I am trying to understand calculating volumes with the use of integrals. Let us consider the area in the first quadrant above $y=x^2$ and below $y=2-x^2$. We revolve it about the y-axis and try to find the volume of the solid by integration. If we apply the shell method (which is based on the areas of cylinders), we get $$\int_0^1 2\pi x (2-2x^2)dx=4\pi\int_0^1 (x-x^3)dx=\pi$$ Now, assuming we don't want to use the shell method but rather the disk method (areas of disks), we see that the solid we get is symmetric. We denote the radius of the disk by ${1 \over 2}(2-2x^2)=1-x^2$ and its area by $\pi(1-2x^2+x^4)$. Now, we have $$\int_{-1}^1 \pi(1-2x^2+x^4)dx=2\pi\int_{0}^1 (1-2x^2+x^4)dx=2\pi\left[x-{2 \over 3 }x^3 + {1\over 5}x^5 \right]_0^1=2\pi(1-{2 \over 3 }+ {1\over 5})={16\over 15}\pi$$ which is not correct. What am I missing here?
Edited to correct the sum of squares.
In the disk method, the integral is to be carried out along $y$, not along $x$: $$ V=\int_0^2 \pi x^2\,dy=2\pi\int_0^1 x^2\,dy. $$ In the last integral you have $y=x^2$ (because $0\le y\le1$), thus: $$ V=2\pi\int_0^1 y\,dy=\pi. $$ As an alternative (convenient when the relation between $y$ and $x$ is more complicated) you can compute the integral changing integration variable from $y$ to $x$: $$ V=2\pi\int_0^1 x^2 y'(x)\,dx=2\pi\int_0^1 2x^3\,dx=\pi. $$