Can someone help me with this question? I don't know where to begin.
We'll find the cross sectional area and multiply by L. If we assume that the circle is centered at the origin then it has the equation
$\displaystyle x^2+y^2 = R^2$
so the area can be found by (we'll use some symmetry)
$\displaystyle 2\int_{-R}^{R} \sqrt{R^2-y^2}\; dy$
This gives $\displaystyle \pi R^2$ but we're not going all the way up, just to a height $\displaystyle h$ from the bottom or $\displaystyle h-R$ from the origin so
$\displaystyle 2\int_{-R}^{h-R} \sqrt{R^2-y^2}\; dy,$
The volume is then
$\displaystyle V = 2L\int_{-R}^{h-R} \sqrt{R^2-y^2}\; dy$.
Well, let's think about it. $\displaystyle h $ is a measure from the bottom of the cylinder to some height $\displaystyle h$. If $\displaystyle h = 0$ then
$\displaystyle
V = 2L\int_{-R}^{-R} \sqrt{R^2-y^2}\; dy = 0
$
If $\displaystyle h > 0$ then $\displaystyle h-R > -R$ so even if $\displaystyle h < R$ then $\displaystyle h-R > -R$. Consider the case where $\displaystyle h = R$ (the bottom half of the circle)
$\displaystyle
V = 2L\int_{-R}^{0} \sqrt{R^2-y^2}\; dy = \left. \frac{y \sqrt{R^2 -y^2}}{2} + \frac{R^2}{2} \sin^{-1} \frac{y}{R} \right| _{-R}^0 = 2L \cdot \frac{\pi R^2}{4} = \frac{\pi R^2 L}{2}
$
which is the area of half a circle times L.