1. ## Really hard Volume Question. Can someone help please?

Can someone help me with this question? I don't know where to begin.

2. Originally Posted by s3n4te
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

$x^2+y^2 = R^2$

so the area can be found by (we'll use some symmetry)

$2\int_{-R}^{R} \sqrt{R^2-y^2}\; dy$

This gives $\pi R^2$ but we're not going all the way up, just to a height $h$ from the bottom or $h-R$ from the origin so

$2\int_{-R}^{h-R} \sqrt{R^2-y^2}\; dy,$

The volume is then

$V = 2L\int_{-R}^{h-R} \sqrt{R^2-y^2}\; dy$.

3. I might be wrong but I think you made a small mistake because your limits of integration don't make sense. It's negative h for the bottom half and then h for the top half. Again, I'm kinda nooby at calculus so I might be wrong...

4. Originally Posted by Kaitosan
I might be wrong but I think you made a small mistake because your limits of integration don't make sense. It's negative h for the bottom half and then h for the top half. Again, I'm kinda nooby at calculus so I might be wrong...
Well, let's think about it. $h$ is a measure from the bottom of the cylinder to some height $h$. If $h = 0$ then

$
V = 2L\int_{-R}^{-R} \sqrt{R^2-y^2}\; dy = 0
$

If $h > 0$ then $h-R > -R$ so even if $h < R$ then $h-R > -R$. Consider the case where $h = R$ (the bottom half of the circle)

$
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.