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Find the volume of the solid generated by the region in the first quadrant bounded above by the 3x+y=6, below by the x-axis, and on the left by the y-axis, about the line x= -2.

  • Calculus - ,

    If we include the area out to x = -2, we have

    v = ∫[0,6] πr^2 dy
    where r = 2+x = 2+(6-y)/3 = 4-y/3
    v = π∫[0,6](4-y/3)^2 dy = 56π

    But we have to subtract out the interior cylinder of radius 2 and height 6, or 24π, leaving us with just 32π generated by the rotating triangle.

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