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

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.

Tags:

Volume

Integration

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