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 alot 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?

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 alot 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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