Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified?

Y = 6-2x goes diagonally down from (0,6) to (3,0) y=4x^2 goes diagonally up (but also curves). First we need to find where the curves intersect: 4x^2 = 6 - 2x 4x^2 + 2x - 6 = 0 2x^2 + x - 3 = 0 (2x+3)(x-1) = 0 So the curves meet at (-1.5, 9) and (1,4) So the region bound by the curves looks a bit like a sail (a straight line at the top, a parabola curving underneath it). It touches the x axis at x=0 but never goes below it.

When this curve is rotated around the x axis, there is sort of a donut shape. Its outer radius is the top line (y=6-2x) while its inner radius is the bottom curve (y=4x^2). To find the volume V, take tiny thin slices of the donut in the x direction.

The volume of each slice is: (pi * outer_radius^2 - pi * inner_radius^2) * width (pi * (6-2x)^2 - pi * (4x^2)^2) * dx So integrate this: V = (integate from -1.5 to 1) (pi * (6-2x)^2 - pi * (4x^2)^2) dx.

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