Multivariable problem involving integrals

• June 18th 2009, 10:59 AM
mjlaz
Multivariable problem involving integrals
I'm having problems with this and was looking for some insight.

"Randy is going to use a porcelain bowl to measure rain. The bowl has the shape $z=x^2+y^2$ from $z=0$ to $z=10$ inches. Randy must calibrate the bowl so that he can precisely measure the amount of rain that falls. What height of water in the bowl corresponds to an inch of rain? That is to say, a circular cylinder would have one inch of rain in it, or that the level ground would have standing water one inch deep."

Are there any ideas?
• June 18th 2009, 12:39 PM
the_doc
Work out the volume $V$ in the bowl up to a height $h$ in cylindrical polars as

$V = \int_0^h \pi r^2 \, \mathrm{d} z$

since the volume of a slice of thickness $dz$ would have volume $\pi r^2 \mathrm{d}z$ where $r^2 = z$ so

$V = \pi \int_0^h z \, \mathrm{d} z = \frac{\pi}{2} h^2$.

The volume, $V_1$ ,representing one inch of rain is the volume in a cylindrical container of the same cross-sectional area as the opening of the bowl. So

$V_1 = \pi R^2 \cdot 1$

where $R$ is the radius at the opening of the bowl and since $z=R^2$ when $z=10$ we have

$V_1 = 10 \pi$.

Equating so $V_1= V$ we have

$10 \pi = \frac{\pi}{2} h^2$,

so

$h = \sqrt{20} \quad \text{inches}$.