<s>[INST] The right way to calculate the volume obtained by rotating the area between 2 graphs around the x axis If i have 2 graphs: $f(x)=x\cdot \frac{\sqrt{1-x^{2}}}{2},\:g(x)=\frac{\sqrt{1-x^{2}}}{2}$
And need to calculate the volume obtained by rotating the area between $f(x)$ and $g(x)$ 
Around the $x$-axis.
I know i need to compute the integral: $\pi \int_{-1}^{1}(g^{2}(x)-f^{2}(x))$
But can i do that direct or do i need to calculate: $\pi( \int_{-1}^{0}(g^{2}(x)-f^{2}(x))+\int_{0}^{1}(g^{2}(x)-f^{2}(x)))$?
Because it doesn't give the same result.
Thanks.
 [/INST] Since you have concerns, we do the calculation. We have $g^2(x)=\frac{1-x^2}{4}$ and $f^2(x)=\frac{x^2(1-x^2)}{4}$, and therefore the difference is $\frac{(1-x^2)^2}{4}$, which is $\frac{1}{4}(1-2x^2+x^4)$. Integrate from $0$ to $1$. An antiderivative is $\frac{1}{4}(x-\frac{2}{3}x^3+\frac{1}{5}x^5)$. 
Do the substitution. We get $\frac{1}{4}\left(1-\frac{2}{3}+\frac{1}{5}\right)$, which should be $\frac{2}{15}$.
To integrate from $-1$ to $0$, we again need to substitute. We get $-\frac{1}{4}\left(-1+\frac{2}{3}-\frac{1}{5}\right)$, the same number.
To integrate if we integrate directly from $-1$ to $1$, we get
$$\frac{1}{4}\left(-1+\frac{2}{3}-\frac{1}{5}\right)-\frac{1}{4}\left(-1+\frac{2}{3}-\frac{1}{5}\right),$$
which is exactly what we get by integrating separately and adding. 
But there was no need to do the calculation for negative $x$. For our integrand is an even function. The integral over any interval $[-a,a]$ is twice the integral from $0$ to $a$. 
Remark: I would almost automatically note the symmetry, integrate from $0$ to $1$, and double. But I am exceptionally bad at handling negative numbers, and really don't like them. They are so $\dots$ negative. 
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