how do you find the volume enclosed by x=3y and x=y^3 rotated about the y-axis? when y=>0

Answer 1

These curves intersect at (0,0) and at (3 sqrt3 , sqrt 3)
in the area enclosed, x = 3y is bigger(to the right of) x = y^3
So we have a horizontal ring of height dy , inner radius y^3, and outer radius 3y
We can find the volume of the outer cone formed by x = 3y from y = 0 to y = sqrt 3 and then subtract the volume formed by x = y^3
First the x = 3y
Area of slice of height dy = pi x^2 dy
= pi *9y^2 dy
so
= 9 pi integral y^2 dy from 0 to sqrt 3
= 9 pi y^3/3 = 3 pi (sqrt 3)^3 = 9 pi sqrt 3 is about 49
Then subtract the x = y^3 contents
Area of slice dy = pi x^2 dy = pi y^6 dy
so
pi integral y^6 dy = pi y^7/7
= (pi/7)(sqrt3)^7 is about 21
so about 49 - 21 = 28

Answer 2

you should have problem calculating that the two graphs intersect at (0,0) and (3√3, √3)

recall that for the volume of a region rotated about the y-axis we would have
volume = pi(integral) x^2 dy from bottom y to top y of the region

so V = pi(integral)[9y^2 - y^6)dy from 0 to √3
= pi[3y^3 - 1/7 y^7] from 0 to √3
= pi(9√3 - 27/7 √3)
= 27pi/7 √3

Check my arithmetic, I am just scribbling on a piece of paper.

Answer 3

ooops, typing error

last line should have been

= 36pi/7 √3

or 27.97 which matches the answer Daman gave