Volume of n-dimensional sphere

[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. set up an integral to allow for use of fubini's theorem in order to find $$V_n(1)$$ in terms of $$V_{n-2}(1)$$.


for a), I assume that $$V_n(r)$$ is proportional to $$r^n$$. So $$V_n(r) = C*r^n$$where C is a constant. $$V_n(1) = C*(1)^n = C$$. we have the equation

$$V_n(1) / V_n(r) = C / C * r^n$$
$$V_n(1) / V_n(r) = 1 / r^n$$
$$V_n(r) = r^n * V_n(1)$$ which completes the proof.

the only problem is, i don't know how to prove the assumption i used - that $$V_n(r)$$ is proportional to $$r^n$$. I know that $$V_1(r) = 2 * r^1 = 2r, V_2(r) = \pi * r^2, and \ V_3(r) = (4/3) \pi r^3$$, which is how i guessed the assumption in the first place, but I don't know how to prove it holds true for $$V_n(r)$$. I tried using induction but I don't know what is $$V_{n+1}(r)$$ in terms of $$V_n(r)$$. My instructor suggested that we set up an integral and use a change of variables of some sort. I was wondering how would I set up an integral to find the volume of a sphere in n-dimensions.


i'm having a lot of trouble understanding b). the bounds of the triple integral would be as follows: the interval for x would be [-1,1] for a sphere centered on the origin, since we're dealing with a radius of 1. the interval for y would be $$[\sqrt{1-x^2}, -\sqrt{1-x^2}]$$. But I don't understand how to derive the bounds for $$B_{n-2}(x,y)$$. Also, how do we find what function over which to integrate?

 

[member 553083]

I'm completely lost. Can someone explain how to solve this problem? For b), we can use the formula for the surface area of a sphere: A_n(r) = \int_{B_n(r)} dA where A_n(r) is the surface area of B_n(r). We can rewrite this as: A_n(r) = \int_{-r}^r \int_{\sqrt{r^2 - x^2}}^{-\sqrt{r^2 - x^2}} \int_{B_{n-2}(x,y)} dV where B_{n-2}(x,y) is the ball in R^{n-2} with radius dependent on x and y. Using Fubini's theorem, we can exchange the order of integration and rewrite the equation as follows: A_n(r) = \int_{B_{n-2}(1)} \int_{-r}^r \int_{\sqrt{r^2 - x^2}}^{-\sqrt{r^2 - x^2}} dV Now we can calculate the volume of B_n(1). Since we are integrating over a ball of radius 1 in R^{n-2}, we have V_n(1) = \int_{B_{n-2}(1)} dV. Now we just need to find the bounds of the integral. We can do this by setting up a spherical coordinate system. The bounds of the integral will be from 0 to 1 for rho and 0 to 2*pi for phi. Thus, we have V_n(1) = \int_0^1 \int_0^{2*pi} \int_{B_{n-2}(1)} dV. We can now use Fubini's theorem again to calculate the volume of B_n(1).

 

[thepatientmental]

To prove that V_n(r) is proportional to r^n, we can use the fact that the volume of a sphere can be calculated by integrating its surface area over the range of its radius. In other words, we can express V_n(r) as the integral of the surface area of the sphere with respect to the radius r. Since the surface area of a sphere with radius r is proportional to r^(n-1), we can write it as V_n(r) = C*r^(n-1) where C is a constant. Then, integrating this over the range of r from 0 to r, we get V_n(r) = C*r^n / n. Since V_n(1) = C, we can see that V_n(r) is indeed proportional to r^n.

Moving on to part b), we can use Fubini's theorem to express the volume of a n-dimensional sphere in terms of the volume of a (n-2)-dimensional sphere. We can write B_n(1) as the product of a line segment I, a two-dimensional disk J(x) and a (n-2)-dimensional ball B_{n-2}(x,y) centered at the origin. So we can express B_n(1) as I * J(x) * B_{n-2}(x,y). Now, using Fubini's theorem, we can write the volume of B_n(1) as a triple integral of the volume of the two-dimensional disk J(x) over the line segment I, and the volume of the (n-2)-dimensional ball B_{n-2}(x,y) over the two-dimensional disk J(x). This can be written as:

V_n(1) = ∫∫∫ J(x) * B_{n-2}(x,y) dI dxdy

Now, we can use a change of variables to express this integral in terms of the volume of a (n-2)-dimensional sphere. Let x = cos(θ) and y = sin(θ), then we have J(x) = cos(θ) and dxdy = dθ. Also, the bounds of the integral for θ would be from 0 to π/2. Substituting these values, we get:

V_n(1) = ∫∫∫ cos(θ) * B_{n-2}(cos(θ