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! $\LaTeX$!
One roundabout way would be to prove the Pythagorean theorem geometrically (many standard proofs exist, generally requiring only the formulas for area of squares and triangles that you're given), and use that to compute the area of an equilateral triangle with side 22cm.
It's straight-forward to see that the semi-circle can be divided into 3 such equilateral triangles, plus the 3 $60^o$ circular segments. Thus, you can find the area of one of the circular segments.
The intersection of the semi-circle and quadrant is obviously 1 equilateral triangle plus 2 circular segments. Subtracting that from the formula for the area of a semi-circle gives you the shaded region.
Best Answer
Hint: Using Pythagoras theorem, $r=6$ units.
Thus area: $\frac{1}{2}\pi r^2 + \frac{1}{2} b \times h$ where $b$=$2r$ and $h=8$ units