Solution a.): Here is a carefully labeled sketch of the region with a shell marked on the $x$-axis at $x$. Note that the height of the shell at $x$ depends on which side of 1 that $x$ lies, so that we have to split the integral into two parts ! The shell has radius $r$, measured from the $y$-axis, and height $h$, taken parallel to the $y$-axis at $x$. It is IMPORTANT to mark ALL of $x$, $r$, and $h$ in the sketch of the region !!!
Thus the total volume of this Solid of Revolution is
$$ Volume = 2 \pi \int_{0}^{\pi} (radius)(height) \ dx = 2 \pi \int_{0}^{\pi} rh \ dx $$
$$ = 2 \pi \int_{0}^{\pi} (x)( \sin x) \ dx $$
Solution b.): IMPORTANT CHANGE: Because we are revolving the region about the $x$-axis, we must mark a shell on the $y$-axis at $y$ !!! The shell has radius $r$, measured from the $x$-axis, and height $h$, taken parallel to the $x$-axis at $y$. It is IMPORTANT to mark ALL of $y$, $r$, and $h$ in the sketch of the region !!!
Thus the total volume of this Solid of Revolution is
$$ Volume = 2 \pi \int_{0}^{1} (radius)(height) \ dy = 2 \pi \int_{0}^{1} rh \ dy $$
$$ = 2 \pi \int_{0}^{1} (y)((\pi - \arcsin y) - \arcsin y) \ dy $$
$$ = 2 \pi \int_{0}^{1} (y)(\pi - 2 \arcsin y) \ dy $$
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