A. To find the volume of water in the bowl, we will integrate the area of cross-section with respect to the depth x.
The cross-section of the bowl is a circle with radius r, and the area of a circle is given by A = πr^2.
Since the bowl is a hemisphere, the radius r is constant and equal to 12 cm.
Let's denote the depth x from the surface of the water. The radius of the cross-section at depth x can be calculated using the Pythagorean theorem as r(x) = √(12^2 - x^2).
The area of the cross-section at depth x is A(x) = πr(x)^2 = π(√(12^2 - x^2))^2 = π(144 - x^2).
To find the volume V of the water in the bowl, we integrate the cross-sectional area from 0 to 9 cm:
V = ∫[0,9] A(x) dx
V = ∫[0,9] π(144 - x^2) dx
V = π ∫[0,9] (144 - x^2) dx
V = π [144x - (1/3)x^3] |[0,9]
V = π [(144 * 9) - (1/3)(*9^3)] - [(144 * 0) - (1/3)(*0^3)]
V = π [1296 - (1/3) * 729]
V = π (1296 - 243)
V = 1035π cm^3
Therefore, the volume of water in the bowl is 1035π cm^3.
B. To find the mass of water in the bowl, we need to multiply the volume by the density.
The density at any point x below the surface is given by ρ(x) = 2e^(0.15x) g/cm^3.
The mass of an infinitesimally small volume element at depth x is dm = ρ(x) * dV = 2e^(0.15x) * dV.
To find the total mass M of the water, we integrate the mass density over the volume of water:
M = ∫[0,9] 2e^(0.15x) * dV
M = 2 ∫[0,9] e^(0.15x) * dV
Substituting the value of V from part A, we have:
M = 2 ∫[0,9] e^(0.15x) * 1035π dx
M = 2070π ∫[0,9] e^(0.15x) dx
M = 2070π (1/0.15) [e^(0.15x)] |[0,9]
M = 2070π (1/0.15) [e^(0.15*9) - e^(0.15*0)]
M = 2070π (1/0.15) [e^1.35 - 1]
M ≈ 29015 g
Therefore, the mass of water in the bowl is approximately 29015 grams.
C. When the bowl is inverted, the water reaches a new equilibrium. The depth of the water in the inverted bowl will be the same as the initial depth of 9 cm.
Therefore, the depth of the water in the inverted bowl is 9 cm.