Calculating the volume of a solid of revolution obtained by rotating two parabolas I'm trying to calculate the volume of a solid of revolution obtained by rotating two parabolas, $y=x^2-x-6$ and $y=-x^2+x+6$.
I want to use the formula: volume $=\pi \int_{a}^{b}(g^2(x)-f^2(x))dx$.
I already calculated $a,b = -2, +3$, which are intersections of the two parabolas.
However, I am not sure which function to put for $g^2$ and which to put for $f^2$. I know that the "upper" function should be $g(x)$, however, which one is upper in this case?
When using the latter function as $g(x)$ (because on the interval it has higher values), I get $0$ as a result of the integral. Not sure what to do now. 
Thanks
 A: 
As shown in Fig. 1, the area bounded by the two parabolae is symmetrical about the $x$-axis. This allows us to choose $x$-axis as the axis of rotation of the generated solid in question. Furthermore, instead of rotating both parabolae simultaneously through an angle of $\pi$, we can rotate one of them through an angle of $2\pi$. We opt for the parabola $y=-x^2+x+6$. Now, consider a yellow strip with height $y$ and width $dx$. Volume $dV$ generated by the strip when it is rotated by an angle $2\pi$ is given by
$$dV=\pi y^2 dx=\pi\left(-x^2+x+6\right)^2dx=\pi\left(x^4-2x^3-11x^2+12x+36\right)dx.$$
The integration that yields the total volume runs along $x$-axis from $x=-2$ to $x=+3$. Therefore, the total volume of the solid of revolution is,
$$V=\int dv=\pi\int^{+3}_{-2}\left(x^4-2x^3-11x^2+12x+36\right)dx,$$
$$V=\pi \left|\frac{x^5}{5}-\frac{x^4}{2}-\frac{11x^3}{3}+6x^2+36x\right|^{+3 }_{-2}=\frac{625}{6}\pi.$$
I have included Fig. 2 to show you that it is also possible to find the required volume by considering a horizontal strip and running the integration in $y$ direction from $y=0$ to $y=6.25$ and finally multiplying the result of the integration by 2. In this instance the axis of rotation is the vertical line given by $x=0.5$.
I assume that you have find out why you got a solid of revolution with a zero volume in your attempt earlier. @Kostya_I has described the reason in her comment.
