We could do this with a memorized formula (skip to the end for the general formula), but here it is in a somewhat detailed form:
We are interested in the area bounded by
color(white)("XXX")color(red)(y=4-x^2)XXXy=4−x2
and the X-axis (where y=0y=0) and the Y-axis.
(Note that y=4-x^2y=4−x2 crosses the positive X-axis at x=2x=2)
![enter image source here]()
If we start by considering the color(green)("Area")Area of a circle formed by rotating color(red)(y=4-x^2)y=4−x2 for any arbitrary point color(green)x in [0,2]x∈[0,2]
![enter image source here]()
Then if we consider a thin disk with a thickness of color(blue)(dx)dx formed from this circular area:
![enter image source here]()
We can (hopefully) see that provided color(blue)(dx)dx is small, the sum of all such disk volumes will be close to the volume of the entire rotated function.
In fact, it would be reasonable to claim that the volume of the rotated image of y=4-x^2y=4−x2 is
color(white)("XXX")lim_(dxrarr0) Sigma_(x=0,"step:"dx)^2color(white)("xx") pi(4-x^2)^2
[Take a moment to consider this.]
This is, of course, just an expansion of the normal notation:
color(white)("XXX")int_0^2 pi(4-x^2)^2 dx
Expanding (4-x^2)^2 to x^4-8x^2+16
and then taking the indefinite integral:
color(white)("XXX")int_0^2 pi(x^4-8x^2+16) =pi((x^5)/5-(8x^3)/3+16x+C)
We have the required volume:
color(white)("XXX")pi((x^5)/5-(8x^3)/3+16x)|color(white)(uarr)_color(brown)0^color(magenta)2
color(white)("XXX")=color(magenta)(pi(32/5-64/3+32))-color(brown)(pi(0-0+0))
color(white)("XXX"256/15 pi
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
...and here is the promised general formula for a function f(x) bounded by the Y-axis and rotated about the positive X-axis:
color(white)("XXX")"Vol"=int_0^a pi(f(x))^2 dx
where a is the x-intercept.
If you followed through with the example, the reason for this general formula should be clear.
