A bead is made by drilling a cylindrical hole of radius 1 mm through a sphere of radius 5 mm. Set up a triple integral in cylindrical coordi

Question

A bead is made by drilling a cylindrical hole of radius 1 mm through a sphere of radius 5 mm. Set up a triple integral in cylindrical coordinates representing the volume of the bead. Evaluate the integral.

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Luna 2 weeks 2021-09-28T14:50:24+00:00 1 Answer 0

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    2021-09-28T14:51:34+00:00

    Answer: Volume = 64π\sqrt{6} mm³

    Step-by-step explanation: A triple integral is generally used to calculate the volume of a region. In this case, the integral will be

    V = \int\limits^a_b {\int\limits^a_b {\int\limits^a_b   \, dV. The limits of the triple integral will be:

    1≤r≤5, in which r is the radius of the sphere;

    0≤θ≤2π, which θ is the angle of the sphere;

    and the limits determined by the bead;

    To find the limits of the bead, we use the cylinfrical coordinates. In it, the sphere is represented by the equation r^{2} + z^{2} = 25

    So, the region of the bead will be:

    r^{2} + z^{2} = 25

    z = ±\sqrt{25 - r^{2} }

    \sqrt{25 - r^{2} } ≤ z ≤ \sqrt{25 - r^{2} }

    Calculating and substituing:

    volume = \int\limits^A_B  \, dV

    volume = ∫∫∫ r dzdθdr

    volume = ∫∫ rz dθdr

    Using the limits for z:

    volume = ∫∫ r·(√25 – r²) + r·(√25 – r²) dθdr

    volume = ∫∫ 2r\sqrt{25-r^{2} }dθdr

    volume = ∫ 2r\sqrt{25-r^{2} }∫dθdr

    Using the limits for r and θ, we have:

    volume = 2π · [\frac{-2}{3}[tex](25 – 5^{2} )^{\frac{3}{2} }[/tex] + \frac{2}{3} (25 - 5^{2} )^{\frac{3}{2} } ]

    volume = 64π\sqrt{6} mm³

    The volume of a bead inside a sphere is 64π\sqrt{6} mm³

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