A spherical balloon with radius 3 inches is partially �filled with water. If the water in
the balloon is 4 inches deep at the deepest point, how much water is released when
the balloon hits the wall and breaks?
not an easy problem
follow these steps:
http://mathworld.wolfram.com/SphericalCap.html
or
http://en.wikipedia.org/wiki/Spherical_cap
or
http://keisan.casio.com/has10/SpecExec.cgi?id=system/2006/1223382199
To determine how much water is released when the balloon hits the wall and breaks, we need to find the volume of the water inside the balloon.
First, let's find the volume of the entire balloon. The formula to calculate the volume of a sphere is:
V_sphere = (4/3) * π * r^3
where V_sphere is the volume, π is a mathematical constant approximately equal to 3.14159, and r is the radius.
Given that the radius of the balloon is 3 inches, we can substitute this value into the formula:
V_sphere = (4/3) * π * 3^3 = (4/3) * 3.14159 * 27 = 113.097 inch^3
Now, let's find the volume of water in the balloon. Since the water is 4 inches deep at the deepest point, we can consider it as a spherical cap.
The formula to calculate the volume of a spherical cap is:
V_cap = (1/3) * π * h^2 * (3r - h)
where V_cap is the volume of the spherical cap, h is the height of the cap, and r is the radius.
In this case, the height of the cap is 4 inches, and the radius is 3 inches.
V_cap = (1/3) * π * 4^2 * (3 * 3 - 4) = (1/3) * 3.14159 * 16 * (9 - 4) = 251.327 inch^3
Finally, to determine the amount of water released when the balloon breaks, we subtract the volume of the cap from the volume of the sphere:
Volume of water released = V_sphere - V_cap = 113.097 inch^3 - 251.327 inch^3 = -138.23 inch^3
The negative value indicates that there is no water released since the volume of the cap is greater than the volume of the sphere.