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Integrals/Related Rates(?)/Volume

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Fluid is flowing in a tube that has a radius of 3 centimeters. Water is flowing through a circular cross section at a rate of (9-r^2) cm/s, where r is the distance from the center of the cross section. What is the total amount of water that flows through the cross section in 4 seconds?

  • Integrals/Related Rates(?)/Volume -

    ok, to get the flow, you integrate area*flow rate through that area.

    so dArea= dr*2PI

    dVolume= flowrate*dArea

    Volume= INT 9r^2*2PI*dr=18PI r^3/3 from r=0 to r=3

    volumeflowrate= 6PI *27 cm^3/sec

    total amount H2O: volumeflowrate*time

  • Integrals/Related Rates(?)/Volume -

    through every ring of radius r and thickness dr and amount per second flows equal to 2 pi r v dr
    dq = 2 pi (9 r - r^3) dr

    integrate that from 0 to 3 and then multiply by 4 seconds

    18 pi (r^2/2) - 2 pi (r^4/4)

    9 pi (9) - (pi/2)(81)

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