# `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.

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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`

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 r**ectangular strip is parallel to x-axis (axis of rotation)**. We can let:

`r=y`

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

`h=y^(1/5)-y^(1/2)`

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 [0]`

`V= (6pi)/55-0`

`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`

or

`V =pi int_a^b ( (f(x))^2 -(g(x))^2) dx`

where:

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*[0]`

`V =pi*[6/55 ]-0`

`V =(6pi)/55` or `0.343` (approximated value)