# `y = sqrt(9-x^2)` 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 find the volume of a solid by revolving the graph of `y =sqrt(9-x^2)` about the x-axis, we consider the bounded region in between the graph and the x-axis. To evaluate this, we apply the **Disk method** by using a *rectangular strip perpendicular to the axis of rotation*. As shown on the attached image, we consider a **vertical rectangular strip with a thickness =dx**.

We follow the formula for the Disk Method in a form of:` V = int_a^b pir^2 dx ` or `V = pi int_a^b r^2 dx` where *r is the length of the rectangular strip*.

In this problem, we let the *length of the rectangular strip*`=y_(above)-y_(below)` .

Then `r =sqrt(9-x^2) -0=sqrt(9-x^2)`

Boundary values of x: `a= -3` to `b=3` .

Plug-in the values on the formula `V = pi int_a^b r^2 dx` , we get:

`V =pi int_(-3)^3 (sqrt(9-x^2))^2 dx`

`V =pi int_(-3)^3 (9-x^2) dx`

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

`V =pi *[ int_(-3)^3 (9) dx- int_(-3)^3(x^2) dx]`

For the integral of `int_(-3)^3 (9) dx` , we apply basic integration property: `int c dx = cx` .

`int_(-3)^3 (9) dx =9x|_(-3)^3`

For the integral of `int_(-3)^3(x^2) dx` , we apply Power rule for integration: `int x^n dx = x^(n+1)/(n+1)` .

`int_(-3)^3(x^2) dx = x^((2+1))/((2+1))|_(-3)^3` .

` =x^3/3|_(-3)^3.`

Then,

`V =pi *[ int_(-3)^3 (9) dx-int_(-3)^3(x^2) dx]`

`V =pi *[ 9x-x^3/3]|_(-3)^3`

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

`V =pi *[ 9(3)-(3)^3/3] -pi *[ 9(-3)-(-3)^3/3]`

`V =pi *[ 27-27/3] -pi *[ -27- (-27)/3]`

`V =pi *[ 81/3-27/3] -pi *[ (-81)/3- (-27)/3]`

`V =pi *[ 54/3] -pi *[ (-54)/3]`

`V =(54pi)/3 - ((-54pi)/3)`

`V =(54pi)/3 + (54pi)/3`

`V =(108pi)/3`

`V =36pi ` or `113.1 ` (approximated value)