Find the volume of the solid formed by revolving the region bounded by y = 16 - x^2 and y = 0 about the x-axis .
To find the volume of the solid formed by revolving the given region about the x-axis, we can use the method of cylindrical shells.
First, let's find the points of intersection of the two curves by setting them equal to each other and solving for x:
16 - x^2 = 0
x^2 = 16
x = ±√16
x = ±4
So, the region is bounded by x = -4 and x = 4.
Next, let's express the equations of the curves in terms of y:
y = 16 - x^2
x^2 = 16 - y
x = ±√(16 - y)
Now, let's imagine a vertical strip of width Δy, located at a distance y from the x-axis. This strip will be a cylindrical shell with inner radius x and height Δy.
The volume of this cylindrical shell is given by:
V = 2πxΔy
To find the total volume, we need to integrate this expression over the region:
V = ∫[from 0 to 16] of (2πxΔy) dy
We can express x in terms of y as ±√(16 - y) and substitute it into the integral:
V = ∫[from 0 to 16] of (2π(±√(16 - y))Δy) dy
Now, we can simplify the expression inside the integral:
V = 2π∫[from 0 to 16] of (√(16 - y)Δy) dy
Taking the integral, we have:
V = 2π∫[from 0 to 16] of (√(16 - y)Δy) dy
V = 2π∫[from 0 to 16] of (16 - y)^(1/2) dy
This integral can be evaluated using the power rule for integrals:
V = 2π[(2/3)(16 - y)^(3/2)]|[from 0 to 16]
V = 2π[(2/3)(16 - 16)^(3/2) - (2/3)(16 - 0)^(3/2)]
V = 2π[(2/3)(0) - (2/3)(16)^(3/2)]
V = 2π(0 - (2/3)(16)^(3/2))
V = -2π(2/3)(16)^(3/2)
V = -2π(2/3)(64√2)
V = -128π√2/3
The volume of the solid formed by revolving the region bounded by y = 16 - x^2 and y = 0 about the x-axis is -128π√2/3.