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

well

44-3-3-3=35

35+4+4+4+4=51

so the ansewr would be 51 degreesat 7: 00