a cylindrical drinking glass until 1/2 the base is exposed

[UPDATE: Just found this gem: 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]

• I would think it natural to still let the axis of the glass be the $x$ axis. Did you try that? – Arthur Jul 1 at 6:01
• My thought was that if I model the triangle and revolve it around the hypotenuse then divide it by 2 I would essentially have the volume of water. I just let the x-axis represent the hypotenuse. The problem is I'm not sure if my logic of revolving the triangle around the hypotenuse and dividing by 2 actually results in the correct volume. – Corey Jul 1 at 6:05
• There's not enough symmetry to do revolution. – DanielV Jul 1 at 6:26
• Step 1 for solving this is just assume the radius and height of the cylinder is $1$. Increasing the height increasing the volume by the same factor, and increasing the radius increases the volume by a square of that factor. So you can do all the scaling after you solve the problem. – DanielV Jul 1 at 6:40

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.

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.

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.