I found this problem in a study guide and I've been stuck at it:
Suppose that a hollow spherical object of outer radius $R_1$ and inner radius $R_2$ is dropped into a fluid such as water or oil. Determine how far the spherical object will sink into the fluid.
Then I found a similar problem (and its solution) in the book, Numerical Methods Using MATLAB, Fourth Edition:
Consider the physical problem that involves a spherical ball of radius $r$ that is submerged to a depth $d$ in water .... Assume that the ball is constructed from a variety of longleaf pine that has a density of $\rho$ .... How much of the ball will be submerged when it is placed in water?
The solution given began as follows:
The mass $M_w$ of water displaced when a sphere is submerged to a depth $d$ is $$M_w = \int_0^d \pi(r^2 - (x - r)^2) dx = \frac{\pi d^2(3r-d)}{3},$$ and the mass of the ball is $M_b = 4 \pi r^3 \rho/3$. Applying Archimedes' law, $M_w = M_b$, produces the following equation that must be solved: $$\frac{\pi(d^3 - 3d^2r + 4r^3\rho)}{3} = 0.$$
What I don't understand is how the formula for $M_w$ was derived. What does the integral there represent?
I would really appreciate your help. Thanks
On paper, draw a circle of radius $r$ and then draw the X and Y axes so that the origin is at the $\it{bottom}$ of your circle. Starting from the X-axis, draw horizontal lines $\Delta x$ apart. Since it's really 3-D, you've made a bunch of disks whose horizontal surface area is given by the integrand for $M_w$.