integrate $\int_{0}^{1}\int_{3y}^{3} e^{x^2}dxdy$ 
$$\int_{0}^{1}\int_{3y}^{3}e^{x^2}dxdy$$

I understand I need to change the integration limits but just changing the order does not help to solve the integral.
$\int_{3y}^{3}\int_{0}^{1}e^{x^2}dydx=\int_{3y}^{3}[ye^{x^2}]_{0}^{1}dx=\int_{3y}^{3}e^{x^2}dx$
How should I approach it? 
 A: The initial limits describe $y$ values ranging from $0$ to $1$ and $x$ values from $3y$ to $3$.  So, the biggest that $x$ can be is $3$ and the smallest that $x$ can be is $0$.  This means that your region is inside a rectangle of height $1$ and width $3$.  Taking the diagonal from the lower left corner to the upper right corner, the region of integration is below that line.  It takes practice to get good at describing these regions.  To get a better idea of how to draw the region, pick a $y$-value (fixed) and observe that the allowed $x$-values range between $3y$ and $3$ (which is the region below the diagonal).
Now, we can change the order of integration.  The $x$ values range from $0$ to $3$, so our outer limits will be from $0$ to $3$.  For any particular (fixed) $x$-value, the possible $y$-values start at $y=0$ and end at $x=3y$ or $y=\frac{x}{3}$.  Therefore, the integral that you get from changing the order of integration is
$$
\int_0^3\int_{0}^{x/3}e^{x^2}dydx.
$$
This is something that you can integrate.
A: $$\int_0^1\int_{3y}^3e^{x^2}\,dx\,dy=\int_{0}^{3}e^{x^2}\int_{0}^{x/3}\,dy\,dx$$
