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$\displaystyle dV=\pi x^2\,dy$

Since we have $\displaystyle x^2=y-1$ we may state:

$\displaystyle dV=\pi(y-1)\,dy$

And by integration, we have:

$\displaystyle V=\pi\int_1^5 y-1\,dy$

Using the shell method, we find:

$\displaystyle dV=2\pi x(5-y)\,dx$

Since $\displaystyle y=x^2+1$, we may state:

$\displaystyle dV=2\pi x(5-(x^2+1))\,dx=2\pi(4x-x^3)\,dx$

And by integration, we have:

$\displaystyle V=2\pi\int_0^2 4x-x^3\,dx$