Reverse the order of integration of the double integral from 0 to 1 and x^2 to 1 of (x^3)sin(y^3) dydx...?

Answer 1

x is going from 0 to 1 and y is going from x^2 to 1, so we are integrating over the area above the curve y=x^2 on [0, 1].

The overall limits of y are from 0 to 1, and for a particular y x extends from 0 to √y (not from √y to 1 as the previous answerer had; this would give the area BELOW the curve). So we get

∫ (0 to 1) ∫ (0 to √y) x^3 sin (y^3) dx dy.

Answer 2

y = x^2 and y = 1, so if you're reversing it, dx would be first, so the limits of integration would be x = sqrt(y) and x = 1, and limits of integration for y would be y = 0 and y = 1 because of the equation y = x^2, so answer is

double integral from 0 to 1 and sqrt(y) to 1 of (x^3)sin(y^3) dxdy

--edit--
sorry, i should have drawn the domain first. this answer would give you the area under the graph, so the other guy is correct