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`...
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.
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