# `y = -x^2 + 1, y=0` Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region about the y-axis.

To be able to use the shell method, a a rectangular strip from the bounded plane region should be parallel to the axis of revolution.

By revolving multiple rectangular strip, it forms infinite numbers of hollow pipes or representative cylinders.

In this method, we follow the formula: `V = int_a^b` (length * height * thickness)

or `V = int_a^b 2pi` * radius*height*thickness

where:

radius `(r)` = distance of the rectangular strip to the axis of revolution

height `(h)` = length of the rectangular strip

thickness = width  of the rectangular strip  as `dx` or `dy` .

For the bounded region, as shown on the attached image, the rectangular strip is parallel to y-axis (axis of rotation). We can a let:

`r=x`

`h=f(x) or h=y_(above)-y_(below)`

`h = -x^2+1 -0`

`h = -x^2+1`

thickness `= dx` with boundary values from `a=0` to `b=1` .

Plug-in the values on `V = int_a^b 2pi` * radius*height*thickness , we get:

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

`V =int_0^1 2pi(-x^3+x)dx`

Apply basic integration property:` intc*f(x) dx = c int f(x) dx` and

`int (u+v) dx=int (u) dx+ int (v) dx` .

`V = 2pi int_0^1 (-x^3+x)dx`

`V = 2pi[ int_0^1 (-x^3)dx +int_0^1 (x)dx]`

Apply power rule for integration: `int x^n dy= x^(n+1)/(n+1)`

`V = 2pi[ -x^(3+1)/(3+1) + x^(1+1)/(1+1)]|_0^1`

`V = 2pi[ -x^4/4 + x^2/2]|_0^1`

Apply definite integration formula: `int_a^b f(y) dy= F(b)-F(a)` .

`V = 2pi[ -(1)^4/4 + (1)^2/2] -2pi[ -(0)^4/4 + (0)^2/2]`

`V =2pi[-1/4+1/2]-2pi[0+0]`

`V =2pi[1/4]-2pi[0]`

`V =pi/2-0`

`V =pi/2 ` or `1.57 ` (approximated value)