A cylindrical beer glass has radius R and height h. It is partially filled with
some beer such that when you tilt the glass just enough for the beer to start
to pour out of the glass, exactly half of the bottom of the glass is still covered
by beer. Calculate the volume of beer in the glass by setting up a triple integral and then converting it to cylindrical coordinates to solve it.
What i tried
I could visualize the cylinder being titled such that half the length of the base of the cylinder is being covered with beer. While the entire height h of the glass is filled with beer.I first picture the z axis cutting through the middle of the cylinder parallel to the height of the cylinder and then looking down from the z axis.The equation of the z axis is $z=hx/R$ .Hence $0<z< hx/R$ for the iterated integral of z.What im stuck at is how to set up the iterated integral What i got was
$$\iiint hx/R\, dxdydz,$$
Could anyone help me with this. Thanks
Best Answer
You mentioned that you needed to solve this in the cylindrical coordinates system. You could try to first work out what the domain for your calculation is going to be. In the Cartesian system we have the function of $z = f(x,y) = \frac{h}{R}x$ bounded by the domain of $D=\left\{x^2+y^2 \le R^2,\, x \ge 0\right\}$. We have $x$ being greater than or equal to 0 because without such; there would be symmetry causing the two side to cancel each other out making the volume be equal to 0.
We now need to translate this into sufficient rules for our calculation in the cylindrical coordinates system. It is first obvious that $0 \le z \le \frac{h}{R}x$ which then translates into $0 \le z \le \frac{h}{R} \rho \cos \phi$. Talking about our problem with symmetry earlier, we can restrict our $\phi$ to only have half of a rotation, the half we want (half with beer). We simply get $-\frac{\pi}2 \le \phi \le \frac{\pi}2$. Lastly we have our $\rho$ which is every-so-exiting being $0 \le \rho \le R$...
Putting all of this into domain $T$ and integrating for our volume, we get $$ \begin{align} V &= \iiint_Tz\,\mathrm{d}z\,\mathrm{d}\rho\,\mathrm{d}\phi \\ V &= \int_{-\pi/2}^{\pi/2}\,\int_{0}^{R}\,\int_{0}^{\frac{h}{R} \rho \cos \phi}\rho\,\mathrm{d}z\,\mathrm{d}\rho\,\mathrm{d}\phi \\ V &= \frac{2 h R^2}{3} \end{align} $$