I tried integrating the 4-volume of a 4-hemisphere, that is, $$\int^{R}_{0} \frac{4}{3} \pi r^3 dw$$ (along w-axis), since ##r## is proportional to ##w##, where ##r=\frac{w}{R} R##, ##r=w##, thus the integral becomes $$\int^{R}_{0} \frac{4}{3} \pi w^3 dw = \frac{\pi}{3} R^4$$ The volume of a 4-D...