`y = 4 - 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 =4-x^2 about the x-axis, we consider the bounded region in between the graph and the x-axis. To evaluate this, we apply 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 = (4-x^2) - 0 = 4-x^2`
Boundary values of x: `a= -2` to `b=2` .
Plug-in the values on the formula `V = pi int_a^b r^2 dx` , we get:
`V =pi int_(-2)^2 (4-x^2)^2 dx`
Expand using FOIL method:`(4-x^2)^2 = (4-x^2)(4-x^2)= 16-8x^2+x^4` .
The integral becomes:
`V =pi int_(-2)^2 (16-8x^2+x^4) dx`
Apply basic integration property:`int (u+-v+-w)dx = int (u)dx+-int (v)dx+-int(w)dx` to be able to integrate them separately using Power rule for integration: `int x^n dx = x^(n+1)/(n+1)` .
`V = pi[ int_(-2)^2(16) dx -int_(-2)^2(8x^2)dx+int_(-2)^2(x^4)dx]`
`V = pi[16x-(8x^3)/3+x^5/5]|_(-2)^2`
Apply definite integration formula:` int_a^b f(y) dy= F(b)-F(a)` .
`V = pi[16(2)-(8(2)^3)/3+(2)^5/5]-pi[16(-2)-(8(-2)^3)/3+(-2)^5/5]`
`V=(512pi)/15 ` or `107.23` (approximated value)