Volume of n-dimensional sphere

In summary, we define B_n(r) as the sphere around the origin with a radius of r in R^n, and V_n(r) as the volume of B_n(r). We can prove that V_n(r) is proportional to r^n through the equation V_n(r) = r^n * V_n(1). However, we do not know how to prove this assumption holds true for all V_n(r). In order to find the volume of B_n(1), we can set up an integral using Fubini's theorem and a change of variables, where the bounds for x are [-1,1] and the bounds for y are [\sqrt{1-x^2}, -\sqrt{1-x^2}
  • #1
rocket
10
0
let [tex]B_n(r) = \{x \epsilon R^n| |x| \le r\} [/tex] be the sphere around the origin of radius r in [tex] R^n. [/tex] let [tex] V_n(r) = \int_{B_n(r)} dV [/tex] be the volume of [tex]B_n(r)[/tex].

a)show that [tex] V_n(r) = r^n * V_n(1) [/tex]
b)write [tex] B_n(1) [/tex] as [tex] I*J(x) * B_{n-2}(x,y), [/tex] where I is a fixed interval for the variable x, J an interval for y dependent on x, and [tex]B_{n-2}(x,y) [/tex] a ball in [tex] R^{n-2} [/tex] with a radius dependent on x and y. set up an integral to allow for use of fubini's theorem in order to find [tex]V_n(1) [/tex]in terms of [tex]V_{n-2}(1)[/tex].


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

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

the only problem is, i don't know how to prove the assumption i used - that [tex]V_n(r) [/tex] is proportional to [tex]r^n[/tex]. I know that [tex]V_1(r) = 2 * r^1 = 2r, V_2(r) = \pi * r^2, and V_3(r) = 4/3 * \pi * r^3[/tex], which is how i guessed the assumption in the first place, but I don't know how to prove it holds true for [tex]V_n(r)[/tex]. I tried using induction but I don't know what is [tex]V_{n+1}(r) [/tex] in terms of [tex]V_n(r) [/tex]. 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 [tex][\sqrt{1-x^2}, -\sqrt{1-x^2}][/tex]. But I don't understand how to derive the bounds for [tex] B_{n-2}(x,y) [/tex]. Also, how do we find what function over which to integrate?
 
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  • #2
rocket said:
the only problem is, i don't know how to prove the assumption i used - that [tex]V_n(r) [/tex] is proportional to [tex]r^n[/tex].

I assume you're defining [tex]V_n(r)[/tex] as an integral over a volume, where r gives the radius. Fix n. Given r, you can recalculate the integral to be a constant times [tex]V_1(r)[/tex] by making a variable change [tex]x_k <= r x_k'[/tex] for [tex]1 \leq k \leq n[/tex]. The constant is your [tex]r^n[/tex] factor.

Carl
 
  • #3



To prove that V_n(r) is proportional to r^n, we can use the formula for the volume of a sphere in n-dimensions, which is V_n(r) = \frac{\pi^{n/2}r^n}{\Gamma(n/2 + 1)}. This formula can be derived using integration and the change of variables method.

For b), we can use the formula for the volume of a sphere in n-dimensions and apply Fubini's theorem to set up an integral. We can write B_n(1) as the product of B_{n-2}(x,y) and the interval I*J(x), where I is the interval for x and J(x) is an interval for y dependent on x. The bounds for x would be [-1,1] and for y, we can use the equation \sqrt{1-x^2}, -\sqrt{1-x^2}. This can be derived using the Pythagorean theorem.

To find the function to integrate, we can use the formula for V_n(r) and substitute 1 for r. Then, we can use the change of variables method to convert the integral into a single variable integral in terms of x and y. This will give us an equation in terms of V_{n-2}(1), which we can solve for to find V_n(1) in terms of V_{n-2}(1).
 

1. What is the formula for finding the volume of a n-dimensional sphere?

The formula for finding the volume of a n-dimensional sphere is V = (π^(n/2) * r^n) / Γ(n/2 + 1), where V is the volume, π is the mathematical constant pi, r is the radius of the n-dimensional sphere, and Γ is the Gamma function.

2. How do you determine the radius of a n-dimensional sphere?

The radius of a n-dimensional sphere can be determined by finding the distance from the center of the sphere to any point on its surface. This is typically denoted as 'r' in the formula for finding the volume of a n-dimensional sphere.

3. What is the difference between a n-dimensional sphere and a n-dimensional ball?

A n-dimensional sphere is the set of all points in n-dimensional space that are equidistant from a given point, also known as the center. A n-dimensional ball, on the other hand, includes all points within a given distance (the radius) from the center. Essentially, a n-dimensional sphere is the boundary of a n-dimensional ball.

4. Can the volume of a n-dimensional sphere be negative?

No, the volume of a n-dimensional sphere cannot be negative. The formula for finding the volume of a n-dimensional sphere always results in a positive value, as the radius and n value are both raised to even powers.

5. How is the volume of a n-dimensional sphere related to its surface area?

The volume of a n-dimensional sphere is directly proportional to its surface area. As the radius of the sphere increases, both the volume and surface area will increase, but the volume will increase at a faster rate. This can be seen in the formula for finding the volume of a n-dimensional sphere, where the radius is raised to a higher power than in the formula for finding the surface area (V = (π^(n/2) * r^n) / Γ(n/2 + 1) vs A = (n * π^(n/2) * r^(n-1)) / Γ(n/2 + 1)).

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