# Calibrating a rain gauge (problem interpreting question)

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1. Apr 24, 2016

### TheSodesa

1. The problem statement, all variables and given/known data

We wish to make a precipitation meter shaped like a paraboloid $z = x^2 + y^2, 0 \leq z \leq 10$. Devise a scale on the z-axis that tells you the amount of precipitation in cm. In other words, at what height $z = h$ is the surface of water in the dish when there has been $a$ cm of rainfall?

correct answer: $h(a) = \sqrt{20}a$

2. Relevant equations

The rain meter exhibits cylindrical symmetry, so we might need to tackle the problem in cylindrical coordinates, where:

\begin{cases}
x = rcos(\theta)\\
y = rsin(\theta)\\
z = z\\
x^2 + y^2 = r^2\\
dV = r \ dz\ dr\ d\theta
\end{cases}

3. The attempt at a solution

This is a case of me seeing what the question is saying, but not understanding the setting. Specifically, what does it mean there has been $a$ cm of rainfall? Am I supposed to compare the volume of the paraboloid to some standard rain gauge with a certain radius?

The article on rain gauges on Wikipedia says that:

If my guess is right, in the case of the standard cylinder that we will call $c$, the volume $V_c$ of the water in $c$ is directly proportional to the height. Specifically

V_c = \pi r^2 h = \pi h,

since the radius $r = 1cm$.

I'm going to try and calculate the volume of the paraboloid dish $V_p$ as a function of R by integrating in cylindrical coordinates, set it equal to the volume gathered by $c$ and solve for $h$. In general, if we ignore the fact that we know $z_{max} = 10\ cm$, the paraboloid is limited by the cylindrical parameters as follows:
\begin{cases}
0 \leq r \leq R\\
0 \leq \theta \leq 2 \pi\\
r^2 \leq z \leq R^2 \text{ ||The surface of water is at some heigh } h = R^2
\end{cases}
Therefore the volume of the paraboloid
\begin{align*}
V_p
&= \iiint_{T}dV\\
&= \int_{0}^{2\pi}\int_{0}^{R}\int_{r^2}^{R^2} r\ dz\ dr\ d\theta\\
&= \int_{0}^{2\pi}\int_{0}^{R} \left[ rz \right]_{r^2}^{R^2}\ dr\ d\theta\\
&= \int_{0}^{2\pi}\int_{0}^{R}rR^2 - r^3 \ dr\ d\theta\\
&=\int_{0}^{2\pi} \left[ \frac{1}{2}r^2 R^2 - \frac{1}{4}r^4\right]_{0}^{R}\ d\theta\\
&= \int_{0}^{2\pi}\frac{1}{2}R^4 - \frac{1}{4}R^4\ d\theta\\
&= \int_{0}^{2\pi} \left[ \frac{1}{2}R^4 \right]_{0}^{2\pi}\\
&= \pi R^4 - \frac{1}{2}\pi R^4\\
&= \frac{1}{2}\pi R^4\ ||\ R^4 \implies cm^4 \text{, but ok...}
\end{align*}

Looking at this result, I immediately see that just setting $V_c = V_p$ is not going to help me solve for $h(a)$. What exactly is $a$, and how do I ""access"" it?

Last edited: Apr 24, 2016
2. Apr 24, 2016

### SammyS

Staff Emeritus
Your guess is incorrect. The paragraph from wiki does not pertain to this situation.

What are the radius and area of the (upper) opening of this rain gauge ?

3. Apr 24, 2016

### TheSodesa

Well, since the projection of the paraboloid onto the xy-plane is a circle with a radius of $R = \sqrt{10}$, the area of the opening is $A_R = \pi \sqrt{10}^2 = 10\pi$. I don't immediately see how this will help me...

Last edited: Apr 24, 2016
4. Apr 24, 2016

### SammyS

Staff Emeritus
If you have $\ a \,$cm of rain, what will be the volume of water in this rain gauge ?

5. Apr 24, 2016

### TheSodesa

This is exactly what I'm not seeing. $a$ is not the actual height of the surface of the water in cm. It is the reading given by the meter. But how is it connected to the rest of the dimensions of the vessel?

I get that the gauge scoops up water based on the size of its opening, so I can see how the area of the opening would be relevant, but...

I think my problem is that I don't know what it means to have $a$ cm of rain. Is it height per the surface area of the top of the gauge?

EDIT: I mean surface area per height?

6. Apr 24, 2016

### SammyS

Staff Emeritus
A rainfall of $\ a\,$cm of rain will deposit a volume of $\ 10\pi a\,$cm3 water in the rain gauge.

What will be the height of this volume of water in this paraboloid rain gauge? The mark at this height should match $\ a$ . It won't literally be at height $\ a\,$, but if $\ a\$ is 1cm, or 2cm, or 3cm, ... then the mark at the water level corresponding to any of those should be marked as 1cm, 2cm, 3cm, ... respectively.

7. Apr 24, 2016

### TheSodesa

If the volume of a paraboloid of radius R is
$$V_p = \frac{1}{2}\pi R^4 = \frac{1}{2}\pi h^2$$
then
$$h^2 = \frac{2V_p}{\pi} = \frac{2*10\pi a}{\pi} = 20a$$
and the height of the surface of the water is
$$h = \sqrt{20a}$$

And there we go. The answer sheet actually stated as much, but I though that it was a typo (because $\sqrt{cm}$ is not a thing as far as I'm aware), so I typed the answer in the OP as $\sqrt{20}a$.

Thank you for the help.

8. Apr 24, 2016

### SammyS

Staff Emeritus
Good.

By the way. It should be that if you go back through the workings, that constant coefficient, 20, under the radical should also have units of $\ \sqrt{\text{cm}}\ .$

Last edited: Apr 24, 2016
9. Apr 24, 2016

### TheSodesa

Of course. I'm just so used to seeing the units included every step of the way (because of physics) that the answer really threw me off.