For two vectors $v_1, v_2\in S^1$, the (1 dimensional) surface area contained between these two vectors is simply $cos^{-1}(v_1\cdot v_2)$. Similarly, three vectors $v_1, v_2, v_3\in S^2$ will define a triangle on the surface of the sphere, and this triangle will enclose a surface area which is essentially given by the formula here. Now, four vectors $v_1, v_2, v_3, v_4\in S^3$ will define some form of a pyramid on the surface of $S^3$. What portion of the (3 dimensional) surface of $S^3$ is contained by this pyramid?
Of course, one immediately asks if this can be extended to higher dimensions, and while such an answer would be extremely interesting (please do provide it if you know it!), I will accept an answer which treats just the case of $S^3$.