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Let R be the region bounded by the following curves. Use the disk​ (washer) method to find the volume of the solid generated when R is revolved about the​ y-axis.
y=x, y=3x, y=6

Set up the integral that gives the volume of the solid. Use increasing limits of integration.
Limits of the integral is 0 to 6. I need help with the integrand. I thought it would be π(y/3)^2-(y^2), however that was incorrect.

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  1. each disc has thickness dy, so stack them up. Since we're integrating on y, express each function as x = f(y)
    v = ∫[0,6] π(R^2-r^2) dy
    where R=y and r=y/3
    v = ∫[0,6] π(y^2-(y/3)^2) dy

    check, using cylinders of thickness dx. You have to break the region into two parts, since the upper boundary changes at (2,6)
    v = (∫[0,2] 2πrh dx) + (∫[2,6] 2πrh dx)
    where (r=x and h=3x-x) and (r=x and h=6-x)
    v = (∫[0,2] 2πx*2x dx) + (∫[2,6] 2πx(6-x) dx)

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  2. Which expressions can be added to find the volume of the solid figure?
    Select all that apply.

    A solid shape is made up of 2 attached rectangular prisms. First rectangular prism has a length of 3 m, height of 1 m, and width of 4 m. Second rectangular has a length of 3 m, height of 7 m, and width of 4 m
    The total length of the shape is 6 meters.

    A.4×3×7 and 1×3×4
    B.4×6×7 and 1×3×3
    C.1×4×7 and 4×3×6
    D.4×6×1 and 6×4×3
    E.4×6×3 and 7×3×3

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