First of all, for the glass half-full, tilting it so that the water just covers the base will also put the water level at the upper rim, as shown in the figure below. Here we also show why $\tan\theta=h/2r$. Thus, we correct your result as follows

$$I=2\pi r^2h\div(2h/r)\times(h/2r)=\frac{\pi r^2h}{2}$$

Two different methods of solution :

**First (analytical) method :**

First of all, in this kind of volume calculation,

a) either you work with a triple integral as @StubbornAtom as done in his answer

b) or (often preferable if it is possible) express your volume as the volume *over* a certain domain $D$ *under* a certain surface with equation $z=f(x,y)$, as a double integral :

$$\int\int_D f(x,y)dxdy$$

This is what we are going to do here with $f(x,y)=z=2-x-y$:

$$\int_{y=0}^{y=2}\left(\int_{x=0}^{x=2-y}(2-x-y)dx\right)dy$$

(please note the parentheses ; bounds of integration can be understood when looking at Fig. 1).

*Fig. 1 : The way the domain of integration $D$ (in plane $xOy$) is swept, for a given $y$, from $x=0$ to $x=2-y$.*

The integrand $(2-y)-x$ of the internal integral has an antiderivative (with respect to variable $x$) :
$$(2-y)x-\frac{x^2}{2} \ \text{to be taken between} \ x=0 \ \text{and} \ x=2-y$$

giving for the final value of the internal integral

$$\frac{(2-y)^2}{2}$$

It now remains to compute : $\int_{y=0}^2 \frac{(2-y)^2}{2}dy$ ; its value is

$$\frac{-(2-y)^3}{6} \ \text{to be taken between} \ y=0 \ \text{and} \ y=2$$

which finally gives the result $\frac43$

**Second (geometrical) method :**

This volume is a tetrahedron $OABC$ that can be considered as well as a three-sided pyramid (lying on one of its sides), with equilateral triangle $ABC$ (with sidelength $2\sqrt{2}$) as its base. The centroid of this base equilateral base has coordinates the mean of coordinates of $A,B,C$:

$$I=\tfrac13((2,0,0)+(0,2,0)+(0,0,2))=(\tfrac23,\tfrac23,\tfrac23).$$

The height of this pyramid is distance $h=OI=2/\sqrt{3}$.

It remains to apply the formula giving the volume of a pyramid knowing the area $S$ of its base and its height $h$ :

$$V=\tfrac13 \times S \times h = \tfrac13 \times 2 \sqrt{3} \times \tfrac{2}{\sqrt{3}}=\tfrac43.$$

as in the first solution.

## Best Answer

The interval can be iteratively splitted up, i.e. first you have \begin{align*} &A' = \{ ( x_{1}, \ldots, x_{n-1})\mid x_{1} + \cdots + x_{n-1}\leqslant 1\} \\ &A_{(x _{1}, \ldots, x_{n-1})} = \{ x_{n}\mid x_{n} \leqslant 1 - x_{1} - \cdots - x_{n-1}\} \end{align*} yielding \begin{align*} \int_{A_{n}}^{} \mathrm{~d}\mu &= \int_{A'}^{}\left( \int_{A( x_{1}, \ldots, x_{n-1})}^{} \mathrm{~d}x _{n} \right)\mathrm{~d}\mu ( x_{n-1}, \ldots, x_{1}) \\ &= \int_{A'}^{} \int_{0}^{1 - x_{1} - \cdots - x_{n-1}} \mathrm{~d}x _{n} \mathrm{~d}\mu ( x_{n-1}, \ldots, x_{1}) .\end{align*} Continuing this procedure with $A'$ one ends up with \begin{align*} \sigma _{n} = \int_{0}^{1} \int_{0}^{1 - x_{1}} \cdots \int_{0}^{1 - x_{1} - \cdots - x_{n - 1}} \mathrm{~d}x _{n} \cdots \mathrm{~d}x _{2}\mathrm{~d}x _{1} .\end{align*} To compute this, we write \begin{align*} f _{n}( r) = \int_{0}^{r} \int_{0}^{r - x_{1}} \cdots \int_{0}^{r - x_{1} - \cdots - x_{n - 1}} \mathrm{~d}x _{n}\cdots \mathrm{~d}x _{2} \mathrm{~d}x _{1} .\end{align*} This allows us to establish a recursion, namely \begin{align*} f _{n+1}( r) &= \int_{0}^{r} \cdots \int_{0}^{r - x_{1} - \cdots - x_{n}} \mathrm{~d}x _{n+1}\cdots \mathrm{~d}x _{1} = \int_{0}^{r} f _{n}( r - x_{1})\mathrm{~d}x _{1} \\ &= -\int_{r}^{0} f _{n}( x_{1}) \mathrm{~d}x _{1} = \int_{0}^{r} f _{n}( x_{1}) \mathrm{~d}x _{1} \\[5pt] &\implies f _{n+1}'( r) = f _{n}( r) .\end{align*} With the base case $ f _{1}( r) = r$ this yields \begin{align*} f _{n}( r) = \frac{r ^{n}}{n!} \implies f _{n}(1) = \frac{1}{n!} .\end{align*}