Volume of a unit 4-ball without calculus A while back, I wanted a way to find the volume of a unit 4-ball without calculus, but I couldn’t find anything. I end

Volume of a unit 4-ball without calculus A while back, I wanted a way to find the volume of a unit 4-ball without calculus, but I couldn’t find anything. I ended up coming up with the approach below, which uses probability. Comments and criticism welcome. We want to find the volume of a 4-dimensional unit ball. Let the center of the 4-ball be the origin. Enclose the 4-ball in an origin-centered, axis-aligned 4-dimensional cube of side length 2. Choose a point uniformly at random inside the 4-cube. What is the probability that the point lies within the 4-ball? If we knew this, we could find the volume of the 4-ball, because P[point inside 4-ball] = (volume of 4-ball)/(volume of 4-cube), and we know that the volume of the 4-cube is 2⁴. We will find P[point inside 4-ball]. Let (a,b,c,d) be the point's coordinates. Since they were picked uniformly at random within the 4-cube and the 4-cube's coordinates range from −1 to 1, we know that a, b, c, and d are each uniformly distributed from −1 to 1 (in symbols: a ~ U(−1,1), b ~ U(−1,1), etc.). The radius of the 4-ball is 1, so the point is inside the 4-ball just when the radius of the point is less than 1. In symbols: P[point inside 4-ball] = P[a² + b² + c² + d² < 1]. Since all four terms sum to less than 1, surely the first two do as well. Therefore, we have: P[point inside 4-ball] = P[a² + b² < 1] P[a² + b² + c² + d² < 1 | a² + b² < 1] The last two terms must also sum to less than 1, so we have: P[point inside 4-ball] = P[a² + b² < 1] P[c² + d² < 1] P[a² + b² + c² + d² < 1 | a² + b² < 1 ∧ c² + d² < 1] Now a lemma: Suppose x ~ U(−1,1) and y ~ U(−1,1). Then P[x² + y² < 1] = π/4. Proof: Since x ~ U(−1,1) and y ~ U(−1,1), P[x² + y² < 1] is the probability that (x,y) lies within an open unit disc, given that it lies within the enclosing 2x2 square. This is simply the area of a unit disc divided by the area of the enclosing 2x2 square, so we have P[a² + b² < 1] = π/4. Similarly for the probability that c² + d² < 1. Using the lemma on the first two factors, we have: P[point inside 4-ball] = (π/4)(π/4)P[a² + b² + c² + d² < 1 | a² + b² < 1 ∧ c² + d² < 1] Now for a second lemma: Suppose x ~ U(−1,1), y ~ U(−1,1), and x² + y² < 1. Then x² + y² ~ U(0,1). Proof: The assumptions can be visualized as throwing a dart randomly at an origin-centered unit disc enclosed in an origin-centered, axis-aligned 2x2 square. If the dart falls outside the disc, we throw again, until it falls within the disc. Let (x,y) be the coordinates of the point, and consider the probability that the point falls inside an origin-centered disc of area A, where 0 ≤ A < π (the area of the outer unit disc). This happens just when the radius of the point (the distance from it to the origin) is less than the radius of the inner circle. The radius of the point is √(x² + y²), and the radius of the inner circle is √(A/π). Therefore, the desired probability is P[√(x² + y²) < √(A/π)]. Squaring both sides yields P[point inside disc of area A] = P[x² + y² < A/π]. Since the area of the inner disc is A and the area of the enclosing unit disc is π, we have P[point inside disc of area A] = A/π, or P[x² + y² < A/π] = A/π, for 0 ≤ A/π ≤ 1. By the definition of a uniform distribution over the reals, P[X < x] = x for 0 ≤ x ≤ 1 if and only if X ~ U(0,1). Hence, we have shown that x² + y² ~ U(0,1). Now back to the main problem. Write u = a² + b² and v = c² + d². Since a ~ U(−1,1), b ~ U(−1,1), and a² + b² < 1, we know, by the second lemma, that a² + b² ~ U(0,1). Similarly, we know that c² + d² ~ U(0,1). Therefore, we can write: P[point inside 4-ball] = (π/4)(π/4)P[u + v < 1 | u ~ U(0,1) ∧ v ~ U(0,1)] The value of the probability expression in that formula is simply the area below the triangle in a square, or 1/2, so we have: P[point inside 4-ball] = (π/4)(π/4)(1/2) = π²/32 Applying this to our original formula yields: (volume of unit 4-ball) = (π²/32)×(2⁴) = π²/2.