Find the volume if the area bounded by the curve y=|1-square roo(1-x)|, x axis and the limits of x=-1 and x=1 is rotated around x axis?

We'll recall the formula of solid of
revolution:

V = pi*Integral [f(x)]^2 dx, where the limits
of integration are x = -1 and x = 1.

Let y = f(x) = |1 -
sqrt(1-x)|

We'll recall the property of absolute
value:

|a|^2 = a^2 => |f(x)|^2 = [f(x)]^2 = [1 -
sqrt(1-x)]^2

V = pi*Int [1 - sqrt(1-x)]^2
dx

We'll expand the square:

[1
- sqrt(1-x)]^2 = 1 - 2sqrt(1-x) + 1 - x

Int [1 -
sqrt(1-x)]^2 dx = Int [1 - 2sqrt(1-x) + 1 - x]dx

Int [1 -
sqrt(1-x)]^2 dx = 2Int dx - 2Int sqrt(1-x)dx - Int
xdx

We'll calculate Int sqrt(1-x)dx using
substitution:

1 - x = t => -dx =
dt

Int sqrt(1-x)dx = -Int t^(1/2)
dt

-Int t^(1/2) dt = -
t^(3/2)/(3/2)

2Int sqrt(1-x)dx =
-4(1-x)^(3/2)/3

Int [1 - sqrt(1-x)]^2 dx = 2x - x^2/2 +
4(1-x)^(3/2)/3

Now, we'll apply Leibniz Newton
formula:

Int [1 - sqrt(1-x)]^2 dx =  F(1) -
F(-1)

F(1) = 2 - 1/2

F(-1) =
-2 - 1/2 + 8sqrt2/3

F(1) - F(-1) = 2 - 1/2 + 2 + 1/2 -
8sqrt2/3

F(1) - F(-1) = 4 - 
8sqrt2/3

The requested volume of the solid of
revolution is V = pi*(4 -  8sqrt2/3).