Consider a hemispherical bowl with a radius of 12 cm. The bowl contains water to a depth of 9 cm. The density of the water at any point x cm below the surface of the water is given by 2e^(0.15x) g/cm^3.
A. Find the volume of the water in the bowl.
B. Find the mass of the water in the bowl.
C. Suppose that the bowl is covered and then inverted and that no water is lost in the process. Find the depth of the water in the inverted bowl.
Show steps if possible please
A. The volume of a spherical cap of height h of a sphere of radius r is
V = π/3 h^2(r-h)
This is a known formula, but you can always verify it by finding the volume made by rotating a circular segment. For example,
V = ∫[h,r] π(r^2-y^2) dy
B. The mass of any small disc of water is its area times its density.
So, integrating again, with x=r-y,
M=∫[h,r] π(r^2-y^2) *2e^(0.15(r-y)) dy
C. Subtract the volume for (A) from 2π/3 r^3 and solve for h in the volume formula.
A. To find the volume of the water in the bowl, we need to integrate the area of the circular cross-sections starting from the surface of the water to the bottom of the bowl.
We can use a variable, let's call it 'x', to represent the depth below the surface of the water. Since the radius of the bowl is 12 cm, the radius at any depth 'x' would be (12 - x) cm.
To find the area of a circular cross-section at depth 'x', we use the formula for the area of a circle: A = πr², where 'r' is the radius. So, the area of a circular cross-section at depth 'x' would be: A(x) = π(12 - x)².
We need to integrate this area function over the range of depths from 0 to 9 cm to find the total volume of the water:
V = ∫[0, 9] (π(12 - x)²) dx.
Evaluating this integral will give us the volume of the water in the bowl.
B. To find the mass of the water in the bowl, we need to integrate the product of the density and the volume over the same range of depths.
Given that the density of the water at any depth 'x' is 2e^(0.15x) g/cm³, the mass of a cylindrical slice at depth 'x' would be: dm = (2e^(0.15x))(π(12 - x)²) dx.
We can integrate this mass function over the range of depths from 0 to 9 cm to find the total mass of the water:
M = ∫[0, 9] (2e^(0.15x))(π(12 - x)²) dx.
Evaluating this integral will give us the mass of the water in the bowl.
C. To find the depth of the water in the inverted bowl, we need to consider that the volume of the water remains constant.
Since the bowl is inverted, the shape changes from a hemisphere to an inverted hemisphere. The entire volume of water needs to fit within this new shape.
The depth of the water in the inverted bowl is the new radius, let's call it 'd'. The volume of the inverted hemisphere is given by:
V = (2/3)πd³.
The volume of the water in the upright bowl should be equal to the volume of the inverted bowl:
V = ∫[0, 9] (π(12 - x)²) dx.
This equation can be solved to find the depth 'd' of the water in the inverted bowl.
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.