For the region bounded by `y=-x+1` revolve about the x-axis, we can also apply the** Shell Method** using a* horizontal rectangular strip parallel to the axis of revolution (x-axis)*.We may follow the formula for Shell Method as:

`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` .

As shown on the attached file, the rectangular strip has:

`r=y`

`h =f(y) ` or `h = x_2-x_1`

`h = (1-y) -0 = 1-y`

Note: `y =-x+1` can be rearrange into `x=1-y` .

*Thickness* `= dy`

Boundary values of y: `a=0` to `b =1` .

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

`V = int_0^1 2pi* y*(1-y)*dy`

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

`V = 2pi int_0^1 y*(1-y)*dy`

Simplify: `V = 2pi int_0^1 (y-y^2)dy`

Apply basic integration property:`int (u-v)dy = int (u)dy-int (v)dy` to be able to integrate them separately using Power rule for integration: `int y^n dy = y^(n+1)/(n+1)` .

`V = 2pi *[ int_0^1 (y) dy -int_0^1 (y^2)dy]`

`V = 2pi *[y^2/2 -y^3/3]|_0^1`

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

`V = 2pi *[(1)^2/2 -(1)^3/3] -2pi *[(0)^2/2 -(0)^3/3]`

`V = 2pi *[1/2 -1/3] -2pi *[0-0]`

`V = 2pi *[1/6] -2pi *[0]`

`V = (2pi )/6 -0`

`V = (2pi )/6` or `pi/3`

We will get the same result whether we use Disk Method or Shell Method for the given bounded region revolve about the x-axis on this problem.

For the region bounded by `y=-x+1` and revolved about the x-axis, we may apply **Disk method**. For the Disk method, we consider a *perpendicular rectangular strip with the axis of revolution*. Note: The bounded region is only from the first quadrant.

As shown on the attached image, the** thickness of the rectangular strip** is "`dx` " with a *vertical orientation perpendicular to the x-axis* (axis of revolution).

We follow the formula for the Disk method:`V = int_a^b A(x) dx ` where disk's base area is `A= pi r^2` with `r =y=f(x)` .

Note:* r = length of the rectangular strip*.

We may apply `r = y_(above)-y_(below)` .

Then `r =(-x+1)-0 = -x+1`

The* boundary values of x* is `a=0` to `b=1` .

Plug-in the `f(x) ` and the boundary values to integral formula, we get:

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

Apply basic integration property: `intc*f(y) dy = c int f(y) dy` .

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

To find the indefinite integral, let `u = -x+1` then `du = -dx` or `(-1) du =dx`

The integral becomes: `V =pi int (u)^2*(-1) du`

Apply Power rule for integration: `int x^n dx = x^(n+1)/(n+1)` .

`V =pi * u^(2+1)/(2+1)*(-1)`

`V =-(piu^3)/3`

Plug-in `u=-x+1` on `V=-(piu^3)/3` , we get:

`V=-(pi(-x+1)^3)/3 or (pi(x-1)^3)/3 ` with boundary values `a=0` to `b=1`

Apply the definite integral formula: `int _a^b f(x) dx = F(b) - F(a)` .

`V =(pi(1-1)^3)/3 -(pi(0-1)^3)/3`

`V =0 - (-pi)/3`

`V = pi/3` or `1.05` (approximated value)

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