# `y = sqrt(x+2) , y = x , 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 x-axis.

## Expert Answers

To be able to use the shell method, the 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 cylinder.

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

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To be able to use the shell method, the 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 cylinder.

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

This can be also expressed as:

`V=int_a^b (2pi*` radius * height *thickness of rectangular strip)

or  `V=int_a^b (2pir *` length of rectangular strip *thickness of rectangular strip) where r = distance of the rectangular strip from the axis of rotation.

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

`radius=y`

`height=y -(y^2-2) = y-y^2+2`

Note: When we squared both sides of `y =sqrt(x+2)` , we get:` y^2 =x+2.`

It can be rearranged into `x= y^2-2.`

thickness`=dy`

the boundary values of y is `a=0` to` b=2`

Then the integral set-up will be:

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

Simplify:`V= int_0^2 2 pi ( y^2-y^3+2y)dy`

Apply the basic integration property: `int c f(x) dx - c int f(x) dx`

`V= 2pi int_0^2 ( y^2-y^3+2y)dy`

Apply basic integration property: `int (u+-v+-w) dx = int (u) dx +- int (v) dx+- int (w) dx.`

`V= 2pi [int_0^2 ( y^2) dy-int_0^2(y^3)dy+int_0^2(2y)dy]`

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

`V= 2pi [y^3/3 - y^4/4+2*y^2/2]|_0^2`

`V= 2pi [y^3/3 - y^4/4+y^2]|_0^2`

Apply the definite integral formula:` int_a^b f(x) dx = F(b) - F(a)` , we get:

`V= 2pi [(2)^3/3 - (2)^4/4+(2)^2]-2pi [(0)^3/3 - (0)^4/4+(0)^2]`

`V =2pi[8/3 -4+4] -2pi[0-0+0]`

`V= (16pi)/3 -0`

`V= (16pi)/3 `  or `16.76 ` ( approximated value)

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