Question

How do you find the volume of the solid obtained by rotating the region bounded by the curves `y=x^2+1` and `y=-x+3` rotated around the x-axis?

Answer 1

I got area is `23.4 pi`. The "How" has no one sentence answer -- see below.

Explanation:

The curves: `y=x^2+1` and `y=-x+3` intersect at `-2` and at `1`.
On the interval `[-2, 1}`, the value of `-x+3` is greater than that of `x^x+1`. (The line is above the parabola.)

We'll use washers and find `int (pi R^2 - pi r^2) dx` (Where the larger radius is `R` and the smaller `r`.

`V = int_-2^1 (pi(-x+3)^2 - pi(x^2+1)^2) dx`

`= pi int_-2^1 ((x-3)^2 - (x^2+1)^2) dx`

`= pi int_-2^1 ((x^2-6x+9) - (x^4+2x^2+1)) dx`

`= pi int_-2^1 (-x^4-x^2-6x+8) dx`

`= pi[-x^5/5 -x^3/3-3x^2+8x]_-2^1`

`= pi[(-1/5-1/3-3+8)-(32/5+8/3-12-16)]`

`=pi[ -33/5-9/3+5+12+16]`

`= pi[-66/10 -3 +5+12+16]`

`= pi [-6.6 + 30]`

`= 23.4 pi`