# 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?

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### 1 Answer

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).**