I am trying to find the volume of the region between the parabola $y^2=8x$ and the line $x=2$ revolved around the $y$ axis.
My intuition is that taking an infinitesimal horizontal slice of the parabola, the width should be $$ 2-\frac{y^2}{8} $$ Which we would square and multiply by $\pi$ and the infinitesimal height $\mathrm dy$, yielding the integral: $$ \pi\int_{-4}^4(2-\frac{y^2}{8})^2\mathrm dy $$ With the bounds coming from the intersection of $y^2=8x$ and $x=2$. However, any way I slice it, this isn't giving me the answer in the book, which is $$ \frac{128\pi}{5} $$ Where am I going wrong? The next part asks me to revolve around $x=2$, so any tips there would be helpful too! Thanks.
The cross-section is obtained by removing a disk of radius $y^2/8$ from a disk of radius $2$. That gives volume $$\int_{-4}^4 \pi \cdot(2^2)\,dy-\int_{-4}^4 \pi \cdot\frac{y^4}{64}\,dy.$$ The first integral gives the volume of a cylinder, and we need not use integration to compute that! And I would prefer to integrate from $0$ to $4$ and double.
Alternately, one could use the Method of Cylindrical Shells. That is somewhat less pleasant in this case.