Birdbath Volume Problem HELP!

[tmillermillert]

A birdbath 4 in deep has the shape of a segment of a sphere of diameter 16 in. It contains a decorative lead cannonball of diameter 6 in. If the birdbath is filled with water to a depth of x inches, how much water does it contain?

This problem was found in the chapter titled Applications of Integration section Volumes: The Disk Method.

The answer in the back of the book is 5 pi x^2 in^3

 

[daon2]

A birdbath 4 in deep has the shape of a segment of a sphere of diameter 16 in. It contains a decorative lead cannonball of diameter 6 in. If the birdbath is filled with water to a depth of x inches, how much water does it contain?

This problem was found in the chapter titled Applications of Integration section Volumes: The Disk Method.

The answer in the back of the book is 5 pi x^2 in^3

Set up axes so that both circles pass through (0,0) and are in a natural upright position. You have the following curves to contend with:

1) The "outer" radius: $\displaystyle R: x=\sqrt{64-(y-8)^2}$

2) The "inner" radius: $\displaystyle r: x=\sqrt{9-(y-3)^2}$

Here you may choose variables appropriate to your question in order to get an answer containing x. Since I used x already, I'll use h for water height. Also note that the implied restriction is that the water level cannot exceed 4 (here, it means 0<=y<=4). If the bird bath was large enough to encompass the cannon ball's diameter, the second curve would need to be piece-wise.

Now use the forumula:

$\displaystyle \displaystyle V = \pi\int_0^h\ R^2 - r^2 \ dy$

 

[tmillermillert]

Got the answer with your help

Thank you! So the problem was about a spherical cap and not a spherical segment. I was assuming that the bottom of the bird bath was flat and not curved.