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calculus *improper integrals*

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A hole of a radius of 1cm is pierced in a sphere of a 4cm radius. Calculate the volume of the remaining sphere.

I know that the volume of a sphere with an exterior radius is of:

V = pie(R^2 - r^2) dH

where R^2 is the exterior surface area and r^2 is the interior surface area. dH is the thickness.

for R and r, do I simply just plug in the values that are given to me?

Thank you.

  • calculus *improper integrals* - ,

    The answer depends upon where the hole is drilled. Is it centered along a diameter of the sphere?

  • calculus *improper integrals* - ,

    yes , the hole goes through the sphere, centered along the diameter

  • calculus *improper integrals* - ,

    So far so good.

    h^2 = R^2 - x^2
    2h dh = -2x dx
    or,
    dh = -x/h dx

    so the integral becomes

    pi (R^2-r^2) (-x/sqrt(R^2 - x^2)) dx
    = -pi (R^2-r^2) x/sqrt(R^2-x^2) dx

    now let u = R^2 - x^2
    du = -2x dx
    and the integral is

    pi/2 (R^2-r^2) u^(-1/2) du

    If I've done things right,

    v = 4/3 pi (R^2-r^2)^(3/2)

    makes sense, since if r=0, you have the whole sphere, and if r=R, there's nothing.

  • calculus *improper integrals* - ,

    I see how it makes sense now, thanks a lot Steve!

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