- #1

brotherbobby

- 669

- 159

- Homework Statement
- A basketball of mass 600 g and radius 12 cm floats at rest on the surface of water. Calculate the depth ##\boldsymbol{d}## of the ball that's under the liquid surface.

- Relevant Equations
- (1) Law of floatation : The mass of a floating body is the mass of liquid displaced - ##m_B=\Delta m_L##.

(2) Equation of a spherical surface : ##x^2+y^2+z^2=a^2##, where ##a## is the radius.

**Attempt :**(Turns out, there is more mathematics in this problem than physics. The crucial part involves the use of vector calculus where one needs to find the volume of a region bounded at the top by a portion of a sphere. That is where am stuck.)

The mass of water displaced by the ball ##\Delta m = 0.6\;\text{kg}##. This amounts to the volume of liquid displaced : ##\Delta V_L = 600\;\text{cm}^3##. This is also the volume of the ball inside liquid : ##\Delta V_B = 600\;\text{cm}^3##.

I make a sketch of the problem situation. I need to find the depth inside water ##d=?##

Hence the volume of the region under water is the difference of the volume of region ##{\color{red}{\stackrel{\large{\frown}}{CD}}}\text{AB}## and ##{\color{blue}{\overline{CD}}}\text{AB}##.

But now to find the volume of the region ##{\color{red}{\stackrel{\large{\frown}}{CD}}}\text{AB}## whose upper serface is a portion of a sphere?

This is where am stuck. I am aware that the equation of the upper surface ##\color{red}{\stackrel{\large{\frown}}{CD}}## is ##\color{red}{z=\sqrt{a^2-x^2-y^2}}##. The lower surface of the same region is a square of dimensions ##\sqrt{4ad-2d^2}##. But how to find its volume?

Given my lack of progress on this crucial but elementary point, any hint would be welcome.

Last edited: