# Problem based learning

• Jan 16th 2006, 07:08 AM
poppy_wolf
Problem based learning
I am not really sure what type of question is this. The question is below. Thank you. :confused:

A hollow hemispherical bowl of radius 4cm has a smaller solid spherical marble of radius 1cm at the bottom. If the bowl is filled with water to a depth of 1.5cm, find the volume of water in the bowl.
• Jan 16th 2006, 07:39 AM
dud
Calculate the volume of a hemispherical object with a radius of 1.5cm - the volume of the ball which is submerged in the water?
• Jan 16th 2006, 03:06 PM
ThePerfectHacker
Quote:

Originally Posted by poppy_wolf
I am not really sure what type of question is this. The question is below. Thank you. :confused:

A hollow hemispherical bowl of radius 4cm has a smaller solid spherical marble of radius 1cm at the bottom. If the bowl is filled with water to a depth of 1.5cm, find the volume of water in the bowl.

I would solve this problem with Calculus. The most important step is to know that given a hemisphere filled with water to depth of $\displaystyle a$ having radius $\displaystyle r$ then its volume is $\displaystyle \pi (\frac{2}{3}r^3-ar^2+\frac{1}{3}a^3)$. Then from that your subtract the volume of the marble inside which in this case has volume $\displaystyle \frac{4}{3}\pi$
• Jan 17th 2006, 08:03 AM
poppy_wolf
Quote:

Originally Posted by ThePerfectHacker
I would solve this problem with Calculus. The most important step is to know that given a hemisphere filled with water to depth of $\displaystyle a$ having radius $\displaystyle r$ then its volume is $\displaystyle \pi (\frac{2}{3}r^3-ar^2+\frac{1}{3}a^3)$. Then from that your subtract the volume of the marble inside which in this case has volume $\displaystyle \frac{4}{3}\pi$

Will the solution be like this..

Volume of bowl = 2/3 (4^3) - (1.5)(4^2) + 1/3 (1.5^3)
= 17 13/24

volume of water = 17 13/24 - $\displaystyle \frac{4}{3}\pi$

= 13.35287
• Jan 17th 2006, 02:13 PM
ThePerfectHacker
Did you forget to multiply by $\displaystyle \pi$ in the first one?
• Jan 18th 2006, 11:42 AM
earboth
Hello,

to make sure that I got your question I've attached a sketch of the situation as I understand it.

The marble has a diametre of 2 cm, that means it will stuck out of the water if there is only a depth of 1.5 cm.
Therefore you have to calculate two parts of spheres:

$\displaystyle V_{part\ of\ sphere}=\frac{\pi}{3}\cdot h^2 \cdot (3\cdot r-h)$

$\displaystyle V_{1}=\frac{\pi}{3}\cdot 1.5^2 \cdot (3\cdot 4-1.5)$

$\displaystyle V_{2}=\frac{\pi}{3}\cdot 1.5^2 \cdot (3\cdot 1-1.5)$

The difference of both volumes is the volume of water you're looking for.

Bye
• Jan 18th 2006, 01:34 PM
ThePerfectHacker
I see what you are saying earboth. I did not take into account the fact that the sphere stuck out of the water!

This is definetly a Calculus problem for volume. It is solved with the formula I posted here before.