I need some help with a multivariable calculus problem.
Your objective is to find the volume V of that four-dimensional sphere (let's call it B4, the set of all points (x, y, z, t) with x²+y²+z²+t²<=rad²), by integrating 1 over its contents. To do this, you'll first want to find a nice parametrization of all the points in that sphere. For any valid parametrization p from any parameter space A to B4, the volume of the sphere is given by V = int
A{det(Dp(x
1, x
2, x
3, x
4))}. (*)
The first and most obvious parametrization you could use is the canonical parametrization
p1: B4 -> B4; (x, y, z, t) |-> (x, y, z, t).
In this case, det(Dp1) = 1, but B4 isn't the most Fubini-friendly set you could integrate over. Let's try and make our parameter space more rectangular, shall we?
So let's try a different parametrization using double polar coordinates:
p2: B2' x [0, 2*pi] x [0, 2*pi] -> B4; (r, u, s, v) |-> (r*cos(s), r*sin(s), u*cos(v), u*sin(v)).
Here B2' is the set of all points (r, u) with r>=0 and u>=0 and r²+u²=rad² (it's a quarter-disk). Here, det(Dp2) = r*u, but our parameter space is much more comfortable to work with. It's still stupid, because r and u are still not independent.
To fix this, let's try a third parametrization based on the previous one:
p3: [0, rad]x[0, pi/2]x[0, 2*pi]x[0, 2*pi] -> B4; (p, q, s, v) |-> (p*cos(q)*cos(s), p*cos(q)*sin(s), p*sin(q)*cos(v), p*sin(q)*sin(v)).
Note that this is very similar to the previous parametrization, we just replaced r and u by p*cos(q) and p*sin(q) respectively. Now our parametrization space is rectangular! Now det(Dp3) = p³*sin(q)*cos(q) = p³*sin(2*q)/2, which is not bad as far as Jacobians go. We'll go with this one to integrate over the hypersphere.
So, let's insert p3 into the formula (*) from above:
V=int
A{det(Dp3(p, q, s, v))}
=int(p³*sin(2*q)/2, p=0..rad, q=0..pi/2, s=0..2*pi, v=0..2*pi)
=pi²*rad
4/2.
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