A few days ago, I was the thinking of the 4-volume (analogue to the volume of a 3-D sphere) and just used analogues to find the 4-volume. I thought about 4 dimensional as something requiring 4 coordinates, instead of 3, in this example, $(x,y,z,w)$.
Say, we have a hypersphere of the equation $x^2 + y^2 + z^2 + w^2 = r^2$, I used the same method we use to find the volume of a 3-D sphere. Considering the cross section of a 4-D sphere would be a 3-D sphere, we have the radius of CS to be $\sqrt{r^2 - w^2}$, and then evaluating its volume by an analogy - since the volume of a 3-D sphere is volume of a 2-D sphere of the cross section (a circle) times the thickness, the 4-volume of a 4-D sphere would be volume of a 3-D sphere times the thickness, in this case, $dw$.
And then evaluate the integral - $\displaystyle \int_{-r}^{r} \dfrac{4}3 \pi {(r^2 - w^2)}^{3/2} dw$ (using trig substitutions, $w = r\sin u $) , which comes out to be $\dfrac{1}2 {\pi}^2 r^4 $ .
And also, I found the surface area again using an analogy, since the volume and surface area of a 3-D sphere are related by $S = \dfrac{dV}{dr}$, the 4-volume and 4-SA(?) have a similar relation - and hence its area is given $S = \dfrac{d}{dr} (\frac{1}2 {\pi}^2 r^4) = 2 {\pi}^2 r^3$.
I have two questions - 1) Is the method of analogies correct? 2) And what does this 4-volume physically mean, EDIT: as in some kind of visualization of the 4-volume of a 4-D sphere?
