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## Homework Statement

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##

## Homework 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}

## 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:

"The standard NWS rain gauge, developed at the start of the 20th century, consists of a funnel emptying into a graduated cylinder, 2 cm in diameter, which fits inside a larger container which is 20 cm in diameter and 50 cm tall. If the rainwater overflows the graduated inner cylinder, the larger outer container will catch it. When measurements are taken, the height of the water in the small graduated cylinder is measured, and the excess overflow in the large container is carefully poured into another graduated cylinder and measured to give the total rainfall"

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

\begin{equation}

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

\end{equation}

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?

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