Question

Check my work:
Find the volume of the solid obtained by rotating the region bounded by the curves y = x and y = x^2 about the y-axis.

Answer 1

y = x -> x = y y = sqrt(y) when y = 0,1
y = x^2 -> x = sqrt(y)

\[\pi \int\limits_{0}^{1} (\sqrt{y})^2 - (y)^2 = \pi/6\]

Answer 2

@experimentX @hartnn

Answer 3

I think it's correct.

Answer 4

i mean the procedure.

Answer 5

what I meant to write was integral (sqrty) -(y)^2 from 0 to 1. Is that good?

Answer 6

@experimentX