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    Three Problems Involving Multiple Integrals

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    Q1a)Find The area of the paraboloid x^2+y^2=z inside the cylinder x^2+y^2=9

    b)write a triple integral in cylindrical coordinates for the volume inside the cone

    z^2=x^2+y^2 and between the planes z=1 ans z=2

    c)Find the moment of inertia of a circular disk (uniform density) about an axis

    through its center and perpendicular to the plane of the disk

    © BrainMass Inc. brainmass.com December 24, 2021, 9:57 pm ad1c9bdddf
    https://brainmass.com/math/integrals/three-problems-involving-multiple-integrals-427446

    SOLUTION This solution is FREE courtesy of BrainMass!

    a) The area is given by the double integral

    A = integral_S(dA) = integral_0^{2 pi}(integral_0^3(rho sec(psi(rho)) drho dphi))

    where rho = sqrt(x^2 + y^2) is the cylindrical radial coordinate, phi is the azimuthal angle, and psi(rho) is the inclination angle of the paraboloid at a given point (which depends only on rho). Now the equation for the paraboloid in cylindrical coordinates is

    z = rho^2

    so we have

    tan(psi(rho)) = dz/drho = 2 rho

    whence

    sec(psi(rho)) = sqrt(1 + tan^2(psi(rho))) = sqrt(1 + 4 rho^2).

    Thus we have

    A = integral_0^{2 pi}(integral_0^3(rho sqrt(1 + 4 rho^2) drho dphi))
    = 2 pi integral_0^3(rho sqrt(1 + 4 rho^2) drho.

    Next we make the substitution

    u = 1 + 4 rho^2

    whence

    du = 8 rho drho

    and

    A = pi/4 integral_1^37(u^(1/2) du)
    = pi/6 [u^(3/2)]_{u=1}^37
    = pi/6[37 sqrt(37) - 1].

    b) In cylindrical coordinates the equation of the cone is |z| = rho, so the integration region is characterized by 0 <= rho <= z, 0 <= phi <= 2 pi and 1 <= z <= 2

    We also have dV = rho drho dphi dz. Thus the volume is given by

    V = integral_1^2(integral_0^{2 pi}(integral_0^z(rho drho dphi dz))).

    c) The moment of inertia is given by the surface integral

    I = integral_S(rho^2 dm)

    where S is the surface of the disk and

    dm = sigma dA = sigma rho drho dphi

    sigma = M/A = M/(pi R^2) being the (uniform) mass density (mass per unit area) of the disk, where M is the mass of the disk and R is its radius. Thus we have

    I = integral_0^{2 pi}(integal_0^R(sigma rho^3 drho dphi))
    = M/(pi R^2)(2 pi) integral_0^R(rho^3 drho)
    = 2M/R^2 [rho^4/4]_{rho=0}^R
    = 2M/R^2 (R^4/4)
    = MR^2/2.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 9:57 pm ad1c9bdddf>
    https://brainmass.com/math/integrals/three-problems-involving-multiple-integrals-427446

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