# Find height of liquid inside spherical container as it is being filled

• Sep 14th 2010, 03:13 PM
surfertmex
Find height of liquid inside spherical container as it is being filled
Here is the problem:

"A spherical container with a radius of 15cm is being filled at a constant rate of 250╥ ml/sec. What is the height of the liquid from the bottom of the container after 4.5 seconds?"

I think you need to use disk integration to solve this one, but I am not sure. I could be over thinking this one.

Also, is it possible to derive an equation representing the height of the liquid as it is being filled vs time?
• Sep 15th 2010, 03:45 AM
HallsofIvy
Yes, treating the sphere as a "solid of revolution" and using the disk method to find the volume is probably the best way to do this.

First start by setting up a coordinate system. You could do this by setting the origin at the center of the sphere but I suspect it will be a little simpler to think about if we set the origin at the bottom of the sphere so that the "height of the liquid" is its z coordinate.. Then the center will be at (0, 0, 15). Looked at from the side, the sphere appears to be a circle, in the xz- plane, with center at (0, 15). Its equation is $\displaystyle x^2+ (z- 15)^2= 15^2= 225$. Each "disk" at a given z, seen from the side as here, appears to be a line segment from -x to + x where $\displaystyle x= \sqrt{225- (z- 15)^2$ from the equation of the circle. Rotated around the z-axis, the disk has area $\displaystyle \pi r^2= \pi (225- (z- 15)^2)$. The volume from z= 0 to z= h is given by $\displaystyle \pi\int_0^h (225- (z- 15)^2)dz$.

Find the volume of water poured in at $\displaystyle 250\pi ml/sec$ for 4.5 seconds, set that integral equal to it and solve for h.
• Sep 15th 2010, 04:27 AM
MathoMan
Quote:

Originally Posted by surfertmex
Here is the problem:

"A spherical container with a radius of 15cm is being filled at a constant rate of 250╥ ml/sec. What is the height of the liquid from the bottom of the container after 4.5 seconds?"

I think you need to use disk integration to solve this one, but I am not sure. I could be over thinking this one.

Also, is it possible to derive an equation representing the height of the liquid as it is being filled vs time?

Liquid inside the sphere would form a spherical cap and the volume of that cap $\displaystyle V_{cap}$ is the volume of the liquid $\displaystyle V_{liquid}$ inside the sphere. Then obviously the height $\displaystyle h$ of the spherical cap is the height of the liquid inside the sphere. Volume of the liquid can be calculated using the information given, and volume of the spherical cap can be calculated using the formula $\displaystyle V_{cap}=\frac{h^2 \cdot Pi}{3}(3R-h)$, where $\displaystyle R$ is the radius and $\displaystyle h$ is the height of the spherical cap. Forming the equation $\displaystyle V_{cap}=V_{liquid}$ makes it possible to find the height $\displaystyle h$.

Easy as $\displaystyle \pi$ and no integration whatsoever.