Why the wrong answer in integral calculus? Calculate the volume of the solid obtained by rotating the region about the $x$-axis, and bounded by $y=x^2+1, y=3-x^2$.

My calculations:
$V=\pi \int_a^b f^2(x) dx; \; V = V_1 + V_2 = \pi \int_{-1}^1 (3-x^2)^2 dx - \pi \int_{-1}^1 (x^2+1)^2 dx;$ 
$V_1 = \pi \int_{-1}^1 (3-x^2)^2 dx = \pi \int_{-1}^1 (9-3x^2-3x^2+x^4) dx = \pi (9x-2x^3+\frac{x^5}{5} \; |_{-1}^1 )=14.4 \pi;$
\begin{align*}
V_2 & = \pi \int_{-1}^1 (x^2+1)^2 dx\\
    & = \pi \int_{-1}^1 (x^2+1)(x^2+1) dx\\ 
    & = \pi \int_{-1}^1 (x^4+x^2+x^2+1) dx\\ 
    & = \pi\left(\frac{x^5}{5}+\frac{2}{3} x^3+x \; \bigg|_{-1}^1\right)\\
    & = \pi\left(\frac{1}{5}+\frac{2}{3}+1+\frac{1}{5}+\frac{2}{3}+1\right)\\             
    & = 2\pi\left(\frac{1}{5}+\frac{2}{3}+1\right)\\
    & =2\pi\left(\frac{3}{15}+\frac{10}{15}+1\right)\\
    & =\left(2+\frac{26}{15}\right)\pi;
\end{align*}
My answer: $V=14.4\pi - (2+\frac{26}{15})\pi;$, but this answer does not fit.
 A: I think it's worth noting that your method, while completely correct (both in result and in detail as far as anyone has found), is far more laborious (and hence more error-prone)  than it needs to be.
In this case, the "washer method" is easier.
For the washer method, we note that a given value of $x$ the disk inside the curve $y=3-x^2$ has area $\pi(3-x^2)^2$ and the disk inside the curve $y = x^2 + 1$ has area $\pi(x^2 + 1)^2$; then we remove the smaller disk from the larger disk, leaving a "washer" (also called an annulus), which is the shape we get when we take the line segment from  $(x,x^2 + 1)$ to $(x,3-x^2)$ and rotate it around the $x$-axis.
The area of the washer is 
$$\pi(3-x^2)^2 - \pi(x^2 + 1)^2 = \pi ((3-x^2)^2 - (x^2 + 1)^2)$$
(in general, $\pi(r_1^2 - r_2^2)$ where $r_1$ is the larger radius and $r_2$ is the smaller),
which we integrate from $-1$ to $1.$ Note that this is exactly equivalent to your method, since
$$ \int_{-1}^1 (\pi(3-x^2)^2 - \pi(x^2 + 1)^2)\,\mathrm dx
= \int_{-1}^1 \pi(3-x^2)^2\,\mathrm dx - \int_{-1}^1 \pi(x^2 + 1)^2\,\mathrm dx. $$
As with the disk method you used, the integral is usually simplified by taking the constant factor $\pi$ outside:
$$ V = \pi \int_{-1}^1 (r_1^2 - r_2^2)\,\mathrm dx
 = \pi \int_{-1}^1 ((3-x^2)^2 -(x^2 + 1)^2)\,\mathrm dx. $$
Now here's the neat thing about this method for this problem. Note that
\begin{align}
(3-x^2)^2 - (x^2 + 1)^2
&= (3^2 - 2(3)(x^2) + (x^2)^2) - ((x^2)^2 + 2(1)(x^2) + 1^2) \\
&= (9 - 6x^2 + x^4) - (x^4 + 2x^2 + 1) \\
&= 9 - 6x^2 + x^4 - x^4 - 2x^2 - 1 \\
&= 8 - 8x^2.
\end{align}
So we just need to integrate
$$ \pi\int_{-1}^1 (8 - 8x^2)\,\mathrm dx, $$
which is a lot less work than two integrals with both $x^4$ and $x^2$ terms in them as well as constant terms.
