Using the method of cylindrical shells, find the volume of the shape created by revolving the region $x^2+(y-5)^2=4$ about $y=-1$.
A cylindrical shell is given by: $2\pi v f(v) \ dv$
I solve $x^2+(y-5)^2=4$ for $y$ such that $y=\pm\sqrt{10y-y^-21}$.
The region bounded by $y=\sqrt{10y-y^-21}$ goes from $5 \leq y \leq 7$, which is the top half of the circle. By its symmetry, multiply by $2$ to find the remaining area.
This implies the volume is given by $4\pi\displaystyle\int_{5}^{7}(y+1)\sqrt{10y-y^-21}{\ dy}$.
However, when evaluated, this is different from what is calculated here.
Where is my error?
You want to use
$\displaystyle V=\int_3^72\pi r(y)h(y)dy=\int_3^7 2\pi(y+1) \cdot2\sqrt{4-(y-5)^2}dy=4\pi\int_3^7(y+1)\sqrt{4-(y-5)^2}dy$
$\hspace{.4 in}$since $x=\pm\sqrt{4-(y-5)^2}\implies h(y)=2\sqrt{4-(y-5)^2}$.
(Notice that we can't double the volume generated by the top half of the circle, since the volume generated by the top half is greater than the volume generated by the bottom half of the circle.)
As a check, by Pappus's Theorem we have $V=A(2\pi\rho)=(\pi\cdot2^2)(2\pi\cdot6)=48\pi^2.$