It looks like you cut the base of the cone with a chord. Then you want to know what the volume of the portion of the cone that lies above one of the base partitions is (doesn't really matter which if you know the volume of the cone).
If the height of an arbitrary point on the upper surface of the cone is h and the maximum of the upper surface of the cone is $H$ then:
$h=f(\sqrt{x^{2}+y^{2}})=H-a\sqrt{x^{2}+y^{2}}$
where a is the gradient at which the cone descends. When $\sqrt{x^{2}+y^{2}} = R$ , where $R$ is the radius of the base of the cone, h equals zero. So, $a = \frac{H}{R}$ , and:
$h = H\left(1 - \frac{\sqrt{x^{2}+y^{2}}}{R}\right)$
The problem should be isotropic so nothing about the line should matter except the minimum distance between the line and the center of the cone's base ($d$ in your second diagram above). To get the volume of the smaller partition integrate:
$\int\limits_{-\sqrt{R^{2}-d^{2}}}^\sqrt{R^{2}-d^{2}} \int\limits_d^\sqrt{R^{2}-x^{2}} H(1-\frac{\sqrt{x^{2}+y^{2}}}{R}) \,dy \,dx$
Sorry about the ugly notation. It's my first post and I don't know how to mark equations up here (figured my answer would get stale if I looked it up first).
Edit: AH! $\LaTeX$!
Place your cylinder so the center of the base is at $(0,0)$, and the apex is at $(0,5)$.
If you imagine looking at the cylinder straight on, it will look like a triangle with base $20$ and height $5$. If you make a horizontal slice at level $y$, then you get a figure with two similar triangles:
^ ^
/ \ |
/ \ |
^ /_____\ 5
| / 2r \ |
y / \ |
| / \ |
V /_____________\ V
|----- 20 -----|
then using similar triangles note that the height of the triangle on the top is $5-y$, and the base is $2r$. So we have
$$\frac{2r}{20} = \frac{5-y}{5}.$$
From this, we can express $r$ in terms of $y$.
This is where you are using the fact that the base of your original cone is $10$ cm.
Best Answer
The volume that you have is not the sum of all those circles, you are actually summing over all those infinite thin disks that have the $2D$ surface resembling the circle. The volume of a disk is the circle's area multiplied by the width of the disk. So, $V_{disk}=\pi r^2dx$ where $dx$ is your infinitely thin width of the disk and r is varying radius of the disk. As you want the entire sum of the volume of the disks, you would have $\int_{0}^{h}\pi r(x)^2dx$ where $h$ is the height of the cone, our infinite widths sum up to the height of the cone. Notice this is not your formula because the upper limit on the integrand and the thin width of disk are different variables from $r$.
To add on, if we let R denote the radius of the cone. We see as we are integrating along the cone, the angle does not change, so in our integration, we always have $\frac{x}{r}=\frac{h}{R}$. Notice h and R are constant properties of the cone where as $x$ and $r$ are variables that change in our integration along the cone. Therefore,$r(x)=\frac{Rx}{h}$ you can substitute in to get $\int_{0}^{h}\pi {(\frac{Rx}{h})}^2dx=\pi \frac{R^2}{h^2} \int_{0}^{h} x^2dx=\frac{\pi R^2h}{3}$