We are contracting you to help design a space capsule for NASA that will orbit the earth and eventually re-enter the atmosphere and splash down in the Indian Ocean. We are seeking your expertise to design the floating space capsule that will be convenient for the unloading astronauts. The capsule is the solid of revolution obtained by rotating around the y-axis, the region bounded by y= -4(x-3)/3, y= 2/3(x^2-9), and y=0. The part above y=0 is a cone; this is where the astronauts sit. The part below y= 0 is a revolved parabola; this is a re-entry shield. The capsule will float with the cone pointing straight up. We want to determine how much of the capsule will be above the water level. As you know, Archimedes' Principle says that the mass of the capsule will be equal to the mass of the water displaced by the capsule (i.e. the mass of the water that used to be where that part of the capsule is now. The density of water is 1Mg/m^3. (1 Mg = 1,000 kg.) We are considering what materials to make the capsule out of to suit our needs, so assume that the density of the capsule is of density of P mg/m^3, where 00 V (y)= (0,y) (-5,3) [(-4(x-3)/3)-(2/3(x^2-9)) = 56.8y Volume = 56.8y Mass of the capsule = 56.8*(32/3)p = 605.9 p 3. Our initial measurements indicate that when the density of the capsule is p9/13Mg/m^3, the water level will be at exactly y=0. Please confirm this. From Archimedes principle 1440 *1* g = (1440 + 605.9)P * g On solving we get p = 9/13 Mg/m^3 4. We will also need to find the water level as a function of the density p. Find a function when 0

9/13 1440+56.8y = (1440 + 605.9)p * g Y = [1440(p-1)+ 605.9p]/56.8 5. The center of buoyancy" is the center of mass displaced by an object. A floating object is most stable when its center of mass exactly below its center of buoyancy. We are worried that our capsule design will be unstable. Please choose some 0p <1 1