Question

How do you find the volume bounded by `y=3-x^2` and y=2 revolved about the y=2?

Answer 1

Volume `=(16pi)/15` cu.units

Explanation:

(Sorry if this is in excessive detail)
The given regions in the Cartesian plane look like:

If this is rotated about `y=2`,
we have:

For ease of calculation, it is convenient if we shift this solid down so the axis of rotation falls along the X-axis:

Note that the radius (relative to the X-axis) of this shifted volume is
equal to `x^2-1` for `x in [-1,+1]`
and we can slice this solid into thin disks, each with a thickness of `delta`,
so that each disk has a volume of `delta * prr^2=pi * delta* (x^2-1)^2`

With very small values of `delta` the sum of the volumes of all such disks will be the volume of the rotated solid.

We can evaluate this sum with `deltararr0` using the integral:
`color(white)("XXX")int_(-1)^(+1) pi * (x^2-1)^2 dx`

`color(white)("XXX")=pi * int_(-1)^(+1) x^4-2x^2+1 dx`

`color(white)("XXX")=pi * ((x^5)/5-(2x^3)/3+x|_(-1)^(+1))`

`color(white)("XXX")=pi* ( (3-10+15)/15-((-3)-(-10)+(-15))/15)`

`color(white)("XXX")=pi * (8/15-(-8)/15)`

`color(white)("XXX")=16/15pi`