Application of Fubini's Theorem to a simple function I'm trying to solve the integral:
$$\int_0^2\int_0^{x/2}xy^2dydx$$
Using both sides of Fubini's Theorem - that is,  doing $dydx$ and then obtaining the right intervals of integration and calculating $dxdy$. For the first part, I get:
$$\int_0^2\int_0^{x/2}xy^2dydx=\frac{4}{15}$$
However, when I invert the order of integration using $y\in[0,1],x\in[0,2y]$, I obtain:
$$\int_0^1\int_0^{2y}xy^2dxdy=\frac{2}{5}$$
Can someone point out my mistake?
Thanks for helping! :D
(I calculated these integrals with wolfram alpha and the results check)
Edit: Thanks for all your answers! I now understand the problems with the areas, albeit I still don't understand how to obtain the equivalent area for the other direction of the integration.
 A: 

The first graph represents the domain you have before applying Fubini's theorem, while the second is the same domain but viewed from a different perspective.
A: The error is in the bounds on your second integral. If you sketch the region of integration described in your first integral, it is the right triangle with hypotenuse $y = \frac{x}{2}$; one leg the interval $[0,2]$ on the $x$-axis, and one leg the vertical line segment connecting $(2,0)$ and $(2,1)$.
When you change the bounds by switching $dy$ and $dx$, you are correct that $y$ is between $0$ and $1$, but it is not true that $x$ is between $0$ and $2y$, as this is a different triangle than the one just described. In fact, $x$ should be between the hypotenuse just described, and the leg connecting $(2,0)$ and $(2,1)$. I will leave it to you to decide how to proceed from here.
A: Your mistake is in this:
$$\int_0^1 \int_{2y}^2 xy^2 \ dx \ dy$$
Limits of integration in yours is $0\ \text{to}  \ 2y$, it should be $2y \ \text{to} \ 2$ as given above.
