4.36. On a sphere of radius 2, there are three pairwise tangent circles of radius $\sqrt{2}$. The part of the sphere's surface located outside the circles forms two curvilinear triangles. Find the areas of these triangles.

## § 6. Radical Plane

Let a line $l$, passing through point $O$, intersect the sphere $S$ at points $A$ and $B$. It is easy to verify that the product of the lengths of segments $O A$ and $O B$ depends only on $O$ and $S$, but not on the choice of line $l$ (for points lying outside the sphere, it equals the square of the tangent drawn from point $O$). This quantity, taken with a "plus" sign for points lying outside $S$, and with a "minus" sign for points lying inside $S$, is called the power of point $O$ relative to the sphere $S$. It is easy to verify that it equals $d^{2}-R^{2}$, where $d$ is the distance from point $O$ to the center of the sphere, and $R$ is the radius of the sphere.
4.36. Consider a cube with edge $2 \sqrt{2}$. A sphere of radius 2 with its center at the center of the cube touches all its edges, and its intersections with the faces are circles of radius $\sqrt{2}$. The surface of the sphere is divided by the surface of the cube into 6 segments and 8 curved triangles. Let $x$ be the area of the segment, and $y$ be the area of the curved triangle. The required areas are $y$ and $16 \pi-y-3 x$ ( $16 \pi$ is the surface area of the sphere of radius 2). Since the height of each segment is $2-\sqrt{2}$, then $x=4 \pi(2-\sqrt{2}), \quad$ hence, $y=(16 \pi-6 x) / 8=$ $=\pi(3 \sqrt{2}-4)$ and $16 \pi-y-3 x=\pi(9 \sqrt{2}-4)$.