How Is the Volume of a Sphere Calculated in Higher Dimensions?

[rocket]

let $$B_n(r) = \{x \epsilon R^n| |x| \le r\}$$ be the sphere around the origin of radius r in $$R^n.$$ let $$V_n(r) = \int_{B_n(r)} dV$$ be the volume of $$B_n(r)$$.

a)show that $$V_n(r) = r^n * V_n(1)$$
b)write $$B_n(1)$$ as $$I*J(x) * B_{n-2}(x,y),$$ where I is a fixed interval for the variable x, J an interval for y dependent on x, and $$B_{n-2}(x,y)$$ a ball in $$R^{n-2}$$ with a radius dependent on x and y. this decomposition should allow for use of fubini's theorem in part c)

c)find $$V_n(1)$$ in terms of $$V_{n-2}(1)$$

d)find $$V_n(1)$$ in terms of only n (eg. find a closed form for $$V_n(1)$$)



for b), is the answer
$$B_n(1) = \{I, J \epsilon R^n, B_{n-2}(x,y) \epsilon R^{n-2} | I \epsilon [0,1], J \epsilon [-\sqrt{1-x^2}}, \sqrt{1-x^2}], B_{n-2}(x,y) \epsilon [0,1]*[0,1]*[0,1]\}$$
not sure about the last part...how do i show that radius of B_{n-2}(x,y) is dependent on x and y?

for c), i found the answer to be $$V_n(1) = V_{n-2}(1) * \int_{0}^{2 \pi} \int_{0}^{1} r dr d{\theta}$$ (using polar coordinates)

i'm stuck on d). what does it mean by "closed form"

 

[HallsofIvy]

Okay, what have you done on this?

 

[thepatientmental]

and how do i go about finding it?

a) To show that V_n(r) = r^n * V_n(1), we can use the change of variables formula for integrals. Let x = ry, where y is a vector in R^n. Then, the integral becomes V_n(r) = \int_{B_n(r)} dV = \int_{B_n(1)} \frac{1}{r^n} dV = \frac{1}{r^n} * V_n(1). Therefore, V_n(r) = r^n * V_n(1).

b) To write B_n(1) as I*J(x) * B_{n-2}(x,y), we can use polar coordinates. Let x = (r, \theta, \phi, ..., \psi), where r is the radius and \theta, \phi, ..., \psi are the angles in R^{n-1}. Then, I = [0,1], J(x) = [-\sqrt{1-r^2}, \sqrt{1-r^2}], and B_{n-2}(x,y) = [0,1]^{n-2}. This decomposition allows us to use Fubini's theorem in part c).

c) Using Fubini's theorem, we can express V_n(1) in terms of V_{n-2}(1) as V_n(1) = \int_{B_n(1)} dV = \int_{I} \int_{J(x)} \int_{B_{n-2}(x,y)} dV = \int_{I} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \int_{B_{n-2}(x,y)} dV dx dy = \int_{I} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} V_{n-2}(1) dx dy = V_{n-2}(1) * \int_{I} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} dx dy = V_{n-2}(1) * \int_{I} 2\sqrt{1-x^2} dx = 2V_{n-2}(1) * \int_{0}^{1} \sqrt{1-x^