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.

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)