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Integral over a sphere

  1. Jul 19, 2005 #1
    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. this decomposition should allow for use of fubini's theorem in part c)

    c)find [tex]V_n(1) [/tex] in terms of [tex] V_{n-2}(1)[/tex]

    d)find [tex] V_n(1) [/tex] in terms of only n (eg. find a closed form for [tex] V_n(1) [/tex])

    for b), is the answer
    [tex] 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]\} [/tex]
    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 [tex] V_n(1) = V_{n-2}(1) * \int_{0}^{2 \pi} \int_{0}^{1} r dr d{\theta} [/tex] (using polar coordinates)

    i'm stuck on d). what does it mean by "closed form"
  2. jcsd
  3. Jul 19, 2005 #2


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    Okay, what have you done on this?
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