I will do the first two for you
1. Vol = pi(integral) y^2 by dx from 0 to 4
= pi (integral) (1 + 2x^1/2 + x)dx from 0 to 4
= pi[x + (4/3)x^3/2 + (1/2)x] from 0 to 4
= pi[4 + 32/3 + 8 - 0]
2. from your y = 1 + √x you will need x^2 since you are rotating about the y=axis
y-1 = √x
(y-1)^4 = x^2
vol = pi (integral) (y-1)^4 dy from 1 to 3
= pi[1/5(y-1)^5] from 1 to 3
= pi/5( 32 - 0]
How about the last one?
You have an ellipse rotated about the x-axis
the vertices are (-2,0) and (2,0)
so because of the symmetry I will find the volume from x=0 to x=2 and double it.
from the equation x^2 = -x^2 /4 + 1
so volume = 2pi(integral)((-x^2)/4 + x)dx from 0 to 2
= 2pi[(-1/12)x^3 + x] from 0 to 2
= 2pi[ -8/12 + 2 - 0]
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