# Find the volume of the solid obtained by rotating the region bounded by the given curves: y=x^2, x=y^2, about y=1. (The answer is `11pi/30`, how come? )

You need to find the limits of integration, such that:

`x^2 = sqrt x => x^4 = x => x^4 - x = 0 => x(x^3 - 1) = 0 => x(x - 1)(x^2 + x + 1) = 0 => {(x = 0),(x - 1 = 0),(x^2 + x + 1 != 0):}`

Hence, evaluating the limits of integration yields `x = 0` and `x = 1` .

You need to use the washer method, hence, you need to find the inner and outer radii, such that:

`R = 1 - x^2` (outer radius)

`r = 1 - sqrt x` (inner radius)

You need to use volume formula, such that:

`V = pi*int_0^1 (R^2(x) - r^2(x))dx`

`V = pi*int_0^1 ((1 - x^2)^2 - (1 - sqrt x)^2)dx`

`V = pi int_0^1 (1 - 2x^2 + x^4 - 1 + 2sqrtx - x) dx`

`V = pi int_0^1 x^4 dx - pi int_0^1 2x^2 dx - pi int_0^1 x dx + pi int_0^1 2sqrt x dx`

`V = pi(x^5/5 - 2x^3/3 - x^2/2 + 4/3xsqrt x)|_0^1`

Using the fundamental theorem of calculus yields:

`V = pi(1/5 - 2/3 - 1/2 + 4/3 - 0/5 + 0/3 + 0/2 - 0)`

Bringing the terms to a common denominator yields

`V = pi(6 - 20 - 15 + 40)/30`

`V = (11pi)/30`

**Hence, evaluating the volume of solid obtained by rotating the region bounded by the given curves, yields `V = (11pi)/30` .**