Let $$ R $$ be the region bounded by $$ y=x^{2}, x=1 $$, and $$ y=0 $$. Use the shell method to find the volume of the solid generated when $$ R $$ is revolved about the line $$ x=16 $$. Set up the integral that gives the volume of the solid using the shell method. Select the correct choice below and fill in the answer boxes to complete your choice. (Type exact answers.) A. $$ \int_{0}(\square) $$ dy The $$ \int_{0}^{1}\left(2 \pi(16-x) x^{2}\right) d x $$ Type ane exact answer.) (Type ane the solid is

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$$
\text{Radius: } 16 - x
$$
$$
\text{Height: } y = x^2
$$
$$
\text{Shell element: } dV = 2\pi (\text{radius})(\text{height})\,dx = 2\pi (16-x)x^2\,dx
$$
$$
\text{Volume: } V = \int_{0}^{1}2\pi (16-x)x^2\,dx
$$
$$ \int_{0}^{1} 2\pi (16 - x) x^2 \, dx $$

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