The region between $y=x^2+1$ and $y=-x+3$ is rotated about the $x$-axis.
I have to compute the volume. The intersection between these two curves are at $x=-2$ and $x=1$.
At first, I thought about $V(x)=\pi\displaystyle\int_{-2}^1(3-x-(x^2+1))^2dx=\frac{81\pi}{10}$, where the radius is the difference between the two curves.
Another approach I made was $V(x)=\pi\displaystyle\int_{-2}^1[(3-x)^2-(x^2+1)^2]dx=\frac{117\pi}{5}$. I'm not sure why both results are different, and which one is correct.
It might be easier to think about it in steps. So, we know the formula for the volume of a solid of revolution about the $x$ axis from $a$ to $b$, right?
$$V = \pi \int_a^b (f(x))^2 dx$$
Think of the $f^2$ as serving the role of a radius, and the integral as contributing the length by summing up over all of the super-small bits of length $dx$ on the interval.
If you have the area between two curves $f$ and $g$, though - let $f \ge g$ here - you can think of it not as revolving the area between $f$ and $g$, but rather as just revolving the outer area $f$ and then cutting a hole from it based on $g$, and you want the volume left over. That is, for instance,
$$V = \underbrace{\pi \int_a^b (f(x))^2 dx}_{\text{original volume}} - \underbrace{\pi \int_a^b (g(x))^2 dx}_{\text{hole you cut out}}$$
Simplifying, we get
$$V = \pi \int_a^b \Big( (f(x))^2 - (g(x))^2 \Big) dx$$
or, if you understand that $f^2(x)$ means $(f(x))^2$, you can make this a little neater to write:
$$V = \pi \int_a^b \Big( f^2(x) - g^2(x) \Big) dx$$
Thus your second approach would be the correct one.