Use the method of cylidrincal shells to find the volume of the solid generated by rotating the region bounded by the curves y=6x-2x^2 and y=x^2 about the y axis.

Find where the two curves intersect:

y = 6x - 2x ²

y = x ²

6x - 2x ² = x ²   →   3x ² - 6x = 3x (x - 2) = 0   →   x = 0 and x = 2

Now, for a shell of radius x units away from the axis of revolution, the height of the shell would be the vertical distance between the upper curve and the lower curve. For 0 ≤ x ≤ 2, we have 6x - x ² ≥ x ², so the height of any given shell is (6x - x ²) - x ² = 6x - 2x ².

Then volume of the solid is

it has 4

step-by-step explanation:

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the answer would be 1 1/72

step-by-step explanation:

you can rewrite the equation by separating the parts and then solving individually:

10-9=1

1/8 - 1/9=

find l c d and then combine:

1 1/72