$($a$)$ Determine the water level as a function of the mass density and a $($p $<mtext></mtext><mn>1</mn>$for this part$)$ as follows. Determine via a definite integral the volume of the capsule from its bottom

y$=-9$a to an arbitrary height y$,$ for $-9$a $$ y $<mtext></mtext><mn>4</mn>$; call this V$($y$).$

Note that you need a different formula for $-9$a $$ y $<mtext></mtext><mn>0</mn>$ and $0$ y If the capsule is submerged up to y$,$ then V$($y$)$ also gives the mass of water displaced $($since the density of

water is $1).$

Thus, the water level is the value of y$,$ call it $0$ y $,$ for which V$($y$)$ equals the total mass of

the capsule. Split into cases depending on whether V$(0)$ is greater than or less than the

total mass of the capsule.

$($b$)$ For p$=0\mathrm{.}5,$ what must a be in order for the water level to be y$\_0=0?$ What about for

p$=0\mathrm{.}8?$ Can you explain the change between these two cases from the physical problem? For fixed a$,$ what happens to the water level y$\_0$ as p approaches $1?$ Does that

make physical sense?

The $``$center plane'$\text{'}$ of the capsule is the value of y $=$ y$\_$c so that half of the capsule's mass is above y $=$ y$\_$c and half below it$.$ Given your function V$($y$),$ you can determine c y from

this definition, or it is probably easier to use the formula

Total Mass of Capsule

where dV$($y$)$ is the volume of the cross section at position y $($with width dy$).$

The $``$center$-$of$-$buoyancy plane'$\text{'}$ of the submerged capsule is the value of y $=$ y$\_$c so that half of the mass of the displaced water is above y $=$ y$\_$b and half below it$.$ The

corresponding integral $($now in terms of the water level $0$ y $)$ is

Total Mass of Water Displaced by the Capsule

where dV$($y$)$ is again the volume of the cross$-$section at position y$.$

Buoyancy theory suggests that the submerged capsule will be most stable if the center

plane is below the center$-$of$-$buoyancy plane.