`y = x^2 , y = x^5` Set up and evaluate the integral that gives the volume of the solid formed by revolving the region about the x-axis.
To be able to use the Shell method, 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 2pirhdy`
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 x-axis (axis of rotation). We can let:
`h= f(y)` or `h =x_2-x_1`
Note: `y = x^5` is expressed as `x = y^(1/5)` and `y = x^2` is expressed as `x = y^(1/2)` .
thickness` = dy`
Boundary values of y: `a=0` to `b=1` .
Plug-in the values on `V = int_a^b 2pirhdy`
`V = int_0^1 2pi*y*(y^(1/5)-y^(1/2)) dy`
Apply basic integration property:` intc*f(x) dx = c int f(x) dx`
V = 2pi int_0^1 y( y^(1/5)-y^(1/2)) dy
Apply Law of Exponent: y^n*x^m = y^((n+m)).
V = 2pi int_0^1 (y^((1/5+1))-y^((1/2+1))) dy
V = 2pi int_0^1 (y^(6/5)-y^(3/2)) dy
Apply basic integration property:`int (u-v)dx = int (u)dx-int (v)dx` .
`V = 2pi [int_0^1 (y^(6/5))dy -int_0^1 (y^(3/2)) dy]`
Apply Power rule for integration: `int y^n dy= y^(n+1)/(n+1).`
`V = 2pi [y^((6/5+1))/((6/5+1)) -y^((3/2+1))/((3/2+1))]|_0^1`
`V = 2pi [y^((11/5))/((11/5)) -y^((5/2))/((5/2))]|_0^1`
`V = 2pi [y^(11/5)*(5/11) -y^(5/2)*(2/5)]|_0^1`
`V = 2pi [(5y^(11/5))/11 -(2y^(5/2))/5]|_0^1`
Apply definite integration formula: `int_a^b f(y) dy= F(b)-F(a)` .
`V = 2pi [(5(1)^(11/5))/11 -(2(1)^(5/2))/5]-2pi [(5(0)^(11/5))/11 -(2(0)^(5/2))/5]`
`V = 2pi [5/11 -2/5]-2pi [0 -0]`
`V= 2pi [3/55]-2pi `
`V= (6pi)/55` or `0.343` (approximated value)
Aside from Shell Method, we can also apply Washer method such that the rectangular strip should be perpendicular to the axis of revolution. As shown in the attached image, we use a vertical rectangular strip with thickness =dx. We may follow the formula for Washer Method in a form of:
`V =int_a^b pi ( (f(x))^2 -(g(x))^2) dx`
`V =pi int_a^b ( (f(x))^2 -(g(x))^2) dx`
where `f(x)` as function of the outer radius
`g(x)` as function of the inner radius
For each radius, we follow the `y_(above) - y_(below)` , we have `y_(below)=0` since it a distance between the axis of rotation and each boundary graph.
For the inner radius, we have: `g(x) = x^5-0=x^5` .
For the outer radius, we have:` f(x) =x^2-0=x^2`
Boundary values of x is `a=0 ` and `b =1` .
Plug-in the values in the formula `V = pi int_a^b( (f(x))^2 -(g(x))^2) dx` , we get:
`V =pi int_0^1 ( (x^2)^2 -(x^5)^2) dx`
Apply Law of Exponent:` (x^n)^m = x^((n*m))` .
`V =pi int_0^1 ( x^((2*2)) -x^((5*2))) dx`
`V =pi int_0^1 ( x^(4) -x^(10)) dx`
Apply the Power rule for integration formula : `int x^n dx =x^(n+1)/(n+1)` .
`V =pi*[ x^(4+1)/(4+1) -x^(10+1)/(10+1)]_0^1`
`V =pi*[x^5/5 -x^11/11]_0^1`
Apply the definite integral formula: `int _a^b f(x) dx = F(b) - F(a)` .
`V =pi*[(1)^5/5 -(1)^11/11]-pi*[(0)^5/5 -(0)^11/11]`
`V =pi*[1/5 -1/11]-pi*[0-0]`
`V =pi*[11/55 -5/55]-pi*`
`V =pi*[6/55 ]-0`
`V =(6pi)/55` or `0.343` (approximated value)